Transcript for:
Faraday's Law and Lenz's Law

in this video we're going to talk about faraday's law of electromagnetic induction and also lens 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 that 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 inducing current and it's not only just moving the magnet into out of the coil if you change the area of the coil 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 magnetic field and 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 unifor area is square meters so when you multiply these two you get the unit of webbers now we know that there's going to be an induced current anytime 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 the 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 see if it's neither parallel or perpendicular to the normal line theta is here it's the angle between a 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 zero and cosine zero is equal to one 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 one 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 induce 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 lenses 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 where 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 i drawn is 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 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 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 they can't be the same so on the right side is 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 now 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 the lenses law to figure this out now the magnetic field is not increasing or in its constant however the area inside the rectangular loop that is exposed to the 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 in 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 lenses 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 axises 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 and 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 a 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 lenses long 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 increase the flux and any time 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 they're current and 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 lens law to find out now if the current in a 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 lenz's law the flux created by the induced current will 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 and 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 lens's law to figure it out so let's focus on the magnetic field created by the wire with the decrease in current if the current is going to the 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 decreased magnetic field and a decrease in magnetic field leads to a decrease in flux so according to lens's law the induced current will be directed in such a way that it's going to try to support the decrease in flux so the flux is decreasing it's going to try to increase it it's going to oppose a change now any time it tries to increase the flux is trying to support the decrease in magnetic field and any time which is supported 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 six volt battery and we have a three ohm resistor so right now the current is zero when the switch is open however once we close the switch the current will become two so it's gonna go from zero to two amps so therefore the current is increasing which means that the flux is increasing and if the flux is increasing according to lens'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 phalaenic field and the external magnetic field and induced magnetic 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 magnetic field and the induced magnetic field will be in opposite directions now we 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 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 and uh counterclockwise direction so on the right side since the current's 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 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 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 and 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 it's parallel to the normal line so the angle theta is the angle between the normal line and the magnetic field therefore the angle is zero degrees and cosine of zero is one 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 zero we know cosine of zero is one point twenty five squared times pi that's point one nine six three five and if you multiply that by the difference of four point eight and one point five which is three point three this will give you the change in flux which is uh 0.648 and the unit for magnetic flux is the weber which is uh 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 one the change in flux is 0.648 and the change in time the magnetic field increased from 1.5 to 4.8 tesla and 23 milliseconds so that's delta t but we need to convert milliseconds into seconds we could do so by dividing by a thousand 23 divided by a thousand is point zero two three so point six four eight divided by point zero two three that's about twenty eight point seventeen 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 two the magnetic flux through a coil of wire containing 20 loops changes from positive two to negative three what is in 425 milliseconds what is the induce emf the induced emf is equal to negative n times the change in the electric flux i mean not the electric flux but the magnetic flux divided by the change in time so in this problem we have 20 loops of wire the flux changes from two to minus three the final flux is negative three the initial flux is positive two so it's negative three minus positive two now we need to convert milliseconds to seconds and we can divide it by a thousand four hundred twenty five milliseconds is point four two five 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 three 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 point 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 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 a normal line and a 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 got to divide by 100 the change in area is going to be the final area minus 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 .05 meters times .08 meters and then let's multiply by cosine of 60 degrees and let's divide everything by point 15 seconds so let's calculate the change in area first point zero seven times point eleven minus point zero five times point zero eight that's point zero zero three seven or three point seven times ten to the minus three next multiply that by cosine sixty and then times one fifty and times twenty five so that's six point nine three seven five and divided by twenty fifteen 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 could find the power dissipated by the circuit it's the voltage or the inducement times the current so it's .46 rather 46.25 that's the inducement 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 watt is one joule per second so we have 21.39 joules per second and if we multiply by point 15 seconds we 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 the circuit so the circuit consumed 3.21 joules within the point 15 seconds in which the area was changing number four a rectangular coil of wire contains 25 loops the angle between a normal line of the coil and the magnetic field changes from 70 to 30 degrees and 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 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 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 c 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 point 15 meters times point 20 meters we have a rectangular surface so the area is simply the length times the width point 15 times point 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 uh 0.085 seconds don't forget to divide 85 milliseconds by a thousand 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 point five to four by point fifteen times point twenty times three times twenty five so that's going to be negative one point one nine and then divided 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 uh 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 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 a 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 zero 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 zero and cosine of zero is one so we don't have to worry about the cosine part and it's formal 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 one so the induced emf is going to be negative n or negative one and since we're looking for the magnitude we don't have to worry about the negative sign this can be 1 times b and delta a is l v 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 in v so it's blv now 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 lenses law to determine the direction of the current in the moving so because the rod is moving to the right the area increases and the flux increases so according to lenses 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 fields 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 instant 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 left of the rod from one point to the other point so to calculate the induced emf using another equation we need to use energy induce 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 clumps 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 v is perpendicular to the surface and so for that formula sine 90 is one so we don't have to worry about the sine part so let's replace f with b q v and d the distance that the charges move when being ex 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 five 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 eight tesla the length of the moving rod is 45 centimeters but we need to convert that to meters divided by 100 that's 0.45 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 could find the electric field in the parallel plate capacitor the electric field in the rod is the voltage in a 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 72 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 one volt per meter is the same as one 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 meters per second what equation can we use to figure this out well since we know the current we could find the force in a rod the magnetic force that acts on an object or basically the magnetic force that acts on a wire with the current is equal to ilb 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 endo rod it's uh 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 area of the rectangular coil it's sine 90 sine 90 is just one so we just got to multiply those three values so it's going to be 0.144 times .45 times eight and so this is equal to point five one eight four newtons so that's the force required to keep the rod moving at a speed of two meters per second once it has 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 six a 60 hertz ac generator rotates in a 0.25 tesla magnetic field the generator consists of a circular coil of radius 10 centimeters with 100 loops what is the angular 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 magnetic field times a the area of each loop times omega which is the angular 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 form 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 a hundred we have a hundred 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 point 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 sign part psi 90 is going to be one 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 the maximum output is 296.1 volts that's going to be the maximum emf generated by this particular generator number seven 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 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 e 2 is going to be n b a times omega 2. 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 angle speed of 2400 rpm 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 3.42 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 a thousand turns because the secondary coil has more turns in 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 uh six volts notice that the ratio between ns and np is ten a thousand divided by hundred ten 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 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 eight a transformer has 50 primary turns and 400 second returns 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 consumed 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 and s is 400. now 400 divided by 50 is eight so the number of turns increases by a factor of eight 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 we can do ns divided by np is equal to vs over vp so np is 50 and s 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 4 800 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 divide both sides by 400. 1200 divided by 400 is 3. so is is 3 amps and then once you have is and vs you can find a power 96 times 3 is 288 or 12 times 24 is also 288 number nine a 200 watt ideal transformer has a primary voltage of 40 volts and the secondary current 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 could 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 divided by 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 40 ms 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 and 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 the 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 factor 4. so 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 4. so it's 20 amps and as you 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 is 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 increase in 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 decrease in 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 decrease in 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 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 mu zero times n squared times a divided by l mu zero is the permeability of free space it's four pi times ten to minus seven 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 two 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 mu zero times n times i i went over this in another video it's the video on magnetism uh 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 flown 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 both equal to the induced emf 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 zero so cosine zero is one 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 mu zero times n times i so let's replace these two so the inductance l is equal to n times mu zero times n times i times a divided by delta i and of course we still have a triangle somewhere so we're going to do is we're going to cancel the current and so we have l is equal to n times mu 0 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 zero times n and then let's replace lowercase n with n over l times a so n 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 centimeters so the radius is always going to be half of the diameter so 8 centimeters divided by two the radius is four centimeters 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 four pi times ten to negative seven the solenoid consists of two hundred 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 henrys now this is equivalent to 1.01 millihenries 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 henrys 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 any time 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 milliamperes but we need to convert it to henry's so it's 150 times 10 to the minus three 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 mu zero 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 lower case 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 mu zero times n times the lowercase n times area the inductance is 150 times ten to the minus three hundreds the permeability of free space that's four pi times ten to the minus seven that's mu zero capital n represents the number of turns there's a total of three hundred turns in this inductor or solenoid the lower case 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. if you can't type in 4 pi times 10 to 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 one times ten to minus three the ten to the minus threes will cancel and so it's simply one fifty over one so lowercase n is a hundred and fifty 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 many how long is it for 300 turns so if one 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 one let's cross multiply so now is equal to capital n which means that the length is equal to the number of loops divided by lowercase n which is the number of turns 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 divided by 150 excuse me that's going to give us uh 2 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 mu zero times n times i mu zero is four pi times ten to minus seven 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 low case u it's b squared divided by two times mu zero 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