In this video, we're going to talk about magnetism. Perhaps you're familiar with bar magnets. And you know that when you place the north pole of a barred magnet with another north pole, these two, they repel.
They're going to push apart. However, let's say if you were to face the north pole of one barred magnet with the south pole of another. they won't repel. These two will feel a force of attraction.
The North Pole is attracted to the South Pole. But if you put the North Pole with another North Pole of another magnet, they will repel. Or if you put the South Pole of one magnet with the South Pole of another, they will also repel.
Every barb magnet has its own magnetic field, which emanates away from the North Pole and it travels towards the South Pole. Likewise, this one has a magnetic field emanating away from the North Pole. Notice that the magnetic field cancels in the middle.
Those two, they repel each other. And, in the case of the other example... The magnetic field leaves the North Pole, but it enters the South Pole. So in the case of attraction, notice that the magnetic field between the North and the South Pole of the two bar magnets, notice that they're in the same direction, and so they're additive. You have that force of attraction.
So what causes magnetic fields? Magnetic fields are created by moving electric charge. And this example can be illustrated if you have a wire.
Whenever there's an electric current that flows through a wire, it creates its own magnetic field. And the magnetic field created by this wire, it looks like this. It's circular. On the left side, the magnetic field is leaving the page, so it's represented by a dot or a circle. And on the right side, it enters the page, which is symbolized by an X.
You need to be familiar with that, by the way. You can use the right hand rule to figure this out. If you take your hand and if you curl it around a pen with your thumb facing the direction of the current, the way your fingers curl around the pen is the way the magnetic field travels around the wire.
Try that. So here's a picture that describes it. As you can see, you want the thumb facing the direction of the current, which is upward.
And as you curl your hand around the wire, notice the way your hand curls around it. On the left side, you can see how it's coming out of the page. and then on the right side your hand curls into the page and so the way your hand curls around that wire is the way the magnetic field created by the moving charge travels around the wire it's out of the page on the left side and it's going into page on the right side Now there's an equation that allows you to calculate the strength of the magnetic field created by such a wire. And here's the equation. B is equal to U0 times I divided by 2 pi R.
So let's say if you want to calculate the magnetic field some distance point A away from the wire. R is the distance between the wire and point A. B is the strength of the magnetic field. field and B is measured in units of Tesla or capital T.
U0 or mu0 is equal to 4 pi times 10 to the minus 7. This is known as the permeability of free space. And the units are tesla times meters per amp. Now notice that the current and the magnetic field are directly related. If you increase the magnitude of the current, the strength of the magnetic field generated by this wire will increase as well.
And the reason for that is because the current is on the top of the fraction. Whenever you increase the numerator of a fraction, the value of the entire fraction will increase. Now R is on the bottom, so that means that R is inversely related to B.
If you increase the distance between... the wire and the magnetic and point of interest I should say the magnetic field at that point will be weaker as you move away from the wire the strength of the magnetic field weakens by the way the number of magnetic lines that you see in a picture is proportional to the strength of the magnetic field So for example, let's say if the magnetic field in this region looks like this, versus... two lines as opposed to three.
The magnetic field on the left is stronger than the one on the right. If you have more lines that are closer together, the strength of the magnetic field is stronger. But make sure you know this relationship. So anytime you increase the electric current in a wire, the strength of the magnetic field will increase.
And as you move away from the wire, the strength of the magnetic field will decrease. Let's work on this problem. A vertical wire carries a current of 45 amps due south. Calculate the magnitude and the direction of the magnetic field 2 centimeters to the right of the wire.
So go ahead and try this problem. So let's say this is the wire, and the current is due south. So it's going down.
And that means that, using the right-hand rule, the magnetic field is going to enter the page on the right side, and it's going to be out of the page. I mean, it enters the page on the left side, but on the right side, it comes out of the page. So when it enters the page, put an X.
And when it leaves the page, put a circle, a closed circle. Now our goal is to find the magnetic field 2 centimeters to the right of the wire. So we already have the direction of the magnetic field. It's out of the page.
All we need to do now is calculate the magnitude. So we can use this equation. B is equal to mu0 times I divided by 2 pi r.
Mu zero is equal to 4 pi times 10 to the minus 7, and the current is 45 amps. R is the distance between the wire and a point of interest. In this case, R is 2 centimeters, but we need to convert that to meters. So we've got to divide by 100. 1 meter is equal to 100 centimeters.
So that's going to be.02 meters. So now all you've got to do is just type this in. And you should get...
4.5 times 10 to the negative 4 Tesla. So that's going to be the strength of the magnetic field 2 centimeters away from the wire. Number 2. A wire carries a current of 10 amps.
At what distance from the wire will a magnetic field of 8 times 10 to the minus 4 Teslas be produced? So we gotta find r in this case. We can use the same equation.
We don't have to worry about the direction, so we don't really need to draw a picture. So let's solve for r. Let's multiply both sides by r.
So on the right side, it's going to cancel. So b times r is equal to mu0 times i divided by 2pi. Now let's multiply both sides of the equation by 1 over b. If we do that, on the left side, b will cancel. So the distance is going to be mu0 times i.
divided by 2 pi times b. So it's 4 pi times 10 to the minus 7 times the current of 10 amps divided by 2 pi times the strength of the magnetic field. So, we can cancel pi.
In fact, 4 pi divided by 2 pi is just 2. So, it's going to be 2 times 10 to the minus 7 times 10 divided by 8 times 10 to the minus 4. And so this is equal to 2.5 times 10 to the minus 3. And the units is meters. So if we want to, we can convert it to millimeters. And to do that, you need to divide, actually multiply by 1,000. There's 1,000 millimeters per meter. If you multiply by 1000, this will give you 2.5 millimeters.
And so that's the answer. That's how far away from the wire that you have to be to measure a magnetic field of 8 times 10 to the minus 4. By the way, if you ever were to place a compass near a wire, whenever there's an electric current flowing through that wire, it will cause the compass to deflect. You should try it. Now let's say if we have a current carrying wire. What's going to happen if we place this wire inside a magnetic field?
Let's say the magnetic field is directed east and the current is moving north. A magnetic field exerts no force on a stationary charge. However, if the electric charge is moving, then the magnetic field will exert a force, specifically a magnetic force. So, whenever you have a wire with an electric current, that means you have moving charges in the wire, the magnetic field will exert a force on the wire.
You can calculate the strength of the magnetic field by calculating the force on the wire. force using this equation F is equal to I LB sine theta so the strength of the magnetic force is proportional to the current if you increase the current the magnetic force will increase The magnetic force is also proportional to the strength of the magnetic field. If you increase the magnetic field, the magnetic force will increase as well. And it's also proportional to the length of the wire. Now, it depends on the angle as well.
So here's one example where the current and the magnetic field are perpendicular. And here's another example when the current and the magnetic field are at an angle. And here's one that's parallel.
Now, theta is the angle between the current and the magnetic field. When they are perpendicular, sine 90 is equal to 1. And 1 basically represents 100%. So, the maximum force occurs when the current and the magnetic field are perpendicular to each other.
Now, when it's at an angle, it's going to be between anywhere from 0 to 100% of its maximum value. So then you can use this equation. When they're parallel, the angle is equal to 0 degrees.
Sine 0 is equal to 0. Therefore, the magnetic field exerts no magnetic force on a moving charge that moves parallel or even anti-parallel to the magnetic field. So for the third example, there's no magnetic force acting on the current. They have to be at an angle with respect to each other. They can't be parallel. Now what about the direction?
Let's go back to our last example. So let's say if the current is due north and the magnetic field is directed east. In what direction is the force? Now the force has to be perpendicular. to the current and to the magnetic field.
So if the current is in the north-south direction, and if the magnetic field is in the west-east direction, then the force is either into the page or out of the page, that is, along the z-axis. So how can we figure this out? Well, we have to use the right-hand rule.
So take your right hand, and you want to extend it. You want your thumb to be in a direction of the current. And you want your other four fingers to be in a direction of the magnetic field.
So this represents B, and this represents the current I. So using your right hand, look at where your right hand opens towards. If you direct it the way it's presented here, it's going to go into the page. The force comes out of the palm of your right hand.
And so it's going to be directed into the page. And that's how you can figure it out. Let's try another example.
So let's say if we have a wire, and a current is directed east, and a magnetic field is directed into the page. In what direction is the force going to be? So if the current is in the east-west direction, and, well, the current's really west to east, and a magnetic field is in the z direction, that is between out of the page and into the page, then the force has to be in a north-south direction.
These three variables have to be perpendicular to each other. So, what you want to do this time, you want to direct your four fingers into the page. And you want your thumb...
directed east, that is in the direction of the current. You want your forefingers to be into the page and the force comes out of the palm of your right hand. So the force will be directed north. If you do that. It's kind of hard to draw the direction of the hand on this video, but hopefully you can visualize it.
Number 3. A 2.5 meter long wire carries a current of 5 amps in the presence of a magnetic field with a strength of 2 times 10 to the minus 3 teslas. Calculate the magnitude of the magnetic force on the wire using the picture shown below. So, feel free to try that.
The equation that we need is F is equal to I LB sine theta. Now let's talk about the angle. So the current is directed east, and the magnetic field is directed 30 degrees relative to the horizontal.
So theta is always going to be the angle between the magnetic field and the current. So you can also use this angle as well, because that angle is between B and I. 180 minus 30 is 150. And it turns out that sine of 150 and sine of 30, they're both equal to 1 half. So it doesn't matter. So whether you choose this angle, which is between I and B, or if you use this angle, the answer will be the same.
So just something to know. So now let's go ahead and calculate F. So it's going to be the current, which is 5, times the length of the wire.
which is 2.5 meters times the strength of the magnetic field which is 2 times 10-3 multiplied by sine of 30. So the magnetic force is very small it's.0125 N And so that's going to be the force exerted on this current carrying wire. Number 4. A current of 35 amps flows due west in a wire that experiences a magnetic force of 0.75 newtons per meter. What is the strength of the magnetic field which is directed due south? So here's the wire, and the current is directed...
West and the magnetic field is directed south Our goal is to find the strength of the magnetic field you need to solve for B. So let's write the equation F is equal to I L B Now because the current and the magnetic field are at right angles to each other because because it's 90 degrees, sine 90 is 1. So we don't need the sine portion of this equation. Now we're given the force per meter. So that's F divided by L. If we divide both sides by L, we're going to get this equation F over L is equal to the current multiplied by the magnetic field.
We have the force per meter. That's 0.75. That value takes care of two of these variables. We have the current, which is 35 amps. So we've got to solve for B.
So it's just going to be 0.75 divided by 35. And so B is equal to 0.0214 Tesla. So anytime you have the force per unit left, or newtons per meter, make sure you understand that it's F divided by L. The entire thing is F over L.
So you might see that expression in this chapter in a few problems. Now what is the direction of the magnetic force? We know the current flows west, the magnetic field is south, so the force is either into the page or out of the page.
So what you want to do is you want to direct your four fingers south and your thumb west, and the force should come out of the page. So let's see if I can draw that. So you want your thumb facing this way, and the four fingers facing this way. Using your right hand, if you do that, I put into the page.
This should be out of the page. The force should come out of the palm of your hand. So make sure the current is aligned with your thumb. The magnetic field is aligned with your four fingers. And then the force should come out of the palm of your hand.
That's out of the page. Now let's say if we have a rectangular metal loop. with a current that flows in the metal loop clockwise.
So in this section, the current is going up, here it's going down, here it's directed towards the right, and here it's directed towards the left. And only a portion of this loop is inside a magnetic field. That is, only the bottom portion. Now let's say the magnetic field is directed into the page. So I'm going to put an X everywhere.
Where will this rectangular loop move if initially it's at rest? Will it move towards the right? Will it begin moving towards the left, up, or down? What would you say? So let's start with this portion of the wire, or of the metal loop.
Let's see what the magnetic force on that portion is directed. So what you want to do is you want to place your thumb facing south. You want the four fingers of your hand to be going into the page.
And the magnetic force should be directed east. So let's see if I can draw that. So you want your four fingers going into the page. You want your thumb going south.
So make sure the magnetic field where your four fingers is going into the page. And your thumb is in a direction of the current. And the force should come out of your hand, and that is out of the palm of your right hand. And it should be directed east, if you do it correctly.
So, make sure you try that, and make sure you can master this right hand rule. Now... For the other side, the left side, everything is the same except the current.
Because the current is in the opposite direction, the force has to be in the opposite direction. Now these two forces are equal in magnitude, and because they're opposite in direction, they will cancel out. So the loop is not going to move towards the left or towards the right.
These forces balance each other out. Now, in the top part of the loop, there's no magnetic field in that region, so therefore there's no force. The net force is going to be based on this portion of the loop.
Because it's not balanced by this portion of the loop. If the entire loop was in the magnetic field, all the forces will cancel. But since it's not, this one will create a net force. So now let's draw another picture. Point your thumb towards the left, and make sure your four fingers are going into the page.
If you do it correctly, the force should be directed south. So what you want to do is, you want your thumb directed west, and you want your four fingers going into the page. So if you do it correctly, the magnetic force should be coming south, out of the palm of your right hand. Now let's move on to another topic. We talked about how to calculate the magnetic force on a current carrying wire.
But what about the magnetic force on a single point charge? Because any moving charge will create a magnetic field. So let's go ahead and come up with an equation.
Let's start with this equation, F is equal to I L B sine theta. Current is defined as the amount of electric charge that flows per unit time. And capital Q is the total amount of charge.
It can be due to many charge particles. So capital Q is going to be... equal to lower case Q which is the magnitude of each charged particle times and which is the number of charged particles and now give you the total charge divided by T Now, distance is equal to the speed multiplied by the time. And length could be thought of as distance.
They're both measured in meters. So, we can replace L with Vt. So, let's replace I with Qn over T. And let's replace L with VT.
So we can cancel T. Now, if you want to find the magnetic force on a single point charge, that means there's only one charge particle, so N is 1. So when N is 1, we can get rid of it. So this leaves us with F is equal to BQV sine theta.
So that's how you can calculate the magnetic force on a moving charged particle when it's inside a magnetic field. So let's say if we have a proton. And let's say it's moving towards the right, and also the magnetic field is directed towards the right. If these two are parallel, sine 0 is equal to 0, so there's going to be no magnetic force. They have to be perpendicular.
Now, if the proton is moving at an angle relative to the magnetic field, And then you can use the equation F is equal to BQV sine theta, where theta is the angle between the magnetic field and the velocity vector. Now let's say if the proton... it's moving perpendicular to the magnetic field. That is, they're at right angles, or 90 degrees relative to each other. Sine 90 is 1. So the magnetic force will have its maximum value at this point, and it's equal to simply BQV.
Now let's say if we have a proton and it's moving towards the right and the magnetic force or rather the magnetic field is directed north. What is the direction of the magnetic force? It has to be in the z-direction. If the velocity is in the x-direction, and if the magnetic field is in the y-direction, the magnetic force has to be in the z-direction.
And you can use the right-hand rule. Think of velocity as the current, so to speak. That's where the charged particles are moving. So you can use the same right hand rule as we've been doing.
You want your four fingers to be in the direction of the magnetic field. And you want your thumb to be in the direction of the velocity. So if you orient your right hand this way, the force should come out of the palm of your right hand.
And so it should come out of the page. So I'm going to put a circle for that. For a proton, the magnetic force is out of the page.
But, for an electron, the magnetic force will be in the opposite direction. That is, it's going to be going into the page. So, for any negatively charged particle, simply reverse the direction of the magnetic force, if you need to find it. For a positively charged particle, it's going to work in the exact same way as the right-hand rule directs it. Number 5. A proton moves east with a speed of 4 times 10 to the 6th meters per second in a magnetic field of 2 times 10 to the minus 4 teslas directed into the page.
What is the magnitude of the magnetic force acting on a proton? So if we need to find the magnitude, we don't have to worry about the direction. All we need to know is that the velocity and the magnetic field are perpendicular to each other.
The proton is moving east in the x direction. The magnetic field is directed into the page, that is, in a negative z direction. So therefore, the magnetic force has to be in the y direction.
So all we need to do is find the magnitude. So we just got to use the equation. F is equal to BQV sine theta.
Now, because the velocity and the magnetic field are perpendicular, that is, the velocity is in the x direction, the magnetic field is in the z direction, the angle has to be 90 degrees. And sine 90 is 1. So F is simply equal to BQV. B is the magnetic field in Tesla.
That's 2 times 10 to the minus 4 Teslas. Q is the charge of just one proton. The charge of a proton is 1.6 times 10 to the negative 19 coulombs.
And that's something you just gotta know. And the speed of the proton is 4 times 10 to the 6 meters per second. So we just got to multiply these three numbers.
And so you should get 1.28 times 10 to the minus 16 newtons. So that's the magnetic force acting on the proton. Now, let's talk about a proton. So let's say if we have a proton, and it's moving towards the right, and the magnetic field is directed everywhere into the page. So let's say there's an X everywhere.
What's going to happen? Where is the magnetic force? So, if you direct your thumb towards the right and your four fingers into the page, the magnetic force will be directed north. Whenever force and velocity are perpendicular to each other, the object will turn. So it's going to go this way.
If force and velocity are in the same direction, the object will speed up. If force and velocity are in the opposite direction, the object slows down. But if they're perpendicular, the object will turn.
And eventually, the particle is going to be moving in a direction of the force. But now that the velocity is directed north, the force is going to change. If you use the right-hand rule again, if you direct your forefinger...
into the page, the velocity north, the force will be directed west. And so what's going to happen is this particle, the proton, is going to move in a circle. I haven't drawn a nice circle because I ran out of space, but you get the picture. When it's at the top, the force is going to be directed south as it moves to the left. And as the proton moves south, the force will be directed east.
So notice that for a moving charged particle, the magnetic force behaves as a centripetal force, or a center secant force. Now what if we have an electron? If we had an electron, the situation would be opposite.
As the electron is moving in the same direction as the proton, it's going to feel a force in the opposite direction. So the proton felt a force that was directed north. The electron will feel a force directed in the opposite direction, that is south. So as the proton moves in the counterclockwise direction, the electron will move in a clockwise direction. So they will move in an opposite direction.
Now how can we calculate the radius of curvature that a proton or electron might travel in a circle? How can we figure out the radius? If you ever get a question like this, what you need to do is set the centripetal force equal to the magnetic force. Let's call FB the magnetic force, FC the centripetal force. The centripetal force is provided by the magnetic force in this particular example.
The centripetal force is equal to mv squared divided by r, which is the radius of the circle. The magnetic force is equal to... Now, because the magnetic field and the velocity are perpendicular, we don't have to worry about the sine portion of this equation.
Now what we're going to do is multiply both sides by 1 over V. So on the right side, V will cancel. On the left side, because we have a V squared, one of them will remain. So MV divided by R is equal to BQ.
So multiplying both sides by R, we have this equation. So make sure you write down this equation, mv is equal to bqr, because in this format, you can solve for anything. So let's say if we want to find the radius of the circle, all we need to do is divide both sides by bq. So the radius of the circle is equal to the mass of the charged particle times the velocity, divided by the magnetic field, times the charge. Number six, a proton moves with a speed of 5 times 10 to the 6 meters per second in a plane perpendicular to a magnetic field of 2.5 Tesla.
Calculate the radius of its circular path. So here's the proton. And if it's moving perpendicular to a magnetic field, it's going to move in a circle.
Our goal is to calculate the radius of that circle. So R is equal to, based on the equation that we had before, it's mv divided by bq. So what is the mass of a proton?
Now the problem doesn't give it to you. which means you can either look it up online or you can look it up in the reference section of your textbook. The mass of a proton is about 1.673 times 10 to the minus 27 kilograms. The speed, which is given, that's 5 times 10 to the 6 meters per second. The strength of the magnetic field is 2.5 Tesla.
And the charge of a proton, which is... The same as that of an electron, but the opposite sign. It's 1.6 times 10 to the negative 19 coulombs.
So if we type these numbers into the calculator, we should get... 0.0209 meters which is equal to 2.09 centimeters so that's the radius of the path that the proton is going to travel in and so that's how you could find it now what about part B what is the energy of the proton and electron volts how can we find the answer to that question Well first, we need to find the energy in Jules. A moving object has kinetic energy.
Any object in motion contains kinetic energy. So we've got to find the kinetic energy of the proton, which is 1 half mv squared. So we know the mass of a proton.
It's 1.673 times 10 to negative 27. And we also have the speed, 5 times 10 to the 6 meters per second squared. so this is equal to 2.09 times 10 to negative 14 joules now once you have the energy in joules you can convert it to electron volts electron volts is basically another unit of energy it's very useful for small particles like protons and electrons One electron volt is equal to 1.6 times 10 to the negative 19 joules. We'll talk about why later, but that's what it's equal to.
And so you just got to convert it to electron volts, and this will give you the answer. which is 130,703 electron volts. So that's it for this problem.
So why is it that 1 electron volt is equal to 1.6 times 10 to the negative 19 joules? Why is that the case? Electric potential, which is measured in volts, is equal to the electric potential energy, which is measured in joules, divided by the charge. So the unit volt, 1 volt, is equal to 1 joule per 1 coulomb.
So therefore... An electron has a charge of 1.6 times 10 to negative 19 coulombs. It's negative, but let's ignore the negative sign.
That's the charge of an electron. And an electron that has 1 volt, or 1 eV, that's 1 electron volt, an electron with 1 volt, will have an energy of 1.6 times 10 to negative 19 joules. Volt is basically the ratio between joules and coulombs. So if you have a charged particle that has one joule and one coulomb, its voltage is one volt. Or the electric potential is one volt.
Voltage is really work per unit charge. Electric potential is energy per unit charge. Now, if we have a charge particle that has an energy of 1.6 times 10 to the negative 19 joules, and a charge of that many coulombs, then these two will cancel, and the electric potential will be 1 volt, which is the case of an electron. An electron has a charge of...
1.6 times 10 to the negative 19 coulombs. And if that electron has an energy of 1 electron volt, its voltage is 1 volt, which means its energy is equal to this number. And that's why 1 electron volt is 1.6 times 10 to the negative 19 joules.
Just in case you were wondering. Now, let's say if we have two wires parallel to each other. Now let's say that there's a current in wire 1 and in wire 2, and these two currents are in the same direction.
Will these two wires attract each other, or will they repel? It turns out that these two wires will attract each other if they have a current in the same direction. Now, if there's two wires... with the opposite current then the situation is different they will not feel a force of attraction rather they will repel each other so why does that happen why do we have a force of attraction in the first case but in the second scenario these two repel What's going on here?
Wire 1 creates a magnetic field because it has a moving charge, it has a current. And that magnetic field exerts a force on wire 2. Now let's focus on wire 1. The current is going north. If you use the first right hand rule that we talked about earlier in the video, where you curl your hand around the wire and your thumb is in the direction of the current, The magnetic force created by wire 1 will be out of the page on the left side, but it's going to be into the page on the right side. So wire 2 is on the right side of wire 1. Therefore, wire 2 sees a magnetic field that's going into the page. Now, using the second right-hand rule that we talked about, what you want to do is you want your fingers to point into the page, but you want your thumb pointing north in the direction of the current.
So here's your thumb. It follows I2. The magnetic field, which is B1, that's created by I1, you want that to be in the page.
And what's going to happen is the force is going to come out of the palm of your hand. And it turns out that force is directed towards wire 1. So it's a force of attraction. And so anytime you have two wires with the current going in the same direction, it's going to create an attractive force.
The two wires will be attracted to each other. If the current is in the opposite direction, then the two wires will repel each other. Now how can we calculate the magnitude of the force between the two wires?
So we said that the first wire, wire 1, creates a magnetic field that causes the second wire to be attracted to the first if they're moving in the same direction. And also, the second wire creates a magnetic field that exerts a force on the first wire, causing the first wire to move towards the second if the currents are in the same direction. So let's start with this equation. Wire 1 creates a magnetic field B1, which is U0 times I1 over 2 pi r, where r is the distance between the two wires.
So wire 1, which produces a current one, will generate a magnetic field. that is at wire 2 and so r is the distance between the two wires. Now we need to use the other equation that is the force that acts on a current carrying wire that's F is equal to I L B sine theta but we're not going to be worried about the angle in this problem. So the force on wire 2 which we'll call F2 is equal to the current on that wire times the length times the magnetic field created by wire 1. So what we're going to do now is replace B1 with mu0 I1 over 2 pi r. So it's going to be I2 times L.
times mu zero I1 over 2 pi r. So that's how you can calculate the force between the two wires. F2 and F1, they have the same magnitude. Number 7. What is the magnitude and direction of the force between two parallel wires that are 30 meters long and 2 centimeters apart, each carrying a current of 50 amps in the same direction?
Well, let's draw a picture. So we have two wires, and they have the same current. So I1 and I2 both equal 50 amps. Now we have the length of the wire, which is L. And the length of both wires is 30 meters.
And the distance between them, which is R, that's 2 centimeters, which is equivalent to 0.02 meters. Now, for these type of problems, there's only two answers for the direction. Either the force is attractive, or they will repel. Because the currents are in the same direction, we have a force of attraction. And so that answers the direction of the force.
It's simply attraction. Now all we need to do is find the magnitude. So let's use the formula.
F is equal to mu0 times... I1 times I2 times L, which is the length of the wires, divided by 2 pi times R. So it's 4 pi times 10 to the minus 7 times 50 times 50, which we can write as 50 squared, times L, which is 30 meters, divided by 2 pi times the radius of 0.02 meters.
Now, 4 pi divided by 2 pi is... it's going to be 2. So it's 2 times 10 to the minus 7 times 50 squared times 30 divided by 0.02. So the force between these two wires is 0.75 Newtons.
And a direction is a force of attraction. These two forces will be pointed towards each other. Now the next thing that we need to talk about is Ampere's Law.
Ampere's Law describes the relationship between the current and the magnetic field produced by that current. Perhaps you've seen this equation. The sum of all the magnetic fields that is parallel to any segments that the magnetic field passes through, that's going to be equal to mu0 times the current enclosed by the path that the magnetic field makes.
It has to be a closed path. A good example is using a wire. So let's say if we have a current that passes through this wire. Now this current will create a magnetic field that travels around the wire in a circular path.
So if we take the magnetic field and multiply it by the path that is parallel to it, let's call it delta L. We can use that to get an equation that will give us the magnetic field created by this wire. So the path that the magnetic field travels is basically the path of a circle. So delta L is...
really 2 pi r if we add up all the small segments. So if we add up this segment plus that segment plus that segment we're going to get the circumference of a circle which is 2 pi r. So it's going to be b times 2 pi r which equals mu is not over i, I mean times i. So the magnetic field if you divide both sides by 2 pi r we get this familiar equation nu 0 times i over 2 pi r.
Now it's important to understand that the current in this equation is the current that is enclosed by this loop. So that loop has an area and... The current that passes through that area, that's this current. It's the current enclosed by that circular loop.
And so that's how you can use Ampere's Law to get the equation of a magnetic field by a current carrying wire. Now let's use Ampere's Law to come up with an equation for a solenoid. A solenoid is basically a device with many loops of wire.
And the reason why it's advantageous to create a solenoid, anytime you create a loop of wire, whenever you have a current, the magnetic field that is at the center of the wire is very strong. And for every loop that you add, you increase the strength of the magnetic field inside the wire. And so solenoids are very useful for creating powerful magnetic fields. Outside of the loop, the magnetic field is weak. So the magnetic field will travel in a circular pattern.
So to calculate the, or to derive the formula for a solenoid, we need to create a path of the magnetic field. So let's draw a rectangular path. So let's say this is A, B, C, and D. So what we need to do is add up the magnetic field that is parallel to each segment.
So that's going to be the magnetic field times the left of segment AC. plus the magnetic field times the length of segment CD plus the magnetic field times the length of segment DB plus the magnetic field times the length of segment BA Now, let's focus on segment BD. Segment BD is outside of the solenoid, and the magnetic field is very weak outside of the solenoid.
So, we can say that the contribution for BL in segment BD is very small. So, it's negligible. Now, BA and DC, they're perpendicular to the magnetic field that is inside the solenoid. That magnetic field is much stronger than the magnetic field on the outside. So, because it's perpendicular to the magnetic field that's inside the solenoid, its contribution is going to be negligible.
So, we can eliminate BA and DC. So therefore, all we have is the segment that is parallel to the magnetic field that is inside the solar node, that's AC. Segment AC is the most important segment because, one, it's parallel to the magnetic field that is inside the solenoid. And that magnetic field is the strongest one compared to the ones that are outside of it. And so, BL in segment AC will have the greatest contribution towards the sum of all the magnetic fields that is parallel to each segment.
So now using Amper's Law, which is basically this equation, we can now replace this term with B times L, where L is simply the length of the solenoid. The length of the solenoid being segment AC, but we'll just call it L. So, BL is equal to mu0 times I.
But this is for if we only have one loop. Let's say if you have a wire with a current of 10. If you have one loop, then the enclosed current is 10 amps. But it turns out that if you add another loop, the current enclosed by the magnetic field It's going to be twice as much.
Even though 10 amps is flowing through the wire, the enclosed current is now 20. And if you add another loop, the enclosed current is 30, even though 10 amps is still flowing through the same wire. So therefore, we need to add n to this equation because the enclosed current increases. So the enclosed current is basically the current that flows in one loop times the number of loops. So now all we need to do is divide both sides by L. So, B is equal to mu0 times n times I divided by L.
Lowercase n is equal to capital N over L. Capital N represents the number of turns or loops. L is the length in meters.
So, lowercase n is the number of loops or turns per meter. So we're going to replace capital N over L with lowercase n. So now we have the equation of a solenoid. B is equal to mu0 times n times I. So the magnetic field produced by a solenoid is proportional to the current that passes through it.
As the current increases, the strength of the magnetic field will increase. The second way to increase the magnetic field is to increase the number of turns. If you can increase the amount of turns per unit left or per meter, the strength of the magnetic field will greatly increase. So you want to increase the number of turns, you want to increase the current, but you also want to decrease the length.
If you can decrease the length, the magnetic field will increase as well. So the magnetic field is directly related to the current, it's related to the number of turns, and it's inversely related to the length. Number 8. A solenoid has a length of 15 centimeters and a total of 800 turns of wire. Calculate the strength of the magnetic field at its center if the solenoid carries a current of 5 amps. So first, let's calculate lowercase n, which is capital N over L.
So there's 800 turns, and the length of the wire is 15 centimeters. But we need to convert that to meters. So let's divide by 100. 15 divided by 100 is 0.15, or 0.15 meters. So 800 divided by 0.15, that's equal to 5,333 turns per meter, or loops per meter. So now to calculate the strength of the magnetic field, it's mu0 times n times i.
So it's 4 pi times 10 to the minus 7 times the number of turns per meter, which is 5,333, times the current, which is 5 amps. So the strength of the magnetic field at the center, it's about 0.0335 Tesla. And so, that's it for this problem. Now what's going to happen if we have a current carrying loop inside a magnetic field? In a magnetic field, the loop is going to rotate.
It's going to produce a torque. So let's say on the left side there's a current, and on the right side there's a current. So the current travels clockwise in this loop. And there's also a magnetic field that is directed east. Now what type of force will we have on the left side of the wire, or the metal loop?
So using the right hand rule, place your thumb going up and your four fingers in the direction of the magnetic field. And if you do that, notice that the palm of the hand opens into the page. And so that's where the force is going to be. So the magnetic field will exert a force on the left side of the loop, going into the page.
Therefore, on the right side of the loop, the current is reversed, so it must be out of the page. So therefore, this... The loop is going to turn. It's going to turn this way, and then into the page on the other side. Now let's talk about how to derive an equation for the torque of this current carrying loop.
F is equal to ILB. It's ILB sine theta, but the force is perpendicular. both to the current and magnetic field.
So sine 90 is 1. So the force acting on the right side and on the left side of the loop is going to be ILB. There's no force on the top section in the bottom section of the loop. the loop. The reason for that is because the current is parallel to the magnetic field. And whenever the current is parallel to the magnetic field, there's going to be no magnetic force.
It has to be perpendicular to it. So there's only a force on the left side and on the right side, not on the top or bottom section of the loop. So, since we're focused on this side, that's where the magnetic force will be exerted on. L represents the length of that section. So, that's this L.
B is the magnetic field. The width of the loop, let's call it W. Let's put that here. Now we need to calculate the torque.
This force will create a torque, and this force will create another torque that's additive. So the net torque is going to be T1 plus T2. And just to refresh you on how to calculate torque, torque is the product of the force times the level arm.
So it's F times R. And if the force is at an angle, it's going to be FR sine theta. So, T1 and T2 is going to be F1 times R1 times sine theta plus F2 times R2 sine theta. So what exactly is R in this example?
We know that R is the distance between where the force is applied and the axis of rotation. That's where the object moves around. So in this particular problem, the axis of rotation is here.
So which means that R... is half of W. So R is W divided by 2. So now we can replace F with ILB and we can replace R with W over 2. And then T2 is going to be the same thing. It's the same current times the same length. They both have the same length.
They're both exposed to the same magnetic field. And R is the same. R1 and R2 is the same. So we no longer need this picture. So what I'm going to do is I'm going to factor out ILB and sine theta.
So T is ILB sine theta W over 2 plus W over 2. Half plus half is a whole. So, T is equal to ILWB sine theta. Now for the rectangular loop that we have, which has a length L and a width W. The area of the loop is basically the length times the width. So A is L times W.
So let's replace LW with A. So the torque of a single loop is the current multiplied by the area. times the strength of the magnetic field, times sine of theta.
Now what if we have many loops? If you have two loops, the torque is going to be twice as strong. Three loops, three times strong. So we need to add n to this equation. for the number of loops.
So it's NIAB sine theta. By the way, the quantity NIA is known as the magnetic dipole moment, represented by capital M. So it's equal to the number of loops times the current times the area. That's the magnetic dipole moment.
But this is the equation that we need to calculate the torque. It's niab sine theta. Now let's say if this is the face of the loop. Let's draw the normal line perpendicular to the face of the loop. So this is the area of the loop, which we could describe it as A, and the red line is the normal line that's perpendicular to A.
And sometimes the magnetic field, it's not going to be parallel or perpendicular to the surface. It can be at an angle. The angle theta is between the magnetic field and the normal line, which is perpendicular to the surface of the coil.
So make sure you understand that. So theta is between B and the normal line. Now, instead of drawing a side view of the current carrying loop, let's analyze it by drawing the top view.
So here's... This is the top view of it. And here's the rest of the loop. If you wish to see it this way.
And this part is perpendicular. to the plane of the loop. Let's say the magnetic field is at an angle of 90 degrees.
When it's at an angle of 90 degrees, you're going to get the maximum torque possible. So make sure you understand that. I'm going to redraw like this. So let's say B is directed east.
We're going to have one force going up and the other force going down. So this part is perpendicular to the area or to the surface of the plane of the loop I should say. And here's the magnetic field. So it's at an angle of 90 degrees.
And so you're going to get the greatest torque in this situation. That's when the magnetic field is perpendicular to the normal line, which means it's parallel to the face of the coil. Now, let's draw another picture when it's at an angle. So here is the normal line, which is perpendicular to the face of the coil.
And let's draw the magnetic field, which is directed east. So this time, the angle theta is less than 90. Here, theta is equal to 90. Now, F1 and F2, they're still directed north and south, but they're much less than divisional values, so the torque is less. And let's see what happens when the angle is 0. In this case, the normal line is parallel to the magnetic field, so therefore the angle between the two is 0. And when this happens, the magnetic field is perpendicular to the face of the coil.
It passes through the coil. And so there's going to be no torque whenever you have this situation. Now let's understand why there's no torque. If you look at the two forces, F1 and F2, Notice that they're parallel to the lever arm. Anytime you have a force that's parallel to the lever arm, it cannot create a torque.
The only way a torque can be created if it's perpendicular. Let's say this is the top view of a door. Here's the lever arm, R.
The only way for you to push a door is to apply a force perpendicular to the lever arm, and the door is going to turn. If you try to push a door from this side, it's not going to move. It won't turn.
And that's what's happening here. The axis of rotation is here. And the way the forces are oriented, since they're parallel to the lever arm, which is R, there's going to be no torque.
There's no rotation. And so, that's why whenever the angle is 0, sine 0 is 0, so the torque is going to be 0. So the maximum torque occurs whenever the angle is 90, that is the angle between the magnetic field and the normal line. So whenever the magnetic field is parallel to the face of the coil, you're going to have maximum torque. Whenever it's parallel to the face of the coil, it's perpendicular to the normal line.
The angle is 90. And whenever the magnetic field passes through the coil, that is when it's parallel to the normal line, there's going to be no torque. The angle is 0. Now notice what happens. Let's just erase a few things. I think I'll keep that.
If we look at the first picture on the left. F1 will create a torque that will cause it to rotate in the clockwise direction. And F2 will also create a torque that will cause the system to rotate in the clockwise direction. Now, once it rotates, it's going to move towards picture 2. It's going to look like this.
And as you can see, F1 will still create a torque that will cause the system to rotate clockwise. And the same is true for F2. Eventually, this... system is going to reach this point and F1 and F2 will still be directed north and south, which we see that in the third diagram.
So once we reach the third diagram, where the angle is zero, Then, the system is at equilibrium. The two forces, they're opposite and they're equal, so they cancel out. There's no net force. In fact, there's no net force for all pictures. In the first one, these two forces, they...
balance each other out. However, there is a net torque for the first and second picture, but in the third diagram, the net torque is zero. This force creates no torque, and the same is true for the second one. So the net torque and the net force is zero for the third diagram, which means it's at equilibrium.
So basically, The loop moves from this position, from an angle of 90, and the magnetic field causes it to rotate to an angle of 0, and then it stops. Now, let's work on some problems. Number 9, a circular coil of wire has a radius of 0. of 30 centimeters and contains 50 loops the current is 8 amps and the coil is placed in a magnetic field of 5 Tesla what is the maximum torque exerted on a coil by the magnetic field so if we draw a picture we're gonna have a coil of wire that is circular and has many loops 50 loops And so it's going to look something like that.
And the radius is 30 centimeters. To calculate the torque, it's equal to n i a b sine theta. Now if we wish to find the maximum torque, then the angle is going to be 90, and sine 90 is equal to 1. So n is the number of loops, n is 50. The current is 8 amps.
To find the area, it's going to be the area of a circle, which is pi r squared. So it's pi times the radius squared. The radius is 30 centimeters, but we need to convert that to meters.
30 centimeters is 0.3 meters. So it's pi times 0.3 squared. So that's the area. The magnetic field is 5 tesla, and sine 90 is 1. So let's start with the area.
0.3 squared times pi, which is 9 pi over 100, that's 0.2827. If we multiply that by 5 times 8 times 50, you should get 565.5 Tesla. So don't forget to square 0.3.
And so that's it for this problem. Actually the unit is not Tesla That's for the magnetic field. The unit for torque is Newtons times meters. So this is it. Number 10. A rectangular coil contains 200 loops and has a current of 15 amps.
What is the magnitude of the magnetic field required to produce a maximum torque of 1200 Newtons times meters? Let's draw a picture. And so here's a rectangular loop and there's 200 loops, but I want to draw it 200 times The area is going to be the width multiplied by the left And so we have the dimensions So the torque is going to be equal to the number of loops times the current times the area times the magnetic field. And since we're looking for, or since we're using the maximum torque, the maximum torque occurs at an angle of 90 degrees.
So this is going to be sine 90. Now this problem, the area of a rectangular coil, is going to be the length times the width. And sine 90 is 1. So the maximum torque is, let's put this on the left, it's 1200 newtons times meters. N is equal to 200 loops.
The current is 15 amps. And then the area, the length times the width, we need to convert 40 centimeters to meters. So that's 0.4 meters times 0.5 meters.
And the magnetic field, we're looking for B, so we've got to solve for it. So let's multiply 200 times 15 times 0.4 times 0.5. So that's equal to 600. And so now all we need to do is divide both sides by 600 So these two cancel 1200 over 600 we can cancel the two zeros, so it's 12 divided by 6 and that's 2 So the magnetic field is 2 Tesla And so that's the answer So that's it for this video. If you liked this video, feel free to subscribe. You can check out my channel for more physics videos.
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