Understanding Faraday's and Lenz's Laws

In this video, we're going to talk about Faraday's Law of Electromagnetic Conduction, and also Lenz's Law. So let's say if we have a coil of wire, and if we have a magnet. What's going to happen if we move the magnet into the coil? If we move the magnet into the coil... There's going to be a current that's generated in the coil. Let's say the coil is connected to some meter. As you move the magnet into the coil, a current will be generated, and this current is going to move in the counterclockwise direction. The magnet produces a magnetic field, which goes into the page. X represents into the page, and a circle represents out of the page. Now what's going to happen, let's say if we have the same coil, but this time, if we move the magnet away from the coil? If we move it away from the coil, then the direction of the current will reverse. The current will no longer flow in the counterclockwise direction, but it's going to flow clockwise. So if you move the magnet into the coil, the current is going to flow counterclockwise. If you move it away from the coil, it will change direction. Now, what about changing the speed? Let's say if we move the magnet slowly. If you move the magnet slowly into the coil, the induced current will be very small. But, if you move the magnet quickly into the coil, the induced current will be larger. If you don't move the magnetic field excuse me if you don't move the magnet into the coil there will be no induced current So it also depends on the speed at which the magnet moves into or out of the coil the greater the speed then the greater the induced current will be now there are other ways to induce in current and it's not only just moving the magnet into out of the coil if you change the area of the core Let's say if you stretch it or bend it, that will induce a current. Also if you change the angle, if you turn the coil relative to the magnetic field produced by the magnet, if you change the angle, an induced current will be created. Now there's an equation that you need to know. This equation is the magnetic flux. The magnetic flux is equal to the product of the magnetic field times the area times cosine of the angle. And the unit for magnetic flux is the Weber. W-E-B-E-R. One Weber is basically one Tesla times one square meter. The unit for magnetic field is Tesla, T, and the unit for area is square meters. So when you multiply these two, you get the unit of Webers. Now, we know that there's going to be an induced current any time the magnetic field is changing, if the area of the coil is changing, or if the angle is changing as well. And so, the induced EMF, or the induced current, is dependent on the rate of change of the flux. Because it also depends on how fast you're moving the bar magnet into or out of the coil. If you move it slowly, the induced current will be small. If the bar magnet is moved quickly into the coil, the induced current will be larger. So, the induced EMF, and therefore the induced current, is dependent on the rate of change of the magnetic flux, which we'll talk about soon. So, let's say if we have a surface, and let's draw the normal line to this surface. which we'll call n. And let's say if the magnetic field is perpendicular to the normal line, which means it's parallel to the plane of the surface, or to the face of the coil. The angle theta is between the normal line and the magnetic field, so the angle is 90. Cosine 90 is equal to zero. So therefore the electric flux will be zero. So if the magnetic field doesn't pass through the face of the coil, if it's parallel to the face of the coil, there will be no electric flux and if there's no electric flux there's no induced EMF. Now let's say if it's neither parallel or perpendicular to the normal line. Theta is here. It's the angle between the normal line and B. So in this situation, the electric flux is simply BA cosine theta. Now the last case. is if the magnetic field is parallel to the normal line. So the magnetic field is perpendicular to the plane of the coil. So in this case, the angle theta is 0, and cosine 0 is equal to 1. So the electric flux is simply equal to B times A. The electric flux is greatest when the magnetic field is parallel to the normal line, or when it's perpendicular to the face of the coil. That's when you're going to have the maximum electric flux. And it's at a minimum if the magnetic field is parallel to the face of the coil. Now, the induced EMF is equal to n, which relates to the number of coils. By the way, if you increase the number of coils, let's say if you have 10 loops compared to 1 loop, The induced EMF and therefore the induced current will be larger when you move the bar magnet into the coil. So the more loops you have, the greater the induced current will be. So the induced EMF is equal to negative N times the change in the electric flux. divided by the change in time. This equation is associated with Faraday's law of induction. So basically it states that the induced EMF is proportional to the rate of change of the electric flux. And the electric flux is BA cosine beta. So if the magnetic field B changes, or if the area changes, or if the angle changes, there's going to be an induced EMF generated. Now the induced EMF, you can treat it like voltage. Voltage and EMF, they both have the same unit volt. So if V is equal to IR, then the induced EMF is equal to IR. So if you know the induced EMF and if you know the resistance, you can therefore calculate the induced current. So if the induced EMF increases, the induced current will increase, because the resistance of the circuit should remain the same. Now before we go into Lenz's Law, let's review a few basic things. So let's say if we have a long straight wire with a current traveling north. Anytime you have a current inside a conductor, it will create a magnetic field. For this particular picture, the magnetic field will enter the page on the right side, and it's going to leave the page, or it's going to travel out of the page on the left side. So out of the page is represented by a circle, and into the page is represented by X. Now to figure this out you can use the right hand rule. So let's say if the pen is the conductor or the wire. If you take your right hand and wrap it around the pen and you want your thumb to face the direction of the current. The way your fingers curl around the pen is the direction of the magnetic field as it travels around the conductor. So let's see if I can draw this. My drone's not the best, so... So here is the conductor. And here's the person's hand wrapped around the conductor. So you want your thumb to face the direction of the current. And notice the way your hands curl around the wire. It comes out of the page on the left side, and it curls into the page on the right side. And so that represents the direction of the magnetic field. So I want you to remember this picture because we're going to use it a lot when trying to figure out the direction of the induced current. So this picture is based on the right-hand rule. Now, if the current is going in the opposite direction, then the magnetic field will change as well. So it's going to be going into the page on the left side, and out of the page on the right side. So everything's going to be reversed. Now, Lenz's Law states that the induced EMF always gives rise to a current whose magnetic field opposes the original change in flux. So, let's apply that to the coil of wire that we had in the beginning. So, let's take this bar magnet and let's move it into the coil. And let's use Lenz's Law to determine the direction of the induced current inside this coil. Now, as we move the magnet into the coil, The magnetic flux, is it increasing or decreasing? Well, first we need to find the direction of the magnetic field. The magnetic field emanates away from the North Pole, enters the South Pole. So, the magnetic field is going into the page, that's the external magnetic field. Now the external magnetic field is increasing, which means that the flux is also increasing. If the flux increases, the induced current will be directed in such a way to decrease the flux. So if you try to increase the flux, the induced current will oppose that change, it's going to decrease it. And if you try to decrease the flux, the induced current will try to support it or increase it. So it's always opposite to what you're trying to do. So since the external magnetic field is increasing, the induced current will create a magnetic field. that will oppose the external magnetic field. The external magnetic field is directed into the page. So if it's going to oppose it, it has to be out of the page. It has to be opposite to it. So I'm going to represent that in blue. So on the right, it's going to be opposite to the magnetic field that is in the center of the loop. And on the left, it has to be X. It can't be the same. So on the right side, it's going to look like this. So if you take any segment of the wire, On the inside of the loop, it's going to have a magnetic field that opposes or is opposite to the external magnetic field inside the loop. Don't focus on the outside part. Focus on the inside part of the loop. Now, what is the direction of the current if we have a wire where the induced magnetic field is going into the page on the right side? I mean, out of the page on the right side, but into the page on the left side. So, hopefully you remember those two pictures that we went over. The induced current has to be going in this direction. That's the only way the magnetic field will be coming out of the page on the right side into the page on the left. And you can use the right hand rule to figure this out. So therefore, what we have is a current traveling in the counterclockwise direction. Now let's work on another example. So let's say if we have a rectangular metal conductor, which looks like this. And we also have a magnetic field confined in this blue region. And let's say the magnetic field is going into the page, everywhere in this region. What is the direction of the induced current? if the conductor is moving into the magnetic field is the current traveling clockwise or counterclockwise in this rectangular loop so feel free to pause the video and use lenses law to figure this out now the magnetic field is not increasing or decreasing its constant however the area inside the rectangular loop that is exposed to magnetic field is increasing so therefore we could say that the magnetic flux is increasing and the induced current will be in such it's going to be in a direction where it's going to try to decrease the flux. Because the magnetic flux is increasing according to Lenz's law, the induced current will try to decrease the flux. And the only way to do that is to direct itself opposite to the magnetic field. Now the external magnetic field at the center of the loop is going into the page. If the induced current is going to oppose it, then the induced magnetic field has to be going out of the page at the center of the rectangular loop. So what you need to do now is select a segment of wire. Let's use this segment. And on the right side, it has to be going out of the page. Whatever we have here, that's going to be the induced magnetic field at the center of the loop. So, if on the inside, or at the center, it's going out of the page, then on the outside, it's into the page. Now, don't forget these two pictures. Anytime you have a segment of wire, when the current is going up, The induced magnetic field will be going out of the page on the right side, into the page on the left. Let's take that back. It's into the page on the right side, out of the page on the left. And if the current is going down, then it's out of the page on the right side, into the page on the left, using the right hand rule. So what we have here is basically this picture. So the current is going down on the left side, which means it's traveling in the counterclockwise direction. And so that's the answer. For the sake of practice, let's work on another example. So this time, the loop is going to move away from the magnetic field. The external magnetic field will be directed into the page again, just like before. And the loop is moving away from it. Go ahead and use Lenz's Law to determine the direction of the current. So the first thing we need to do is determine if the flux is increasing or decreasing. Now the external magnetic field is constant, but the coil or the rectangular loop is moving away from the magnetic field. Therefore, the area that is exposed to the magnetic field is decreasing. And so the flux is decreasing as the coil moves away from the magnetic field. Now, if the magnetic flux is decreasing, an induced current will be created in such a way to increase the flux. If the flux is decreasing, according to Lenz's Law, it wants to oppose the change in flux. So, it wants to increase the flux. The only way to increase the flux is to support the magnetic field. The external magnetic field is directed into the page. As we can see by the X's here. And so in order for the induced current to support it, it has to go in the same direction, not the opposite direction as we did in the last example. So it's going to be going into the page as well. So let's choose a segment of wire that we should focus on. Let's choose this segment. Now, at the center of the loop, it's going to be going into the page. Which means, on the right side, it will be out of the page. So, using this information, what is the direction of the current? Well, anytime you have a segment of wire, where on the right side, if the magnetic field is going into the page, and on the left side, the induced magnetic field, I keep mixing up. On the right side, it's going out of the page, and on the left side, into the page. In this case, using the right-hand rule, the current is going to travel down. So if it's traveling down here, then it's going to be to the left, and on the left side it's going up, and at the top it's going to the right. So therefore, it's traveling in the clockwise direction. Which makes sense. So as the coil moves into the magnetic field, the current is traveling in the counterclockwise direction, and as in this example, as it moves away from the magnetic field, it's going to reverse, traveling in the clockwise direction. So you have to be careful with every step. Hopefully these three examples gives you a good idea of the process that you have to use in order to determine the direction of the current. Let's work on a new example. So let's say if we have a coil of wire and it's placed inside a magnetic field. And this is the external magnetic field which is directed everywhere into the page. Now let's say we take this coil and we shrink it. We're going to make it smaller. So we're going to decrease the area of the coil. such that it looks like this. Determine the direction of the current inside that coil. So because the area is decreasing, the magnetic flux is decreasing as well. Now, if the flux is decreasing, according to Lenz's Law, the induced current will be in such a direction that it's going to try to oppose the change in flux. If the flux is decreasing, the induced current will try to... the flux. And anytime it tries to increase the flux, that means that it's going to create an induced magnetic field that is in the same direction as the external magnetic field. The external magnetic field is going into the page. So therefore, in order for the induced current to support the failing magnetic field, it has to be in the same direction. So it has to be into the page as well. So let's draw our small wire. And let's choose a segment of the wire. Let's focus on the right side. We can choose any segment, but I like to choose a segment that looks like this because we know the two possibilities already. If the current is going up, we know that the magnetic field, that is the magnetic field created by this current, the induced magnetic field, is going to be going into the page on the right side and out of the page on the left side. And if we reverse it, let me draw this somewhere else. If the current is going down, Then the magnetic field is going to be going into the page on the right side. I mean, out of the page on the right side. Into the page on the left. X is into the page, this dot is out of the page. Now going back to this, the induced current has to be going into the page at the center. So relative to this segment of wire, the center is on the left side. So there's going to be an X on the left side, which means a dot on the right side. So these two look similar, which means that the current in that segment of wire is going down. So therefore, the current in this wire is traveling in the clockwise direction. And so that's the answer. Let's say if we have a wire and the current in this wire is increasing. Determine the direction of the induced current in the circular wire. Is it going to be clockwise or counterclockwise? Feel free to pause the video and use Lenz's Law to find out. Now, if the current in the straight wire, if it's increasing, then that means the magnetic field is increasing, which means the flux generated by that wire is increasing. and that is the magnetic field that is produced by this wire that is in the center of this loop that's going to increase which will increase the flux so therefore the induced current in the circular wire is going to create a flux that opposes the original flux so if the original flux is increasing according to the lenses long the flux created by the induced current we'll try to decrease the increase in flux. So it's going to oppose it. It's going to be opposite to it. Now, if it's going to oppose it, then the magnetic fields have to be in the opposite direction, not in the same direction. So let's focus on the external magnetic field. So using the right hand rule, whenever you have a current going in the upward direction, the magnetic field is going to be going out of the page on the left side and into the page on the right side. So it's going to be going in this general direction. This is how it's going to look like. Now the coil of wire, the circular coil, is on the right side, and the magnetic field generated by the straight wire on the right side is X. So the external magnetic field is into the page at the location of the circular wire. Now because the induced current wants to oppose the increase in flux, The induced magnetic field has to be opposite to the external magnetic field. So it has to be out of the page, at the center. Let's use a different color. Let's use blue. Now let's focus on this segment. So at the center, or on the left side, it has to be going out of the page. And on the right side, into the page. So notice that we have the same direction. On the left side we have a dot, on the right side we have an x. Therefore the current must be going up in that segment of wire. Which means that the current is traveling in the counterclockwise direction. So that's the answer. Now let's say if we have a wire, and there's a current traveling towards the left. And that current is decreasing, and we have a circular wire below it. What is the direction of the induced current in the circular wire? Go ahead, pause the video, use the lens as a lot to figure it out. So, let's focus on the magnetic field created by the wire with the decreasing current. If the current is going to... left. If you take your hand and wrap it around a pen with your thumb pointing towards the left in the direction of the current, you'll see that your fingers will curl into the page at the top and they're going to come out of the page below the pen. So I'm not going to draw it but that's how it should look like. So the external magnetic field is out of the page below the wire. And that's what we want to focus on because that's where the circular wire is located. It's below this wire. Now the flux is decreasing because the current is decreasing. A decrease in current will produce a decrease in magnetic field. And a decrease in magnetic field leads to a decrease in flux. So according to Lenz's law, the induced current will be directed in such a way that it's going to try to support the decreasing flux. So if the flux is decreasing, it's going to try to increase it. It's going to oppose the change. Now... Any time it tries to increase the flux, it's trying to support the decreasing magnetic field. And any time it wishes to support it, the two magnetic fields have to be in the same direction. Now at the center, the external magnetic field, which is this one, is directed out of the page. And because the induced magnetic field has to be in the same direction as the external magnetic field, because it wants to support it, this one also has to be out of the page. Now let's focus on this segment of the wire. So the induced magnetic field is out of the page, and the center is on the left side, so it's going to be out of the page on the left, into the page on the right. So therefore, this will create a current that is going upward. Or, if we focus on the left side instead of the right side, The center is going to be out of the page. The center is on the right side of that segment. And then the left side is going to be going into the page. And so the current has to be going down. So it really doesn't matter where you decide to focus on. As we can see, the current is traveling in the counterclockwise direction. And so that's the answer for this one. Now here's another example for you. So let's say if we have a circular coil of wire, which is attached to a battery, and that's attached to a resistor, and there's an open switch, and inside that coil, there is another coil of wire. When the switch is closed, what is the direction of the current in the coil of wire that is represented by the gray color? So let's say it's a 6 volt battery and we have a 3 ohm resistor. So right now the current is 0 when the switch is open. However, once we close the switch, the current will become 2. So it's going to go from 0 to 2 amps. So therefore, the current is increasing. Which means that the flux is increasing. And if the flux is increasing, according to Lenz's Law, the induced current will create a flux that opposes the original change in flux. The flux is increasing, so the induced current will create a flux that will try to oppose the other one, so it's going to be decreasing. Anytime the induced current creates a decrease in flux, the external magnetic field and the induced magnetic field will have, or will be in the opposite direction. If the induced current creates an increase in flux, it's going to try to support the flux. the philomonetic field. And the external monetic field and the induced monetic field will be in the same direction in that case. But in this particular example, since the flux is increasing, the induced current will try to oppose the flux or decrease it. So the external monetic field and the induced monetic field will be in opposite directions. Now we'll need to determine the direction of the external magnetic field. That is the magnetic field created by the white circular wire. So let's focus on this segment of the wire. So we'll need to determine the direction of the current. The current will flow away from the positive terminal towards the negative terminal. So once the switch is closed, the current will be going in the counterclockwise direction. So on the right side, since the current is going up, using the right hand rule, It's going to be going out of the page on the left side into the page on the right side. So that's the external magnetic field. So at the center, the external magnetic field is out of the page. That's at the center of the gray loop. So that's what we're going to put here. Now, because the induced current wants to oppose the increase in flux, the external magnetic field and the induced magnetic field will be in opposite directions. So, the induced magnetic field will be going into the page at the center of the loop. So, I'm going to represent that in blue. So, let's focus on... this segment of the wire. So at the center, which is to the right of that segment, it's going into the page, which means to the left, it must be going out of the page. And therefore, it must be opposite to this one, because the direction of the x and the dot, actually no, it's in the same direction. Because both cases, the x is on the right side, and the dot is on the left side. So therefore, this must be going up. So the current is traveling in the clockwise direction. So that's the direction of the induced current. Now let's work on some problems. Number one, a single circular loop of wire is perpendicular to a magnetic field, which increases from 1.5 Tesla to 4.8 Tesla in 23 milliseconds. Part A, calculate the change in magnetic flux. So let's begin with that. The change in flux... is equal to the change in the magnetic field times the area times cosine of the angle. Now let's draw a picture. So we have a single circular loop and here's the normal line which is perpendicular to the surface. Here's the radius and the magnetic field is perpendicular to the circular loop which means that is parallel to the normal line. So the angle theta is the angle between the normal line and a magnetic field. Therefore the angle is 0 degrees. And cosine of 0 is 1. So now we can calculate the change in electric flux. So the change in B is going to be the final magnetic field, which is 4.8 Tesla, minus the initial magnetic field, which is 1.5 Tesla. The area is constant, and the area is the area of a circle, pi r squared. The radius is 25 centimeters, which is 0.25 meters squared, and then times cosine of 0. We know cosine of 0 is 1. 0.25 squared times pi, that's 0.19635. And if you multiply that by the difference of 4.8 and 1.5, which is 3.3, this will give you the change in flux. which is 0.648. And the unit for magnetic flux is the Weber, which is teslas times square meters. So that's the answer for part A. It's positive 0.648. Part B. What is the induced EMF? To calculate the induced EMF, we can use this formula. It's negative n times the change in the flux divided by the change in time. n is the number of loops, and we have a single circular loop, so n is 1. The change in flux is 0.648, and the change in time, the magnetic field increased from 1.5 to 4.8 Tesla in 23 milliseconds. So that's delta T. But we need to convert milliseconds into seconds. We could do so by dividing by 1,000. 23 divided by 1,000 is 0.023. So 0.648 divided by 0.023, that's about 28.17 volts. So that's the induced EMF. Now the next thing we need to do is calculate the current. The current is the induced EMF divided by the resistance. So it's 28.17 volts divided by 20 ohms. And this is equal to 1.41 amps. And so that's the answer for part C. Number 2. The magnetic flux through a coil of wire Containing 20 loops changes from positive 2 to negative 3 Webers in 425 milliseconds. What is the induced EMF? The induced EMF is equal to negative n times the change in the magnetic flux divided by the change in time. So in this problem, we have 20 loops of wire. The flux changes from 2 to minus 3. The final flux is negative 3. The initial flux is positive 2. So it's negative 3 minus positive 2. Now we need to convert milliseconds to seconds. We can divide it by 1,000. 425 milliseconds is 0.425 seconds. Negative 3 minus 2 is negative 5 negative 20 times negative 5 is positive 100 and if we divide that by 0.425 The induced emf is equal to 235.3 volts Now that we have the induced EMF, we can calculate the resistance of the coil. So E equals I times R, just as V equals I times R. And solving for R, R is the induced EMF divided by the current. So it's going to be 235.3 volts divided by a current of 5.12 amps. And so the resistance is approximately 46 ohms. Now, let's say that we wanted to calculate the power absorbed by the resistance of this coil. How much power is dissipated in the coil? Power is... equal to the voltage times current or the induced EMF times the current. So it's going to be 235.3 volts times a current of 5.12 amps. So that's about 1205 watts. Number 3. A flexible rectangular coil of wire with 150 loops is stretched in such a way that its dimension changes from 5 by 8 square centimeters to 7 by 11 square centimeters in.15 seconds. In a magnetic field of 25 tesla, that is 30 degrees relative to the plane of the coil. Calculate the induced EMF. The induced EMF is equal to negative n times the change in electric flux divided by the change in time. And the change in electric flux, it's going to be the magnetic field, which is constant, times the area. Since the area is not constant, we're going to multiply it by the change in area times cosine theta divided by delta t. Now let's talk about cosine theta. So let's say this is the rectangular coil of wire. And this is the normal line. And this line is parallel to the plane of the coil. And here is the magnetic field. The magnetic field is 30 degrees relative to the plane of the coil. But theta is the angle between the normal line and the magnetic field. So theta in this problem is 60. It's 90 minus 30. N is the number of loops, which is 150. The magnetic field is 2.5, rather 25 Tesla. Now the change in area. The area of a rectangle is basically the length times the width. But we need to convert centimeters to meters. So we've got to divide by 100. The change in area is going to be the final area minus the width. the initial area. The final area is 7 centimeters by 11 centimeters or 0.07 meters times 0.11 meters. That's the final area. The initial area is 0.05 meters times 0.08 meters. And then let's multiply by cosine of 60 degrees. Let's divide everything by 0.15 seconds. So let's calculate the change in area first. 0.07 times 0.11 minus 0.05 times 0.08. That's 0.0037, or 3.7 times 10 to the minus 3. Next, multiply that by cosine 60, and then times 150, and times 25. So that's 6.9375 and divided by 0.15 seconds. Your final answer should be negative 46.25 volts. So that's the answer to part A. That is the induced EMF. Now let's move on to part B. So let's get rid of a few things. How much energy in joules was dissipated in the circuit if the total resistance is 100 ohms? So before we can find the energy, we need to calculate the power. And before we can find that, we need to find the current. So let's find the current first. The current is equal to the induced EMF divided by the resistance. So it's negative 46.25 volts divided by 100 ohms. Now you really don't need to worry about the negative sign, but you can keep it there if you want to. So let's just ignore the negative sign for now. So the current is going to be 0.4625 amps. Energy is going to be positive, so we're going to make the final answer positive. Now that we have the current, we can find the power dissipated by the circuit. It's the voltage, or the inducing MF, times the current. So it's 0.46, rather, 46.25. That's the inducing MF times the current, which is 0.4625. And so that's going to be 21.39 watts. Energy is equal to power multiplied by time. 1 watt is 1 joule per second, so we have 21.39 joules per second. And if we multiply it by 0.15 seconds, we can see that the unit seconds will cancel, giving us the energy in joules. So the final answer is 3.21 joules. So that's how much energy is dissipated in a circuit. So the circuit consumed 3.21 joules within the 0.15 seconds in which the area was changing. Number 4. A rectangular coil of wire contains 25 loops. The angle between the normal line of the coil and the magnetic field changes from 70 to 30 degrees in 85 milliseconds. Calculate the induced EMF. So let's say this is the rectangular coil of wire. And here is the normal line perpendicular to the surface. So initially, the magnetic field is at an angle of 70 degrees relative to the normal line. And then, after some time, the magnetic field, let's call this B final and B initial. Well, the magnetic field is constant, but the angle changes, so it's really theta final and theta initial. But after some time, the magnetic field is going to be 30 degrees relative to the normal line. So that's the picture that we have in this problem. How can we use this information to calculate the induced EMF? By the way, there's two ways the angle could change. Either the magnetic field is constant, and the coil changes relative to the magnetic field, or the coil is held in place, and the magnetic field changes direction relative to the coil. Now in this problem, it really doesn't matter which one changes. All we need to know is just the final angle and the initial angle. The induced EMF is equal to negative n times the change in flux divided by the change in time. And the flux is going to be B. B is constant. The area is constant, but the angle changes, so it's going to be the change in cosine theta divided by delta t. And the number of loops is 25. The magnetic field B is 3 tesla. The area, we need to convert centimeters to meters, so it's going to be 0.15 meters times 0.20 meters. We have a rectangular surface, so the area is simply the length times the width, 0.15 times 0.20, and then multiplied by the change in cosine. The initial angle is 70, the final is 30. So final minus initial, cosine 30 minus cosine 70, divided by the change in time. which is 0.085 seconds. Don't forget to divide 85 milliseconds by 1,000 to convert it to seconds. Now let's plug this in. Let's start with the cosine part first. Cosine 30 minus cosine 70. That's 0.524. And make sure your calculator is in degree mode. Now let's multiply 0.524 by 0.15 times 0.20 times 3 times 25. So that's going to be negative 1.179. And then divide it by 0.085. So the induced EMF is negative 13.87 volts. And so that's it for this problem. So now you know how to use Faraday's Law of Induction to calculate the induced EMF. So Faraday's Law of Electromagnetic Induction is associated with this formula. Now let's say if we have a moving conductor that can slide freely along the metal rails. And this conductor is moving with a speed v. And the rectangular coil that it forms is perpendicular to the magnetic field. The magnetic field is directed into the page. And the rod has a length l. How can we come up with an equation that will help us to calculate the magnitude of the induced EMF? Let's start with Faraday's Law of Induction. The induced EMF is equal to negative n times the change in the magnetic flux divided by the change in time. Now, as the rod moves, the area that is exposed to the magnetic field will increase. So let's draw a new picture. And let's say the rod is now in this area. So notice that the flux is greater, since the area exposed by the magnetic field contained in that coil is greater. So this rod is going to move some distance of D. And that distance is equal to the speed multiplied by the change in time. So the increase in area is basically... Let me use a different color. This region here represents the increase in area. And the area of that region is basically the length times the width, which is d. So the change in area is L times d. And d is basically the velocity times the change in time. Now, going back to the first equation, let's replace the flux. with the magnetic field which is constant times the change in area and the angle is 0 because the magnetic field is perpendicular to the plane of the coil it's parallel to the normal line which means the angle between the normal line and the magnetic field is 0 and cosine of 0 is 1 so we don't have to worry about the cosine part in this formula so we just have this now let's replace Delta a with LV delta T. And by the way, we only have a single loop, so n is equal to 1. So the induced EMF is going to be negative n, or negative 1. And since we're looking for the magnitude, we don't have to worry about the negative sign. So this is going to be 1 times B, and delta A is LV delta T, divided by delta T. And so we can cancel the change in time. So therefore, the induced EMF is equal to the strength of the magnetic field, B, times the length of the moving rod, L, times the speed of the moving rod, V. So it's BLV. Now, it turns out that there's another way in which we could derive that same equation. But first, let's talk about the current in this circuit. As the rod moves towards the right, the area is increasing, which means the flux is increasing. By the way, if you want to, feel free to pause the video and use Lenz's Law to determine the direction of the current in the moving rod. So because the rod is moving to the right, the area increases and the flux increases. So according to Lenz's law, the induced current will be directed in such a way as to oppose the increase in flux. So it's going to try to decrease the flux. And since it's in opposition to it, the external magnetic field, which is directed into the page, is going to be opposite to the induced magnetic field. So if the external magnetic field is into the page, the induced magnetic field is going to be out of the page. So to the left of the rod, which is the center of the coil, the induced magnetic field will be out of the page, which means to the right it's going to be into the page. And so using the right-hand rule, the current has to be traveling in this direction. Electric current flows from a high potential to a low potential. So from a positive side to a negative side. That means the bottom of the conductor has a higher electric potential and the top part has a lower electric potential. Current represents the flow of positive charge. And so the electric field inside the conductor is in the same direction as the current. So if you have a conductor with an electric field directed north, Any positive charges will fill a force that will accelerate it in the direction of the electric field. But in a metal, the protons are not free to move. The electrons are. The electrons are the charge carriers. The electrons will fill a force that will accelerate them opposite to the direction of the electric field. And the electrons will move a distance, L, which is the length of the rod, from one point to the other point. So to calculate the induced EMF using another equation, we need to use energy. Induced EMF, or voltage, is basically the ratio between work and charge. One volt is one joule per coulomb. The unit of work is joules, the unit of charge is coulombs. Work is basically force times distance. And the magnetic force on a moving charge, we know it's BQV. And it's BQV sine theta, but B is perpendicular to the surface. And so for that formula, sine 90 is 1, so we don't have to worry about the sine part. So let's replace F with BQV. And D, the distance that the charges move when being acted on by a force, the magnetic force, so the electrons are going to move a distance D in a direction of the force. which is basically the same as L. So let's replace D with L. Notice that Q cancels, and we can get the same formula. So the induced EMF is equal to the magnetic field times the length of the moving rod times the speed of the moving rod. So those are just two ways in which you can derive this equation. Now let's work on this problem. Number 5. A moving rod 45 centimeters long slides to the right with a speed of 2 meters per second in a magnetic field of 8 Tesla. What is the induced EMF? So let's use the formula EMF is equal to BLV. So the magnetic field is 8 Tesla. The length of the moving rod is 45 centimeters, but we need to convert that to meters. Divided by 100. That's.45 meters. And then we need to multiply it by the speed, which is moving at 2 meters per second. So let's just multiply these three numbers. 8 times 2 is 16. 16 times 0.45 is 7.2. So that's the EMF for this particular problem. That's all you need to do for Part A. Now, Part B. What is the electric field in the rod? To find the electric field, it's simply equal to the voltage divided by the distance. That's how you can find the electric field in the parallel plate capacitor. The electric field in the rod is the voltage in the rod divided by the distance, or the length of the rod. So it's going to be... The voltage is basically the emf. The EMF is 7.2 volts and the length of the rod is 0.45 meters. 7.2 divided by 0.45, that's going to be 16. So it's 16 volts per meter, or you can describe the electric field in terms of newtons per coulomb. 1 volt per meter is the same as 1 newton per coulomb. So that's the answer for part B. Now, part C. What is the current in the rod? The electric current is going to be equal to the voltage, or the EMF, divided by the resistance. So it's 7.2 volts divided by a resistance of 50 ohms. So let's go ahead and divide those two numbers. 7.2 divided by 50 is equal to 0.144 amps, which is equivalent to 144 milliamps. So that is the electric current in the rod. Now part D. What force is required to keep the rod moving to the right at a constant speed of 2 ms? What equation can we use to figure this out? Well, since we know the current, we can find the force in the rod. The magnetic force that acts on an object, or basically, the magnetic force that acts on a wire with a current is equal to I l b sine theta. We can use the same equation to find out the force required to move this rod, because that force will generate a current. The current end of the rod is 0.144. The length of the rod is 0.45 meters. The magnetic field is 8 Tesla. And because the magnetic field is perpendicular to the end of the rod, to the area of the rectangular coil. It's sine 90, sine 90 is just one. So we just gotta multiply those three values. So it's gonna be 0.144 times 0.45 times eight. And so this is equal to 0.5184 N. So that's the force required to keep the rod moving at a speed of 2 m per second. Once it halves that speed, it's going to generate a current of 0.144 amps. And whenever you have a rod with that current, in the presence of a magnetic field, this is going to be the magnetic force that's acting on the rod. And that magnetic force is equal to the force required to keep it moving at that speed. It just works out that way. Number 6. A 60 Hz AC generator rotates in a 0.25 Tesla magnetic field. The generator consists of a circular coil of radius 10 cm with 100 loops. What is the angle of velocity? And calculate the induced EMF. The induced EMF of a generator can be found using this equation. It's equal to n, which represents the number of loops, times b, the strength of the emf. for the magnetic field, times A, the area of each loop, times omega, which is the angle of velocity, times sine omega t. I'm not going to focus on deriving this equation, but this is the formula that you need to know for an AC generator if you want to calculate the induced EMF for it. To find the angular velocity omega, it's simply equal to 2 pi f, where f is the frequency measured in hertz. The frequency is 60 hertz, so omega is going to be 2 pi times 60 hertz. 2 pi times 60. that's equal to 120 pi which as a decimal is about 376.99 s to minus 1 or radians per second So that's omega. That's how you can find the angular velocity. Now let's calculate the induced EMF using the first equation. So n is 100. We have 100 loops. The strength of the magnetic field is 0.25 tesla. The area is going to be the area of a circle, since we have a circular coil. That's pi r squared. The radius is 10 centimeters, so that's 0.10 meters. And we need to square it. Times omega, which is 376.99. By the way, I should have stated this earlier, but we're looking for the maximum EMF. The maximum EMF, you don't need to worry about the sine part. Sine 90 is going to be 1, and so the maximum EMF is simply... NBA times omega. You don't have to worry about the sine function. So let's go ahead and calculate the maximum EMF. So this is going to be the peak output. It's going to be 100 times 0.25 times pi times 0.1 squared times 376.99. So the maximum output is 296.1 volts. That's going to be the maximum EMF generated by this particular generator. Number 7. A generator produces an EMF of 12 volts at 700 rpm. What is the induced EMF of the generator at an angular velocity of 2400 rpm? We know that the maximum EMF is equal to the number of loops times the strength of the magnetic field times the area times the angular velocity. In this equation, you can clearly see that the angular velocity and the induced EMF are proportional. If you increase the angular velocity, the EMF will increase. They're directly related. So if you double the number of RPMs, the induced voltage will double as well. So let's see if we can write an equation that describes E and W, or E and omega. E2 is going to be nBa times omega2. E1 is going to be... nba times omega 1. So if we're using the same generator, it's going to have the same number of loops, which means we can cancel n, the same magnetic field, and the same area. The only difference is the rotation speed, or the angular velocity. So this is the equation that we need. E2 over E1 is equal to omega 2 over omega 1, or epsilon 2 over epsilon 1. So, epsilon 1, the voltage is 12 volts at an angular speed of 700 RPMs. What is the induced EMF at an angular speed of 2400 RPMs? So, these units will cancel. All we need to do is multiply both sides by 12. So, epsilon 2 is going to be equal to 12 times 2400 divided by 700. 2400 divided by 700 is about 3.429. If you multiply that by 12, this will give us an induced EMF of 41.1 volts. So, that's the answer. Our next topic of discussion is the transformer. A transformer is made up of two sets of coils. The coil on the left is known as the primary coil, and on the right, the secondary coil. These coils are wrapped around an iron core. Now let's say that the primary coil has 100 turns. And let's say that the secondary coil, on the right, has 1000 turns. Because the secondary coil has more turns than the primary coil, this is going to be a step-up transformer. A step-up transformer is used to increase the AC voltage. So let's say that the input voltage at the primary, or at the primary coil rather, is let's say 6 volts. Notice that the ratio between ns and np is 10. 1000 divided by 100 is 10, so the voltage will increase by a factor of 10. The voltage at the secondary coil will be 60. When the voltage goes up, the current goes down, such that the power will be the same for an ideal transformer, if it's 100% efficient. So let's say that the input current at the primary coil is 20 amps. The output current at the secondary coil is going to decrease by a factor of 10. If the voltage goes up by a factor of 10, the current will decrease by a factor of 10. So the current is going to be 2 amps. And notice that the power is the same. The power in the primary coil is just the voltage times the current. So it's 6 volts times 20 amps, and that's equal to 120 watts. The power at the secondary coil is going to be Vs times Is, voltage times current. 60 volts times 2 amps is 120 watts. And it makes sense. Energy is conserved. The amount of energy that's being transferred on the left side should be equal to the energy being transferred out of the right side. The amount of energy that is going into the system should equal the amount of energy coming out of the system for an ideal transformer. So because the input power is the same as the output power, this transformer is 100% efficient. Now most real life transformers are about 99% efficient. So. So for the most part, you can assume that these two are virtually the same. But if you ever need to calculate the percent efficiency, it's going to be equal to the output power divided by the input power. The output power can only be equal to or less than the input power. It cannot be more. So it's the output divided by the input power times 100%. That's how you can calculate the percent efficiency. So the equations that you need for transformers are as follows. Ns over Np is equal to Vs over Vp, which is equal to Ip over Is. And the power is voltage times current. Vs times Is is equal to Vp times Ip, if it's 100% efficient. These two will equal each other. The input power is always going to be Vp times Ip. The output power is equal to Vs times Is. But at 100% efficiency, the output power equals the input power. Number 8. A transformer has 50 primary turns and 400 secondary turns. The input voltage is 12 volts and the input current is 24 amps. What is the voltage and current at the secondary coil? And also, how much power is... by the primary coil. So feel free to pause the video and try this problem. So let's draw a picture. This is going to be the primary side and this is going to be the secondary side. Now is this a step up or step down transformer? Since NS is greater than NP, this is going to be a step up transformer. The voltage is going to increase. Now the input voltage is 12 volts, so that's a VP. The input current is 24 amps, so that's IP. And there's 50 turns in the primary coil and 400 turns in the secondary coil. So, NP is 50, NS is 400. Now, 400 divided by 50 is 8. So, the number of turns increases by a factor of 8 as we move from the primary coil to the secondary coil. Which means the voltage has to increase by 8 and the current should decrease by 8. So 12 times 8 is equal to 96 volts. And 24 divided by 8 is 3 amps. And let's make sure the power is the same, assuming it's an ideal transformer. So to calculate the power at the primary coil... The power is equal to VP times IP. So that's 12 volts times 24 amps. So that's 288 watts. And if we calculate it at the secondary coil, it's 96 volts times 3 amps, which is also 288. watts. Now for those of you who want to use an equation to calculate Vs and Is, here's what we can do. Ns divided by Np is equal to Vs over Vp. So Np is 50, Ns is 400. Vp is 12, and let's solve for Vs. Let's cross multiply. So this is going to be 50 vs is equal to 12 times 400. 12 times 400 is 4800. And to solve for vs, divide by 50. 4800. divided by 50 is 96 so that's how you can use the formula to find vs because sometimes the numbers may not be as nice as the whole number that we have here now let's use the formula to calculate is NS over NP is inversely related to the current, so it's going to be IP over IS. Notice that the subscripts, they're opposite relative to each other. So NP is 50, NS is 400. We have IP, which is 24, and we need to solve for IS. So let's cross multiply. 50 times 24. is 1200 and that's equal to 400 times is to find is divided both sides by 400 1200 divided by 400 is 3 so is is 3 amps and then once you have is in VS you can find the power 96 times 3 is 288 or 12 times 24 is also 288. Number 9 a 200 watt ideal transformer has a primary voltage of 40 volts and a secondary current of of 20 amps. Calculate the input current and output voltage. Is this a step up or step down transformer? If there are 80 turns in a primary coil, how many turns are there in a secondary coil? So let's begin. So the power is equal to 200 watts. And we know that the primary voltage, which is the input voltage, Vp, that's 40 volts. And the secondary current, Is, is 20 amps. So what is the secondary voltage, and what is the primary current? So power is equal to voltage times current. The power is 200 watts. The voltage at the primary coil is 40. So we can solve for the primary current. So we need to divide both sides by 40. 200 divided by 40 is 50. So the primary current is 5 amps. Now let's calculate the secondary voltage. Power is equal to Vs times Is. So 200 watts is equal to Vs times the current of 20 amps. 200 watts by 20. 20 is 10. So the voltage at the secondary coil is 10 volts. 20 times 10 is 200. 40 times 5 is 200. Now is this a step up or a step down transformer? What would you say? Is the voltage increasing or decreasing? Notice the voltage went down from 40 to 10. So the voltage is decreasing, which means that this is a step-down transformer. Now, if there are 80 turns in the primary coil, how many turns are there in the secondary coil? Well, we can use the equation ns over np is equal to vs over vp. So the primary coil has... 80 turns, that's NP. We're looking for NS, the number of turns in a secondary coil. VP is 40, VS is 10. So let's cross multiply. So 80 times 10 is 800, and 40 times NS is just 40NS. So to solve for NS, let's divide both sides by 40. So we could cancel with zero. And so, 800 over 40 is the same as 80 over 4, and if 8 divided by 4 is 2, 80 divided by 4 is 20. So there's 20 turns in the secondary coil. Now, let's make sense of this. Let's draw a picture. So the primary coil has more turns than the secondary coil. Primary coil has 80 turns. The secondary coil has 20. The primary voltage is 40 volts. and the secondary voltage is 10 volts. Because the voltage decreased by a factor of 4, the number of turns must also decrease by a factor of 4. So it went down from 80 to 20. Now the current in the primary coil is 5 amps. And so the current in the secondary coil has to increase by a factor of 4, so it's 20 amps. And as we can see, the power is the same. 10 times 20 is 200 watts. 40 times 5 is 200 watts. So hopefully, this setup helps you to understand how transformers are used, and how you can solve them conceptually. You really don't need the formulas if you understand it. But you could use it if the numbers are not whole numbers like we have here. If you have decimal values, I suggest using the formulas. Let's talk about inductance. Let's say if we have a battery attached to a coil of wire, which is also known as a solenoid. When the current is constant, there's not going to be any induced EMF. But if the current is changing, there is going to be an induced EMF. So let's say if the current is increasing, the induced EMF will be negative, which means that it's going to oppose the change in current. If the current's increasing, the induced EMF will try to decrease the current. So therefore, the induced current is going to be in the opposite direction to the increasing current. Now let's say if we have the same circuit, and we have an inductor, or a solenoid, which is basically a coil of wire, and the current is flowing in the same direction. However, this current is decreasing, as opposed to increasing. If the current is decreasing, the induced current generated will try to support the decreasing current, so the EMF will be positive. So the regular current is flowing clockwise, and the induced current will be in the same direction as the decreasing current, and so the induced current will be clockwise. So anytime the current of a circuit is increasing, the induced current will be opposite to the direction of that current. If the current is decreasing, the induced current will be in the same direction as the main current in the circuit. The induced EMF can be calculated using this equation. It's negative L times the change in current divided by the change in time. Notice the negative sign. When the current is increasing, the change in current is positive. If delta I is positive, then the EMF will be negative, as we can see here. Now, if the current is decreasing, the change in current will be negative. So delta I is negative plus the negative sign on the outside, so the EMF will be positive. Now, sometimes you may need to calculate L in this equation. So what exactly is L? L is the inductance, which is measured in units of Henry's, or capital H. And to calculate the inductance of a solenoid, here's the equation that you can use. L is equal to mu0 times n squared times a divided by L. Mu0 is the permeability of free space. It's 4 pi times 10 to the minus 7. L is basically the length of the solenoid. A is the area of the coil. So typically it's circular, so the area is going to be pi r squared. And just keep in mind, whenever you have a circle, the radius is half of the diameter. This is the diameter, and the radius is just one half of that. So r is d over 2. N represents the number of turns in the coil. So let's talk about how to derive this particular formula. So first, we know that the magnetic field created by a solenoid is equal to mu0 times N times I. I went over this in another video. It's the video on magnetism, magnetic fields. Perhaps you've watched that video before this one. So that's where you can get this equation from. Lowercase n represents the number of turns per meter. So it's capital N, which is the number of turns, divided by the length of the solenoid. Now the induced EMF is equal to negative n, the number of turns, times the change in the magnetic flux, divided by the change in time. Anytime you have a current flowing in a wire, it creates a magnetic field. And if you have a change in current, it's going to produce a change in magnetic field, which produces an EMF. Now we also said that the induced EMF is equal to negative L times the change in current divided by the change in time. So therefore, we can set these two equal to each other. So negative n delta flux, magnetic flux, over delta T is equal to negative L delta I over delta T, since they're both equal to the inducing MF. Now what we're going to do is we're going to multiply both sides by negative delta T. so T will cancel and also the negative signs will cancel So now what we have left over is n times the change in magnetic flux, and that is equal to L times the change in the current. So our goal is to solve for the inductance of the inductor, or the solenoid. So L is basically equal to the number of loops times the change in flux. divided by the change in current. So in this equation, we divided both sides by delta I. If you want to show your work, here's what we did. So these two cancel. So now what should we do next? We know that the change in flux is equal to the magnetic field times the area, times cosine theta. But let's assume that the angle is 0, so cosine 0 is 1. So the change in flux is going to be equal to the change in B times A. Now we said that B is equal to mu0 times n times i. So let's replace these two. So the inductance L is equal to n times mu0 times n times i times a divided by delta i. And of course we still have a triangle somewhere. So what we're going to do is we're going to cancel the current. And so we have L is equal to N times mu0 times lowercase n times A. Now lowercase n is equal to capital N. over L. So let's replace lowercase n with this. So the inductance is equal to mu 0 times n and then let's replace lowercase n with n over L times a. So n times times n is n squared. So now we have this equation. L is equal to the permeability of free space times the square of loops times the area of the coil divided by L. And so that's how you can calculate the inductance of a solenoid. Now some other equations that you may need to know is the potential energy stored in an inductor. It's one-half L times I squared. So whenever a current flows through an inductor, energy is stored in the magnetic field of that inductor. Just as when you charge up a capacitor, the energy can be stored in the electric field of the capacitor. Sometimes you may need to calculate the energy density, which is represented by lowercase u, and that's equal to the energy capital U divided by the volume. The potential energy is measured in joules. The energy density is going to be joules per cubic meter. The energy density is equal to b squared over 2 mu zero. That's how you could find the energy density relative to the magnetic field. Number 10. A solenoid consists of 200 loops of wire and has a length of 25 centimeters. Calculate the inductance of the solenoid if it has a diameter of 8 cm. So the radius is always going to be half of the diameter. So 8 cm divided by 2, the radius is 4 cm. Now the inductance of a solenoid is equal to mu zero times n squared times the area divided by the length of the solenoid. Mu zero, the permeability of free space, is 4 pi times 10. to negative 7. The solenoid consists of 200 loops and the area which is usually the solenoid is usually made up of a circular coil of wire it's going to be pi times R squared. So I should have mentioned circular coil of wire but if you see the keyword diameter it's associated with a circle. So this is going to be 0.04 meters squared divided by the length of the solenoid, which is 25 centimeters or 0.25 meters. So let's go ahead and type this in. So this will give you an inductance of 1.01 times 10 to the minus 3 Henry's. Now this is equivalent to 1.01 milliHenry's. Now let's calculate the induced EMF. This is equal to negative L times the change in current divided by the change in time. So we have the value of L. It's 1.01 times 10 to the minus 3 Henry's. The change in current. The final current is 30 minus the initial current of 8 divided by delta T. 240 milliseconds is 0.24 seconds. So 30 minus 8, that's 22, times negative 1.01 times 10 to minus 3, divided by 0.24. This is equal to negative 0.0926 volts. So that's the induced EMF, which is equivalent to negative 92.6 millivolts. Now because the EMF is negative, that means that the induced current is opposite in direction to the original current that created it. So let's say if this is the solenoid, and this is the original current. That current increases from 8 to 30. Because the current increases, the change in current is positive. So therefore, if delta I is positive, a positive value times a negative sign will give us a negative EMF. And anytime the current increases, an induced current will be created in the coils that will oppose the increasing current that created it. So whenever you have a negative EMF, the induced current is opposite to the direction of the original current. So if the original current is flowing clockwise, the induced current will be counterclockwise. It's just going to be opposite to what it is. So that's the direction of the induced EMF and the induced current. Number 11. A solenoid has an inductance of 150 millihenries with 300 turns of wire and a circular area of 2.653. square meters. What is the potential energy stored in the inductor when a current of 20 amps passes through it? So what equation can we use to calculate the potential energy stored in this inductor? Here's the equation that you need. It's one half times the inductance times the square of the current. So the inductance is 150 millihenries, but we need to convert it to henries. So it's 150 times 10 to the minus 3 henries. The current is equal to 20 amps. And let's not forget to square it. So let's go ahead and plug this in. So therefore, the potential energy stored in this inductor is equal to 30 joules. So that's the answer to part A. Now let's move on to part B. How many turns per meter does the solenoid have? So how can we figure this out? Well, we know that the inductance is equal to mu0 times n squared times the area divided by the length of the inductor. So let's separate N squared and write it as N times N over A. Now we know that N over L is equal to lowercase n. And lowercase n represents the number of turns per meter. So we need to use this form of the equation. So L is equal to mu0 times n times the lowercase n times the area. The inductance is 150 times 10 to the minus 3 Henry's. The permeability of free space, that's 4 pi times... times 10 to the minus 7, that's mu0. Capital N represents the number of turns. There's a total of 300 turns in this inductor, or solenoid. The lowercase n is what we're looking for. And the area is given to us. The area is 2.653 square meters. so first let's multiply these three numbers so 4 pi times 10 to the minus 7 times 300 times 2.653 that's about 1.0 times 10 to minus 3 Now, if you can't type in 4 pi times 10 to the minus 7 in your calculator, type in pi first and then multiply that by 4 times 10 to the minus 7. You might find that helpful. Now let's divide both sides by 1 times 10 to minus 3. The 10 to minus 3's will cancel, and so it's simply 150 over 1, so lowercase n is 150. So that's how many turns there are per meter. Now I know this question is not listed in the problem, but how long is the solenoid? So if there are 150 turns per 1 meter, how long is it for 300 turns? So if 1 meter contains 150 turns, then 2 meters will contain 300 turns. So therefore, the solenoid is 2 meters long. You can also use the... equation to get the same answer. We know that lowercase n is equal to capital N divided by L. So let's put this over 1 and let's cross multiply. So nL is equal to capital N, which means means that the left is equal to the number of loops divided by lowercase n which is the number of times per meter so there's 300 turns in the entire solenoid and there's 150 turns per meter so 150 divided by 300 I mean 300 by 150 excuse me that's going to give us two meters so that's the length of the solenoid Now let's move on to part C. What is the magnetic field when the current is 20? The magnetic field of a solenoid can be calculated using this equation. It's mu0 times n times i. Mu zero is 4 pi times 10 to the minus 7. Lowercase n is 150 turns per meter. And the current is 20 amps. So if we multiply these three things, this is going to be 3.77 times 10 to the minus 3 Tesla. So that's the answer for part C. Now what about part D? What is the energy density of this magnetic field? To calculate the energy density, which is equal to lowercase u, it's b squared divided by 2 times mu0. So we have b already, and we just got to divide it by 2 and the permeability of free space. 5.655 joules per cubic meter. So that is the energy density of the magnetic field that is in the solenoid. So that's how you can find it. So that's it for this video. We've covered a lot of topics. And so thanks for watching. And if you like this video, feel free to subscribe. And have a great day.

Let's say the coil is connected to some meter. As you move the magnet into the coil, a current will be generated, and this current is going to move in the counterclockwise direction. The magnet produces a magnetic field, which goes into the page.

X represents into the page, and a circle represents out of the page. Now what's going to happen, let's say if we have the same coil, but this time, if we move the magnet away from the coil? If we move it away from the coil, then the direction of the current will reverse. The current will no longer flow in the counterclockwise direction, but it's going to flow clockwise.

So if you move the magnet into the coil, the current is going to flow counterclockwise. If you move it away from the coil, it will change direction. Now, what about changing the speed?

Let's say if we move the magnet slowly. If you move the magnet slowly into the coil, the induced current will be very small. But, if you move the magnet quickly into the coil, the induced current will be larger.

If you don't move the magnetic field excuse me if you don't move the magnet into the coil there will be no induced current So it also depends on the speed at which the magnet moves into or out of the coil the greater the speed then the greater the induced current will be now there are other ways to induce in current and it's not only just moving the magnet into out of the coil if you change the area of the core Let's say if you stretch it or bend it, that will induce a current. Also if you change the angle, if you turn the coil relative to the magnetic field produced by the magnet, if you change the angle, an induced current will be created. Now there's an equation that you need to know.

This equation is the magnetic flux. The magnetic flux is equal to the product of the magnetic field times the area times cosine of the angle. And the unit for magnetic flux is the Weber. W-E-B-E-R.

One Weber is basically one Tesla times one square meter. The unit for magnetic field is Tesla, T, and the unit for area is square meters. So when you multiply these two, you get the unit of Webers. Now, we know that there's going to be an induced current any time the magnetic field is changing, if the area of the coil is changing, or if the angle is changing as well.

And so, the induced EMF, or the induced current, is dependent on the rate of change of the flux. Because it also depends on how fast you're moving the bar magnet into or out of the coil. If you move it slowly, the induced current will be small.

If the bar magnet is moved quickly into the coil, the induced current will be larger. So, the induced EMF, and therefore the induced current, is dependent on the rate of change of the magnetic flux, which we'll talk about soon. So, let's say if we have a surface, and let's draw the normal line to this surface.

which we'll call n. And let's say if the magnetic field is perpendicular to the normal line, which means it's parallel to the plane of the surface, or to the face of the coil. The angle theta is between the normal line and the magnetic field, so the angle is 90. Cosine 90 is equal to zero.

So therefore the electric flux will be zero. So if the magnetic field doesn't pass through the face of the coil, if it's parallel to the face of the coil, there will be no electric flux and if there's no electric flux there's no induced EMF. Now let's say if it's neither parallel or perpendicular to the normal line.

Theta is here. It's the angle between the normal line and B. So in this situation, the electric flux is simply BA cosine theta.

Now the last case. is if the magnetic field is parallel to the normal line. So the magnetic field is perpendicular to the plane of the coil.

So in this case, the angle theta is 0, and cosine 0 is equal to 1. So the electric flux is simply equal to B times A. The electric flux is greatest when the magnetic field is parallel to the normal line, or when it's perpendicular to the face of the coil. That's when you're going to have the maximum electric flux. And it's at a minimum if the magnetic field is parallel to the face of the coil. Now, the induced EMF is equal to n, which relates to the number of coils.

By the way, if you increase the number of coils, let's say if you have 10 loops compared to 1 loop, The induced EMF and therefore the induced current will be larger when you move the bar magnet into the coil. So the more loops you have, the greater the induced current will be. So the induced EMF is equal to negative N times the change in the electric flux. divided by the change in time.

This equation is associated with Faraday's law of induction. So basically it states that the induced EMF is proportional to the rate of change of the electric flux. And the electric flux is BA cosine beta. So if the magnetic field B changes, or if the area changes, or if the angle changes, there's going to be an induced EMF generated.

Now the induced EMF, you can treat it like voltage. Voltage and EMF, they both have the same unit volt. So if V is equal to IR, then the induced EMF is equal to IR.

So if you know the induced EMF and if you know the resistance, you can therefore calculate the induced current. So if the induced EMF increases, the induced current will increase, because the resistance of the circuit should remain the same. Now before we go into Lenz's Law, let's review a few basic things. So let's say if we have a long straight wire with a current traveling north. Anytime you have a current inside a conductor, it will create a magnetic field.

For this particular picture, the magnetic field will enter the page on the right side, and it's going to leave the page, or it's going to travel out of the page on the left side. So out of the page is represented by a circle, and into the page is represented by X. Now to figure this out you can use the right hand rule. So let's say if the pen is the conductor or the wire.

If you take your right hand and wrap it around the pen and you want your thumb to face the direction of the current. The way your fingers curl around the pen is the direction of the magnetic field as it travels around the conductor. So let's see if I can draw this.

My drone's not the best, so... So here is the conductor. And here's the person's hand wrapped around the conductor.

So you want your thumb to face the direction of the current. And notice the way your hands curl around the wire. It comes out of the page on the left side, and it curls into the page on the right side.

And so that represents the direction of the magnetic field. So I want you to remember this picture because we're going to use it a lot when trying to figure out the direction of the induced current. So this picture is based on the right-hand rule.

Now, if the current is going in the opposite direction, then the magnetic field will change as well. So it's going to be going into the page on the left side, and out of the page on the right side. So everything's going to be reversed. Now, Lenz's Law states that the induced EMF always gives rise to a current whose magnetic field opposes the original change in flux.

So, let's apply that to the coil of wire that we had in the beginning. So, let's take this bar magnet and let's move it into the coil. And let's use Lenz's Law to determine the direction of the induced current inside this coil.

Now, as we move the magnet into the coil, The magnetic flux, is it increasing or decreasing? Well, first we need to find the direction of the magnetic field. The magnetic field emanates away from the North Pole, enters the South Pole. So, the magnetic field is going into the page, that's the external magnetic field. Now the external magnetic field is increasing, which means that the flux is also increasing.

If the flux increases, the induced current will be directed in such a way to decrease the flux. So if you try to increase the flux, the induced current will oppose that change, it's going to decrease it. And if you try to decrease the flux, the induced current will try to support it or increase it. So it's always opposite to what you're trying to do.

So since the external magnetic field is increasing, the induced current will create a magnetic field. that will oppose the external magnetic field. The external magnetic field is directed into the page.

So if it's going to oppose it, it has to be out of the page. It has to be opposite to it. So I'm going to represent that in blue. So on the right, it's going to be opposite to the magnetic field that is in the center of the loop. And on the left, it has to be X.

It can't be the same. So on the right side, it's going to look like this. So if you take any segment of the wire, On the inside of the loop, it's going to have a magnetic field that opposes or is opposite to the external magnetic field inside the loop.

Don't focus on the outside part. Focus on the inside part of the loop. Now, what is the direction of the current if we have a wire where the induced magnetic field is going into the page on the right side? I mean, out of the page on the right side, but into the page on the left side. So, hopefully you remember those two pictures that we went over.

The induced current has to be going in this direction. That's the only way the magnetic field will be coming out of the page on the right side into the page on the left. And you can use the right hand rule to figure this out.

So therefore, what we have is a current traveling in the counterclockwise direction. Now let's work on another example. So let's say if we have a rectangular metal conductor, which looks like this. And we also have a magnetic field confined in this blue region.

And let's say the magnetic field is going into the page, everywhere in this region. What is the direction of the induced current? if the conductor is moving into the magnetic field is the current traveling clockwise or counterclockwise in this rectangular loop so feel free to pause the video and use lenses law to figure this out now the magnetic field is not increasing or decreasing its constant however the area inside the rectangular loop that is exposed to magnetic field is increasing so therefore we could say that the magnetic flux is increasing and the induced current will be in such it's going to be in a direction where it's going to try to decrease the flux. Because the magnetic flux is increasing according to Lenz's law, the induced current will try to decrease the flux. And the only way to do that is to direct itself opposite to the magnetic field.

Now the external magnetic field at the center of the loop is going into the page. If the induced current is going to oppose it, then the induced magnetic field has to be going out of the page at the center of the rectangular loop. So what you need to do now is select a segment of wire. Let's use this segment.

And on the right side, it has to be going out of the page. Whatever we have here, that's going to be the induced magnetic field at the center of the loop. So, if on the inside, or at the center, it's going out of the page, then on the outside, it's into the page.

Now, don't forget these two pictures. Anytime you have a segment of wire, when the current is going up, The induced magnetic field will be going out of the page on the right side, into the page on the left. Let's take that back. It's into the page on the right side, out of the page on the left.

And if the current is going down, then it's out of the page on the right side, into the page on the left, using the right hand rule. So what we have here is basically this picture. So the current is going down on the left side, which means it's traveling in the counterclockwise direction.

And so that's the answer. For the sake of practice, let's work on another example. So this time, the loop is going to move away from the magnetic field. The external magnetic field will be directed into the page again, just like before.

And the loop is moving away from it. Go ahead and use Lenz's Law to determine the direction of the current. So the first thing we need to do is determine if the flux is increasing or decreasing.

Now the external magnetic field is constant, but the coil or the rectangular loop is moving away from the magnetic field. Therefore, the area that is exposed to the magnetic field is decreasing. And so the flux is decreasing as the coil moves away from the magnetic field. Now, if the magnetic flux is decreasing, an induced current will be created in such a way to increase the flux.

If the flux is decreasing, according to Lenz's Law, it wants to oppose the change in flux. So, it wants to increase the flux. The only way to increase the flux is to support the magnetic field.

The external magnetic field is directed into the page. As we can see by the X's here. And so in order for the induced current to support it, it has to go in the same direction, not the opposite direction as we did in the last example. So it's going to be going into the page as well. So let's choose a segment of wire that we should focus on.

Let's choose this segment. Now, at the center of the loop, it's going to be going into the page. Which means, on the right side, it will be out of the page. So, using this information, what is the direction of the current?

Well, anytime you have a segment of wire, where on the right side, if the magnetic field is going into the page, and on the left side, the induced magnetic field, I keep mixing up. On the right side, it's going out of the page, and on the left side, into the page. In this case, using the right-hand rule, the current is going to travel down.

So if it's traveling down here, then it's going to be to the left, and on the left side it's going up, and at the top it's going to the right. So therefore, it's traveling in the clockwise direction. Which makes sense.

So as the coil moves into the magnetic field, the current is traveling in the counterclockwise direction, and as in this example, as it moves away from the magnetic field, it's going to reverse, traveling in the clockwise direction. So you have to be careful with every step. Hopefully these three examples gives you a good idea of the process that you have to use in order to determine the direction of the current. Let's work on a new example. So let's say if we have a coil of wire and it's placed inside a magnetic field.

And this is the external magnetic field which is directed everywhere into the page. Now let's say we take this coil and we shrink it. We're going to make it smaller. So we're going to decrease the area of the coil. such that it looks like this.

Determine the direction of the current inside that coil. So because the area is decreasing, the magnetic flux is decreasing as well. Now, if the flux is decreasing, according to Lenz's Law, the induced current will be in such a direction that it's going to try to oppose the change in flux.

If the flux is decreasing, the induced current will try to... the flux. And anytime it tries to increase the flux, that means that it's going to create an induced magnetic field that is in the same direction as the external magnetic field.

The external magnetic field is going into the page. So therefore, in order for the induced current to support the failing magnetic field, it has to be in the same direction. So it has to be into the page as well. So let's draw our small wire.

And let's choose a segment of the wire. Let's focus on the right side. We can choose any segment, but I like to choose a segment that looks like this because we know the two possibilities already.

If the current is going up, we know that the magnetic field, that is the magnetic field created by this current, the induced magnetic field, is going to be going into the page on the right side and out of the page on the left side. And if we reverse it, let me draw this somewhere else. If the current is going down, Then the magnetic field is going to be going into the page on the right side.

I mean, out of the page on the right side. Into the page on the left. X is into the page, this dot is out of the page.

Now going back to this, the induced current has to be going into the page at the center. So relative to this segment of wire, the center is on the left side. So there's going to be an X on the left side, which means a dot on the right side.

So these two look similar, which means that the current in that segment of wire is going down. So therefore, the current in this wire is traveling in the clockwise direction. And so that's the answer.

Let's say if we have a wire and the current in this wire is increasing. Determine the direction of the induced current in the circular wire. Is it going to be clockwise or counterclockwise? Feel free to pause the video and use Lenz's Law to find out. Now, if the current in the straight wire, if it's increasing, then that means the magnetic field is increasing, which means the flux generated by that wire is increasing.

and that is the magnetic field that is produced by this wire that is in the center of this loop that's going to increase which will increase the flux so therefore the induced current in the circular wire is going to create a flux that opposes the original flux so if the original flux is increasing according to the lenses long the flux created by the induced current we'll try to decrease the increase in flux. So it's going to oppose it. It's going to be opposite to it. Now, if it's going to oppose it, then the magnetic fields have to be in the opposite direction, not in the same direction.

So let's focus on the external magnetic field. So using the right hand rule, whenever you have a current going in the upward direction, the magnetic field is going to be going out of the page on the left side and into the page on the right side. So it's going to be going in this general direction. This is how it's going to look like.

Now the coil of wire, the circular coil, is on the right side, and the magnetic field generated by the straight wire on the right side is X. So the external magnetic field is into the page at the location of the circular wire. Now because the induced current wants to oppose the increase in flux, The induced magnetic field has to be opposite to the external magnetic field. So it has to be out of the page, at the center. Let's use a different color.

Let's use blue. Now let's focus on this segment. So at the center, or on the left side, it has to be going out of the page.

And on the right side, into the page. So notice that we have the same direction. On the left side we have a dot, on the right side we have an x.

Therefore the current must be going up in that segment of wire. Which means that the current is traveling in the counterclockwise direction. So that's the answer. Now let's say if we have a wire, and there's a current traveling towards the left. And that current is decreasing, and we have a circular wire below it.

What is the direction of the induced current in the circular wire? Go ahead, pause the video, use the lens as a lot to figure it out. So, let's focus on the magnetic field created by the wire with the decreasing current. If the current is going to...

left. If you take your hand and wrap it around a pen with your thumb pointing towards the left in the direction of the current, you'll see that your fingers will curl into the page at the top and they're going to come out of the page below the pen. So I'm not going to draw it but that's how it should look like.

So the external magnetic field is out of the page below the wire. And that's what we want to focus on because that's where the circular wire is located. It's below this wire. Now the flux is decreasing because the current is decreasing. A decrease in current will produce a decrease in magnetic field.

And a decrease in magnetic field leads to a decrease in flux. So according to Lenz's law, the induced current will be directed in such a way that it's going to try to support the decreasing flux. So if the flux is decreasing, it's going to try to increase it. It's going to oppose the change. Now...

Any time it tries to increase the flux, it's trying to support the decreasing magnetic field. And any time it wishes to support it, the two magnetic fields have to be in the same direction. Now at the center, the external magnetic field, which is this one, is directed out of the page. And because the induced magnetic field has to be in the same direction as the external magnetic field, because it wants to support it, this one also has to be out of the page.

Now let's focus on this segment of the wire. So the induced magnetic field is out of the page, and the center is on the left side, so it's going to be out of the page on the left, into the page on the right. So therefore, this will create a current that is going upward. Or, if we focus on the left side instead of the right side, The center is going to be out of the page. The center is on the right side of that segment.

And then the left side is going to be going into the page. And so the current has to be going down. So it really doesn't matter where you decide to focus on. As we can see, the current is traveling in the counterclockwise direction. And so that's the answer for this one.

Now here's another example for you. So let's say if we have a circular coil of wire, which is attached to a battery, and that's attached to a resistor, and there's an open switch, and inside that coil, there is another coil of wire. When the switch is closed, what is the direction of the current in the coil of wire that is represented by the gray color? So let's say it's a 6 volt battery and we have a 3 ohm resistor. So right now the current is 0 when the switch is open.

However, once we close the switch, the current will become 2. So it's going to go from 0 to 2 amps. So therefore, the current is increasing. Which means that the flux is increasing. And if the flux is increasing, according to Lenz's Law, the induced current will create a flux that opposes the original change in flux. The flux is increasing, so the induced current will create a flux that will try to oppose the other one, so it's going to be decreasing.

Anytime the induced current creates a decrease in flux, the external magnetic field and the induced magnetic field will have, or will be in the opposite direction. If the induced current creates an increase in flux, it's going to try to support the flux. the philomonetic field. And the external monetic field and the induced monetic field will be in the same direction in that case. But in this particular example, since the flux is increasing, the induced current will try to oppose the flux or decrease it.

So the external monetic field and the induced monetic field will be in opposite directions. Now we'll need to determine the direction of the external magnetic field. That is the magnetic field created by the white circular wire.

So let's focus on this segment of the wire. So we'll need to determine the direction of the current. The current will flow away from the positive terminal towards the negative terminal.

So once the switch is closed, the current will be going in the counterclockwise direction. So on the right side, since the current is going up, using the right hand rule, It's going to be going out of the page on the left side into the page on the right side. So that's the external magnetic field. So at the center, the external magnetic field is out of the page.

That's at the center of the gray loop. So that's what we're going to put here. Now, because the induced current wants to oppose the increase in flux, the external magnetic field and the induced magnetic field will be in opposite directions. So, the induced magnetic field will be going into the page at the center of the loop. So, I'm going to represent that in blue.

So, let's focus on... this segment of the wire. So at the center, which is to the right of that segment, it's going into the page, which means to the left, it must be going out of the page. And therefore, it must be opposite to this one, because the direction of the x and the dot, actually no, it's in the same direction. Because both cases, the x is on the right side, and the dot is on the left side.

So therefore, this must be going up. So the current is traveling in the clockwise direction. So that's the direction of the induced current. Now let's work on some problems. Number one, a single circular loop of wire is perpendicular to a magnetic field, which increases from 1.5 Tesla to 4.8 Tesla in 23 milliseconds.

Part A, calculate the change in magnetic flux. So let's begin with that. The change in flux...

is equal to the change in the magnetic field times the area times cosine of the angle. Now let's draw a picture. So we have a single circular loop and here's the normal line which is perpendicular to the surface.

Here's the radius and the magnetic field is perpendicular to the circular loop which means that is parallel to the normal line. So the angle theta is the angle between the normal line and a magnetic field. Therefore the angle is 0 degrees. And cosine of 0 is 1. So now we can calculate the change in electric flux.

So the change in B is going to be the final magnetic field, which is 4.8 Tesla, minus the initial magnetic field, which is 1.5 Tesla. The area is constant, and the area is the area of a circle, pi r squared. The radius is 25 centimeters, which is 0.25 meters squared, and then times cosine of 0. We know cosine of 0 is 1. 0.25 squared times pi, that's 0.19635.

And if you multiply that by the difference of 4.8 and 1.5, which is 3.3, this will give you the change in flux. which is 0.648. And the unit for magnetic flux is the Weber, which is teslas times square meters.

So that's the answer for part A. It's positive 0.648. Part B. What is the induced EMF?

To calculate the induced EMF, we can use this formula. It's negative n times the change in the flux divided by the change in time. n is the number of loops, and we have a single circular loop, so n is 1. The change in flux is 0.648, and the change in time, the magnetic field increased from 1.5 to 4.8 Tesla in 23 milliseconds. So that's delta T.

But we need to convert milliseconds into seconds. We could do so by dividing by 1,000. 23 divided by 1,000 is 0.023. So 0.648 divided by 0.023, that's about 28.17 volts. So that's the induced EMF.

Now the next thing we need to do is calculate the current. The current is the induced EMF divided by the resistance. So it's 28.17 volts divided by 20 ohms. And this is equal to 1.41 amps. And so that's the answer for part C.

Number 2. The magnetic flux through a coil of wire Containing 20 loops changes from positive 2 to negative 3 Webers in 425 milliseconds. What is the induced EMF? The induced EMF is equal to negative n times the change in the magnetic flux divided by the change in time.

So in this problem, we have 20 loops of wire. The flux changes from 2 to minus 3. The final flux is negative 3. The initial flux is positive 2. So it's negative 3 minus positive 2. Now we need to convert milliseconds to seconds. We can divide it by 1,000. 425 milliseconds is 0.425 seconds. Negative 3 minus 2 is negative 5 negative 20 times negative 5 is positive 100 and if we divide that by 0.425 The induced emf is equal to 235.3 volts Now that we have the induced EMF, we can calculate the resistance of the coil.

So E equals I times R, just as V equals I times R. And solving for R, R is the induced EMF divided by the current. So it's going to be 235.3 volts divided by a current of 5.12 amps. And so the resistance is approximately 46 ohms. Now, let's say that we wanted to calculate the power absorbed by the resistance of this coil.

How much power is dissipated in the coil? Power is... equal to the voltage times current or the induced EMF times the current. So it's going to be 235.3 volts times a current of 5.12 amps.

So that's about 1205 watts. Number 3. A flexible rectangular coil of wire with 150 loops is stretched in such a way that its dimension changes from 5 by 8 square centimeters to 7 by 11 square centimeters in.15 seconds. In a magnetic field of 25 tesla, that is 30 degrees relative to the plane of the coil.

Calculate the induced EMF. The induced EMF is equal to negative n times the change in electric flux divided by the change in time. And the change in electric flux, it's going to be the magnetic field, which is constant, times the area.

Since the area is not constant, we're going to multiply it by the change in area times cosine theta divided by delta t. Now let's talk about cosine theta. So let's say this is the rectangular coil of wire.

And this is the normal line. And this line is parallel to the plane of the coil. And here is the magnetic field.

The magnetic field is 30 degrees relative to the plane of the coil. But theta is the angle between the normal line and the magnetic field. So theta in this problem is 60. It's 90 minus 30. N is the number of loops, which is 150. The magnetic field is 2.5, rather 25 Tesla. Now the change in area.

The area of a rectangle is basically the length times the width. But we need to convert centimeters to meters. So we've got to divide by 100. The change in area is going to be the final area minus the width. the initial area. The final area is 7 centimeters by 11 centimeters or 0.07 meters times 0.11 meters.

That's the final area. The initial area is 0.05 meters times 0.08 meters. And then let's multiply by cosine of 60 degrees. Let's divide everything by 0.15 seconds.

So let's calculate the change in area first. 0.07 times 0.11 minus 0.05 times 0.08. That's 0.0037, or 3.7 times 10 to the minus 3. Next, multiply that by cosine 60, and then times 150, and times 25. So that's 6.9375 and divided by 0.15 seconds.

Your final answer should be negative 46.25 volts. So that's the answer to part A. That is the induced EMF. Now let's move on to part B.

So let's get rid of a few things. How much energy in joules was dissipated in the circuit if the total resistance is 100 ohms? So before we can find the energy, we need to calculate the power. And before we can find that, we need to find the current. So let's find the current first.

The current is equal to the induced EMF divided by the resistance. So it's negative 46.25 volts divided by 100 ohms. Now you really don't need to worry about the negative sign, but you can keep it there if you want to.

So let's just ignore the negative sign for now. So the current is going to be 0.4625 amps. Energy is going to be positive, so we're going to make the final answer positive.

Now that we have the current, we can find the power dissipated by the circuit. It's the voltage, or the inducing MF, times the current. So it's 0.46, rather, 46.25. That's the inducing MF times the current, which is 0.4625. And so that's going to be 21.39 watts.

Energy is equal to power multiplied by time. 1 watt is 1 joule per second, so we have 21.39 joules per second. And if we multiply it by 0.15 seconds, we can see that the unit seconds will cancel, giving us the energy in joules. So the final answer is 3.21 joules. So that's how much energy is dissipated in a circuit.

So the circuit consumed 3.21 joules within the 0.15 seconds in which the area was changing. Number 4. A rectangular coil of wire contains 25 loops. The angle between the normal line of the coil and the magnetic field changes from 70 to 30 degrees in 85 milliseconds. Calculate the induced EMF. So let's say this is the rectangular coil of wire.

And here is the normal line perpendicular to the surface. So initially, the magnetic field is at an angle of 70 degrees relative to the normal line. And then, after some time, the magnetic field, let's call this B final and B initial.

Well, the magnetic field is constant, but the angle changes, so it's really theta final and theta initial. But after some time, the magnetic field is going to be 30 degrees relative to the normal line. So that's the picture that we have in this problem. How can we use this information to calculate the induced EMF?

By the way, there's two ways the angle could change. Either the magnetic field is constant, and the coil changes relative to the magnetic field, or the coil is held in place, and the magnetic field changes direction relative to the coil. Now in this problem, it really doesn't matter which one changes.

All we need to know is just the final angle and the initial angle. The induced EMF is equal to negative n times the change in flux divided by the change in time. And the flux is going to be B. B is constant.

The area is constant, but the angle changes, so it's going to be the change in cosine theta divided by delta t. And the number of loops is 25. The magnetic field B is 3 tesla. The area, we need to convert centimeters to meters, so it's going to be 0.15 meters times 0.20 meters.

We have a rectangular surface, so the area is simply the length times the width, 0.15 times 0.20, and then multiplied by the change in cosine. The initial angle is 70, the final is 30. So final minus initial, cosine 30 minus cosine 70, divided by the change in time. which is 0.085 seconds.

Don't forget to divide 85 milliseconds by 1,000 to convert it to seconds. Now let's plug this in. Let's start with the cosine part first.

Cosine 30 minus cosine 70. That's 0.524. And make sure your calculator is in degree mode. Now let's multiply 0.524 by 0.15 times 0.20 times 3 times 25. So that's going to be negative 1.179.

And then divide it by 0.085. So the induced EMF is negative 13.87 volts. And so that's it for this problem. So now you know how to use Faraday's Law of Induction to calculate the induced EMF.

So Faraday's Law of Electromagnetic Induction is associated with this formula. Now let's say if we have a moving conductor that can slide freely along the metal rails. And this conductor is moving with a speed v. And the rectangular coil that it forms is perpendicular to the magnetic field.

The magnetic field is directed into the page. And the rod has a length l. How can we come up with an equation that will help us to calculate the magnitude of the induced EMF? Let's start with Faraday's Law of Induction. The induced EMF is equal to negative n times the change in the magnetic flux divided by the change in time.

Now, as the rod moves, the area that is exposed to the magnetic field will increase. So let's draw a new picture. And let's say the rod is now in this area. So notice that the flux is greater, since the area exposed by the magnetic field contained in that coil is greater.

So this rod is going to move some distance of D. And that distance is equal to the speed multiplied by the change in time. So the increase in area is basically...

Let me use a different color. This region here represents the increase in area. And the area of that region is basically the length times the width, which is d.

So the change in area is L times d. And d is basically the velocity times the change in time. Now, going back to the first equation, let's replace the flux.

with the magnetic field which is constant times the change in area and the angle is 0 because the magnetic field is perpendicular to the plane of the coil it's parallel to the normal line which means the angle between the normal line and the magnetic field is 0 and cosine of 0 is 1 so we don't have to worry about the cosine part in this formula so we just have this now let's replace Delta a with LV delta T. And by the way, we only have a single loop, so n is equal to 1. So the induced EMF is going to be negative n, or negative 1. And since we're looking for the magnitude, we don't have to worry about the negative sign. So this is going to be 1 times B, and delta A is LV delta T, divided by delta T.

And so we can cancel the change in time. So therefore, the induced EMF is equal to the strength of the magnetic field, B, times the length of the moving rod, L, times the speed of the moving rod, V. So it's BLV. Now, it turns out that there's another way in which we could derive that same equation.

But first, let's talk about the current in this circuit. As the rod moves towards the right, the area is increasing, which means the flux is increasing. By the way, if you want to, feel free to pause the video and use Lenz's Law to determine the direction of the current in the moving rod. So because the rod is moving to the right, the area increases and the flux increases. So according to Lenz's law, the induced current will be directed in such a way as to oppose the increase in flux.

So it's going to try to decrease the flux. And since it's in opposition to it, the external magnetic field, which is directed into the page, is going to be opposite to the induced magnetic field. So if the external magnetic field is into the page, the induced magnetic field is going to be out of the page.

So to the left of the rod, which is the center of the coil, the induced magnetic field will be out of the page, which means to the right it's going to be into the page. And so using the right-hand rule, the current has to be traveling in this direction. Electric current flows from a high potential to a low potential.

So from a positive side to a negative side. That means the bottom of the conductor has a higher electric potential and the top part has a lower electric potential. Current represents the flow of positive charge. And so the electric field inside the conductor is in the same direction as the current.

So if you have a conductor with an electric field directed north, Any positive charges will fill a force that will accelerate it in the direction of the electric field. But in a metal, the protons are not free to move. The electrons are. The electrons are the charge carriers.

The electrons will fill a force that will accelerate them opposite to the direction of the electric field. And the electrons will move a distance, L, which is the length of the rod, from one point to the other point. So to calculate the induced EMF using another equation, we need to use energy.

Induced EMF, or voltage, is basically the ratio between work and charge. One volt is one joule per coulomb. The unit of work is joules, the unit of charge is coulombs.

Work is basically force times distance. And the magnetic force on a moving charge, we know it's BQV. And it's BQV sine theta, but B is perpendicular to the surface.

And so for that formula, sine 90 is 1, so we don't have to worry about the sine part. So let's replace F with BQV. And D, the distance that the charges move when being acted on by a force, the magnetic force, so the electrons are going to move a distance D in a direction of the force.

which is basically the same as L. So let's replace D with L. Notice that Q cancels, and we can get the same formula.

So the induced EMF is equal to the magnetic field times the length of the moving rod times the speed of the moving rod. So those are just two ways in which you can derive this equation. Now let's work on this problem.

Number 5. A moving rod 45 centimeters long slides to the right with a speed of 2 meters per second in a magnetic field of 8 Tesla. What is the induced EMF? So let's use the formula EMF is equal to BLV.

So the magnetic field is 8 Tesla. The length of the moving rod is 45 centimeters, but we need to convert that to meters. Divided by 100. That's.45 meters.

And then we need to multiply it by the speed, which is moving at 2 meters per second. So let's just multiply these three numbers. 8 times 2 is 16. 16 times 0.45 is 7.2. So that's the EMF for this particular problem.

That's all you need to do for Part A. Now, Part B. What is the electric field in the rod? To find the electric field, it's simply equal to the voltage divided by the distance. That's how you can find the electric field in the parallel plate capacitor. The electric field in the rod is the voltage in the rod divided by the distance, or the length of the rod. So it's going to be...

The voltage is basically the emf. The EMF is 7.2 volts and the length of the rod is 0.45 meters. 7.2 divided by 0.45, that's going to be 16. So it's 16 volts per meter, or you can describe the electric field in terms of newtons per coulomb. 1 volt per meter is the same as 1 newton per coulomb.

So that's the answer for part B. Now, part C. What is the current in the rod? The electric current is going to be equal to the voltage, or the EMF, divided by the resistance. So it's 7.2 volts divided by a resistance of 50 ohms.

So let's go ahead and divide those two numbers. 7.2 divided by 50 is equal to 0.144 amps, which is equivalent to 144 milliamps. So that is the electric current in the rod.

Now part D. What force is required to keep the rod moving to the right at a constant speed of 2 ms? What equation can we use to figure this out?

Well, since we know the current, we can find the force in the rod. The magnetic force that acts on an object, or basically, the magnetic force that acts on a wire with a current is equal to I l b sine theta. We can use the same equation to find out the force required to move this rod, because that force will generate a current. The current end of the rod is 0.144.

The length of the rod is 0.45 meters. The magnetic field is 8 Tesla. And because the magnetic field is perpendicular to the end of the rod, to the area of the rectangular coil.

It's sine 90, sine 90 is just one. So we just gotta multiply those three values. So it's gonna be 0.144 times 0.45 times eight. And so this is equal to 0.5184 N.

So that's the force required to keep the rod moving at a speed of 2 m per second. Once it halves that speed, it's going to generate a current of 0.144 amps. And whenever you have a rod with that current, in the presence of a magnetic field, this is going to be the magnetic force that's acting on the rod.

And that magnetic force is equal to the force required to keep it moving at that speed. It just works out that way. Number 6. A 60 Hz AC generator rotates in a 0.25 Tesla magnetic field.

The generator consists of a circular coil of radius 10 cm with 100 loops. What is the angle of velocity? And calculate the induced EMF.

The induced EMF of a generator can be found using this equation. It's equal to n, which represents the number of loops, times b, the strength of the emf. for the magnetic field, times A, the area of each loop, times omega, which is the angle of velocity, times sine omega t. I'm not going to focus on deriving this equation, but this is the formula that you need to know for an AC generator if you want to calculate the induced EMF for it.

To find the angular velocity omega, it's simply equal to 2 pi f, where f is the frequency measured in hertz. The frequency is 60 hertz, so omega is going to be 2 pi times 60 hertz. 2 pi times 60. that's equal to 120 pi which as a decimal is about 376.99 s to minus 1 or radians per second So that's omega. That's how you can find the angular velocity.

Now let's calculate the induced EMF using the first equation. So n is 100. We have 100 loops. The strength of the magnetic field is 0.25 tesla.

The area is going to be the area of a circle, since we have a circular coil. That's pi r squared. The radius is 10 centimeters, so that's 0.10 meters. And we need to square it. Times omega, which is 376.99.

By the way, I should have stated this earlier, but we're looking for the maximum EMF. The maximum EMF, you don't need to worry about the sine part. Sine 90 is going to be 1, and so the maximum EMF is simply...

NBA times omega. You don't have to worry about the sine function. So let's go ahead and calculate the maximum EMF. So this is going to be the peak output. It's going to be 100 times 0.25 times pi times 0.1 squared times 376.99.

So the maximum output is 296.1 volts. That's going to be the maximum EMF generated by this particular generator. Number 7. A generator produces an EMF of 12 volts at 700 rpm. What is the induced EMF of the generator at an angular velocity of 2400 rpm?

We know that the maximum EMF is equal to the number of loops times the strength of the magnetic field times the area times the angular velocity. In this equation, you can clearly see that the angular velocity and the induced EMF are proportional. If you increase the angular velocity, the EMF will increase.

They're directly related. So if you double the number of RPMs, the induced voltage will double as well. So let's see if we can write an equation that describes E and W, or E and omega. E2 is going to be nBa times omega2.

E1 is going to be... nba times omega 1. So if we're using the same generator, it's going to have the same number of loops, which means we can cancel n, the same magnetic field, and the same area. The only difference is the rotation speed, or the angular velocity.

So this is the equation that we need. E2 over E1 is equal to omega 2 over omega 1, or epsilon 2 over epsilon 1. So, epsilon 1, the voltage is 12 volts at an angular speed of 700 RPMs. What is the induced EMF at an angular speed of 2400 RPMs? So, these units will cancel.

All we need to do is multiply both sides by 12. So, epsilon 2 is going to be equal to 12 times 2400 divided by 700. 2400 divided by 700 is about 3.429. If you multiply that by 12, this will give us an induced EMF of 41.1 volts. So, that's the answer. Our next topic of discussion is the transformer. A transformer is made up of two sets of coils.

The coil on the left is known as the primary coil, and on the right, the secondary coil. These coils are wrapped around an iron core. Now let's say that the primary coil has 100 turns. And let's say that the secondary coil, on the right, has 1000 turns. Because the secondary coil has more turns than the primary coil, this is going to be a step-up transformer.

A step-up transformer is used to increase the AC voltage. So let's say that the input voltage at the primary, or at the primary coil rather, is let's say 6 volts. Notice that the ratio between ns and np is 10. 1000 divided by 100 is 10, so the voltage will increase by a factor of 10. The voltage at the secondary coil will be 60. When the voltage goes up, the current goes down, such that the power will be the same for an ideal transformer, if it's 100% efficient. So let's say that the input current at the primary coil is 20 amps.

The output current at the secondary coil is going to decrease by a factor of 10. If the voltage goes up by a factor of 10, the current will decrease by a factor of 10. So the current is going to be 2 amps. And notice that the power is the same. The power in the primary coil is just the voltage times the current.

So it's 6 volts times 20 amps, and that's equal to 120 watts. The power at the secondary coil is going to be Vs times Is, voltage times current. 60 volts times 2 amps is 120 watts.

And it makes sense. Energy is conserved. The amount of energy that's being transferred on the left side should be equal to the energy being transferred out of the right side. The amount of energy that is going into the system should equal the amount of energy coming out of the system for an ideal transformer.

So because the input power is the same as the output power, this transformer is 100% efficient. Now most real life transformers are about 99% efficient. So.

So for the most part, you can assume that these two are virtually the same. But if you ever need to calculate the percent efficiency, it's going to be equal to the output power divided by the input power. The output power can only be equal to or less than the input power.

It cannot be more. So it's the output divided by the input power times 100%. That's how you can calculate the percent efficiency.

So the equations that you need for transformers are as follows. Ns over Np is equal to Vs over Vp, which is equal to Ip over Is. And the power is voltage times current.

Vs times Is is equal to Vp times Ip, if it's 100% efficient. These two will equal each other. The input power is always going to be Vp times Ip. The output power is equal to Vs times Is.

But at 100% efficiency, the output power equals the input power. Number 8. A transformer has 50 primary turns and 400 secondary turns. The input voltage is 12 volts and the input current is 24 amps.

What is the voltage and current at the secondary coil? And also, how much power is... by the primary coil. So feel free to pause the video and try this problem. So let's draw a picture.

This is going to be the primary side and this is going to be the secondary side. Now is this a step up or step down transformer? Since NS is greater than NP, this is going to be a step up transformer. The voltage is going to increase.

Now the input voltage is 12 volts, so that's a VP. The input current is 24 amps, so that's IP. And there's 50 turns in the primary coil and 400 turns in the secondary coil. So, NP is 50, NS is 400. Now, 400 divided by 50 is 8. So, the number of turns increases by a factor of 8 as we move from the primary coil to the secondary coil.

Which means the voltage has to increase by 8 and the current should decrease by 8. So 12 times 8 is equal to 96 volts. And 24 divided by 8 is 3 amps. And let's make sure the power is the same, assuming it's an ideal transformer. So to calculate the power at the primary coil...

The power is equal to VP times IP. So that's 12 volts times 24 amps. So that's 288 watts.

And if we calculate it at the secondary coil, it's 96 volts times 3 amps, which is also 288. watts. Now for those of you who want to use an equation to calculate Vs and Is, here's what we can do. Ns divided by Np is equal to Vs over Vp.

So Np is 50, Ns is 400. Vp is 12, and let's solve for Vs. Let's cross multiply. So this is going to be 50 vs is equal to 12 times 400. 12 times 400 is 4800. And to solve for vs, divide by 50. 4800. divided by 50 is 96 so that's how you can use the formula to find vs because sometimes the numbers may not be as nice as the whole number that we have here now let's use the formula to calculate is NS over NP is inversely related to the current, so it's going to be IP over IS. Notice that the subscripts, they're opposite relative to each other. So NP is 50, NS is 400. We have IP, which is 24, and we need to solve for IS.

So let's cross multiply. 50 times 24. is 1200 and that's equal to 400 times is to find is divided both sides by 400 1200 divided by 400 is 3 so is is 3 amps and then once you have is in VS you can find the power 96 times 3 is 288 or 12 times 24 is also 288. Number 9 a 200 watt ideal transformer has a primary voltage of 40 volts and a secondary current of of 20 amps. Calculate the input current and output voltage. Is this a step up or step down transformer?

If there are 80 turns in a primary coil, how many turns are there in a secondary coil? So let's begin. So the power is equal to 200 watts. And we know that the primary voltage, which is the input voltage, Vp, that's 40 volts.

And the secondary current, Is, is 20 amps. So what is the secondary voltage, and what is the primary current? So power is equal to voltage times current. The power is 200 watts.

The voltage at the primary coil is 40. So we can solve for the primary current. So we need to divide both sides by 40. 200 divided by 40 is 50. So the primary current is 5 amps. Now let's calculate the secondary voltage.

Power is equal to Vs times Is. So 200 watts is equal to Vs times the current of 20 amps. 200 watts by 20. 20 is 10. So the voltage at the secondary coil is 10 volts.

20 times 10 is 200. 40 times 5 is 200. Now is this a step up or a step down transformer? What would you say? Is the voltage increasing or decreasing? Notice the voltage went down from 40 to 10. So the voltage is decreasing, which means that this is a step-down transformer. Now, if there are 80 turns in the primary coil, how many turns are there in the secondary coil?

Well, we can use the equation ns over np is equal to vs over vp. So the primary coil has... 80 turns, that's NP.

We're looking for NS, the number of turns in a secondary coil. VP is 40, VS is 10. So let's cross multiply. So 80 times 10 is 800, and 40 times NS is just 40NS.

So to solve for NS, let's divide both sides by 40. So we could cancel with zero. And so, 800 over 40 is the same as 80 over 4, and if 8 divided by 4 is 2, 80 divided by 4 is 20. So there's 20 turns in the secondary coil. Now, let's make sense of this. Let's draw a picture. So the primary coil has more turns than the secondary coil.

Primary coil has 80 turns. The secondary coil has 20. The primary voltage is 40 volts. and the secondary voltage is 10 volts. Because the voltage decreased by a factor of 4, the number of turns must also decrease by a factor of 4. So it went down from 80 to 20. Now the current in the primary coil is 5 amps. And so the current in the secondary coil has to increase by a factor of 4, so it's 20 amps.

And as we can see, the power is the same. 10 times 20 is 200 watts. 40 times 5 is 200 watts.

So hopefully, this setup helps you to understand how transformers are used, and how you can solve them conceptually. You really don't need the formulas if you understand it. But you could use it if the numbers are not whole numbers like we have here.

If you have decimal values, I suggest using the formulas. Let's talk about inductance. Let's say if we have a battery attached to a coil of wire, which is also known as a solenoid. When the current is constant, there's not going to be any induced EMF. But if the current is changing, there is going to be an induced EMF.

So let's say if the current is increasing, the induced EMF will be negative, which means that it's going to oppose the change in current. If the current's increasing, the induced EMF will try to decrease the current. So therefore, the induced current is going to be in the opposite direction to the increasing current.

Now let's say if we have the same circuit, and we have an inductor, or a solenoid, which is basically a coil of wire, and the current is flowing in the same direction. However, this current is decreasing, as opposed to increasing. If the current is decreasing, the induced current generated will try to support the decreasing current, so the EMF will be positive. So the regular current is flowing clockwise, and the induced current will be in the same direction as the decreasing current, and so the induced current will be clockwise. So anytime the current of a circuit is increasing, the induced current will be opposite to the direction of that current.

If the current is decreasing, the induced current will be in the same direction as the main current in the circuit. The induced EMF can be calculated using this equation. It's negative L times the change in current divided by the change in time. Notice the negative sign. When the current is increasing, the change in current is positive.

If delta I is positive, then the EMF will be negative, as we can see here. Now, if the current is decreasing, the change in current will be negative. So delta I is negative plus the negative sign on the outside, so the EMF will be positive.

Now, sometimes you may need to calculate L in this equation. So what exactly is L? L is the inductance, which is measured in units of Henry's, or capital H. And to calculate the inductance of a solenoid, here's the equation that you can use. L is equal to mu0 times n squared times a divided by L.

Mu0 is the permeability of free space. It's 4 pi times 10 to the minus 7. L is basically the length of the solenoid. A is the area of the coil. So typically it's circular, so the area is going to be pi r squared.

And just keep in mind, whenever you have a circle, the radius is half of the diameter. This is the diameter, and the radius is just one half of that. So r is d over 2. N represents the number of turns in the coil.

So let's talk about how to derive this particular formula. So first, we know that the magnetic field created by a solenoid is equal to mu0 times N times I. I went over this in another video. It's the video on magnetism, magnetic fields.

Perhaps you've watched that video before this one. So that's where you can get this equation from. Lowercase n represents the number of turns per meter. So it's capital N, which is the number of turns, divided by the length of the solenoid. Now the induced EMF is equal to negative n, the number of turns, times the change in the magnetic flux, divided by the change in time.

Anytime you have a current flowing in a wire, it creates a magnetic field. And if you have a change in current, it's going to produce a change in magnetic field, which produces an EMF. Now we also said that the induced EMF is equal to negative L times the change in current divided by the change in time.

So therefore, we can set these two equal to each other. So negative n delta flux, magnetic flux, over delta T is equal to negative L delta I over delta T, since they're both equal to the inducing MF. Now what we're going to do is we're going to multiply both sides by negative delta T.

so T will cancel and also the negative signs will cancel So now what we have left over is n times the change in magnetic flux, and that is equal to L times the change in the current. So our goal is to solve for the inductance of the inductor, or the solenoid. So L is basically equal to the number of loops times the change in flux.

divided by the change in current. So in this equation, we divided both sides by delta I. If you want to show your work, here's what we did. So these two cancel.

So now what should we do next? We know that the change in flux is equal to the magnetic field times the area, times cosine theta. But let's assume that the angle is 0, so cosine 0 is 1. So the change in flux is going to be equal to the change in B times A.

Now we said that B is equal to mu0 times n times i. So let's replace these two. So the inductance L is equal to n times mu0 times n times i times a divided by delta i. And of course we still have a triangle somewhere. So what we're going to do is we're going to cancel the current.

And so we have L is equal to N times mu0 times lowercase n times A. Now lowercase n is equal to capital N. over L.

So let's replace lowercase n with this. So the inductance is equal to mu 0 times n and then let's replace lowercase n with n over L times a. So n times times n is n squared.

So now we have this equation. L is equal to the permeability of free space times the square of loops times the area of the coil divided by L. And so that's how you can calculate the inductance of a solenoid.

Now some other equations that you may need to know is the potential energy stored in an inductor. It's one-half L times I squared. So whenever a current flows through an inductor, energy is stored in the magnetic field of that inductor.

Just as when you charge up a capacitor, the energy can be stored in the electric field of the capacitor. Sometimes you may need to calculate the energy density, which is represented by lowercase u, and that's equal to the energy capital U divided by the volume. The potential energy is measured in joules. The energy density is going to be joules per cubic meter.

The energy density is equal to b squared over 2 mu zero. That's how you could find the energy density relative to the magnetic field. Number 10. A solenoid consists of 200 loops of wire and has a length of 25 centimeters.

Calculate the inductance of the solenoid if it has a diameter of 8 cm. So the radius is always going to be half of the diameter. So 8 cm divided by 2, the radius is 4 cm. Now the inductance of a solenoid is equal to mu zero times n squared times the area divided by the length of the solenoid.

Mu zero, the permeability of free space, is 4 pi times 10. to negative 7. The solenoid consists of 200 loops and the area which is usually the solenoid is usually made up of a circular coil of wire it's going to be pi times R squared. So I should have mentioned circular coil of wire but if you see the keyword diameter it's associated with a circle. So this is going to be 0.04 meters squared divided by the length of the solenoid, which is 25 centimeters or 0.25 meters.

So let's go ahead and type this in. So this will give you an inductance of 1.01 times 10 to the minus 3 Henry's. Now this is equivalent to 1.01 milliHenry's. Now let's calculate the induced EMF. This is equal to negative L times the change in current divided by the change in time.

So we have the value of L. It's 1.01 times 10 to the minus 3 Henry's. The change in current. The final current is 30 minus the initial current of 8 divided by delta T. 240 milliseconds is 0.24 seconds.

So 30 minus 8, that's 22, times negative 1.01 times 10 to minus 3, divided by 0.24. This is equal to negative 0.0926 volts. So that's the induced EMF, which is equivalent to negative 92.6 millivolts.

Now because the EMF is negative, that means that the induced current is opposite in direction to the original current that created it. So let's say if this is the solenoid, and this is the original current. That current increases from 8 to 30. Because the current increases, the change in current is positive.

So therefore, if delta I is positive, a positive value times a negative sign will give us a negative EMF. And anytime the current increases, an induced current will be created in the coils that will oppose the increasing current that created it. So whenever you have a negative EMF, the induced current is opposite to the direction of the original current.

So if the original current is flowing clockwise, the induced current will be counterclockwise. It's just going to be opposite to what it is. So that's the direction of the induced EMF and the induced current. Number 11. A solenoid has an inductance of 150 millihenries with 300 turns of wire and a circular area of 2.653. square meters.

What is the potential energy stored in the inductor when a current of 20 amps passes through it? So what equation can we use to calculate the potential energy stored in this inductor? Here's the equation that you need. It's one half times the inductance times the square of the current.

So the inductance is 150 millihenries, but we need to convert it to henries. So it's 150 times 10 to the minus 3 henries. The current is equal to 20 amps. And let's not forget to square it.

So let's go ahead and plug this in. So therefore, the potential energy stored in this inductor is equal to 30 joules. So that's the answer to part A.

Now let's move on to part B. How many turns per meter does the solenoid have? So how can we figure this out?

Well, we know that the inductance is equal to mu0 times n squared times the area divided by the length of the inductor. So let's separate N squared and write it as N times N over A. Now we know that N over L is equal to lowercase n.

And lowercase n represents the number of turns per meter. So we need to use this form of the equation. So L is equal to mu0 times n times the lowercase n times the area. The inductance is 150 times 10 to the minus 3 Henry's.

The permeability of free space, that's 4 pi times... times 10 to the minus 7, that's mu0. Capital N represents the number of turns. There's a total of 300 turns in this inductor, or solenoid. The lowercase n is what we're looking for.

And the area is given to us. The area is 2.653 square meters. so first let's multiply these three numbers so 4 pi times 10 to the minus 7 times 300 times 2.653 that's about 1.0 times 10 to minus 3 Now, if you can't type in 4 pi times 10 to the minus 7 in your calculator, type in pi first and then multiply that by 4 times 10 to the minus 7. You might find that helpful. Now let's divide both sides by 1 times 10 to minus 3. The 10 to minus 3's will cancel, and so it's simply 150 over 1, so lowercase n is 150. So that's how many turns there are per meter. Now I know this question is not listed in the problem, but how long is the solenoid?

So if there are 150 turns per 1 meter, how long is it for 300 turns? So if 1 meter contains 150 turns, then 2 meters will contain 300 turns. So therefore, the solenoid is 2 meters long.

You can also use the... equation to get the same answer. We know that lowercase n is equal to capital N divided by L.

So let's put this over 1 and let's cross multiply. So nL is equal to capital N, which means means that the left is equal to the number of loops divided by lowercase n which is the number of times per meter so there's 300 turns in the entire solenoid and there's 150 turns per meter so 150 divided by 300 I mean 300 by 150 excuse me that's going to give us two meters so that's the length of the solenoid Now let's move on to part C. What is the magnetic field when the current is 20?

The magnetic field of a solenoid can be calculated using this equation. It's mu0 times n times i. Mu zero is 4 pi times 10 to the minus 7. Lowercase n is 150 turns per meter. And the current is 20 amps.

So if we multiply these three things, this is going to be 3.77 times 10 to the minus 3 Tesla. So that's the answer for part C. Now what about part D? What is the energy density of this magnetic field? To calculate the energy density, which is equal to lowercase u, it's b squared divided by 2 times mu0.

So we have b already, and we just got to divide it by 2 and the permeability of free space. 5.655 joules per cubic meter. So that is the energy density of the magnetic field that is in the solenoid. So that's how you can find it.

So that's it for this video. We've covered a lot of topics. And so thanks for watching.

And if you like this video, feel free to subscribe. And have a great day.

Transcript for:Understanding Faraday's and Lenz's Laws

Transcript for:
Understanding Faraday's and Lenz's Laws