Understanding Magnetic Forces and Fields

Okay, so today we're going over magnetic force on a current carrier and conductor. Now, magnetic force says between two parallel wires are attractive if the current is moving in the same directions. and between anti-parallel currents if they're moving in opposite directions the forces between them are repulsive Now the right hand rule for current carrying wire if you place your thumb your right hand in the direction of the current, your thumb of your right hand in the direction of the current and your fingers curl around the circle in the direction of the magnetic field lines and as you can see here, the magnetic field strength or vector at any given time is always tangent to the circle So imagine that there are concentric circles around the wire you place your thumb in the direction of the current and your fingers are going to curl in the direction of the magnetic field lines magnetic field. Again, what we see here, if you place your thumb in the direction of the current, your fingers of your right hand are going to be in the direction of the magnetic field, any distance away from the wire. Now, now we have the direction of the B field between the two wires points out for current in opposite directions. Now, if we assume that the current that is moving in this wire is going this way, right? I is going this way and then the current that's going in this wire is going this way. Now if you place your thumb in the direction of the current for this wire, for this wire, you can see that your fingers are going to curl and then it's going to curl all the way way in the middle between the two wires your fingers are going up so the magnetic field is going up. And then if we do the same thing for this wire, and then you place your thumb in the direction of the current and then you can see that your fingers are going up again in the direction between the wires. So here the resultant magnetic field is going in the up direction. Now to determine the magnetic field strength, we are going to use the magnetic for a straight long wire carrying a current. So, B is equal to some constant mu zero times the current over 2 pi r. And then the reason for that is because if we have a wire, we know that the magnetic field lines are moving in a circle or concentric circles around the wire. So, the radius is the radius at any distance away from the wire. So, r is the radius distance in meters, i is the current and amps in the wire, and mu zero times the current. it is a constant. This is equal to 4 pi times 10 to the negative 7 and this is tesla times meters per amps. And then this is called a magnetic constant. Now, if we replace this constant within this equation, what we get is that the pi cases out with the pi. 4 divided by 2 will get rid of this 2, and then I have 2 over here. So what we are left with is that the magnetic field for a long straight wire is simply equal to 2 times 10 to the negative 7 times I over the distance away from the wire to where you want to find the magnetic field strength. Now, if we have a ring, okay, now for this type of a ring, so you want to place your fingers in the direction of the current. So, in this case, if you place your fingers in the direction of the current, your thumb will point in the direction of the magnetic field. case here your fingers in direction of the current and your thumb is actually in a direction of magnetic field so this is the magnetic field direction for a ring okay okay first by the right hand rule fingers in direction of current and thumb points in the direction of the magnetic field. Now for a ring, we only care about the magnetic field at the center of the ring because this is the only place it is uniform. Okay, so we only care about the center of the ring. So the B field at the center of the ring is B center is equals to mu zero times I over 2R and R would be the radius of the ring. Okay, and then if we have N loops, if we have more than one loops, N number of loops, then it simply becomes because the b at the center is equal to n times u0 times i over 2r, where n is the number of turns, and r is the radius of the loop. Now, for this one we have a magnetic field due to a solenoid. Now, once we set up the current in a solenoid, and the current in this case, you can see that the current is going this way. It's going down this way around the ring of the solenoid. Now, again, if you place your right hand, your fingers of your right hand in the direction of the current, so in this case, fingers... and direction of current, okay? For this, so fingers in the direction of the current, and then your thumb will be pointing in the direction of the magnetic field. So, in this case, you can see that the magnetic field, B, will be in this direction, along the positive X direction. Okay. Your fingers place in the direction of the current. Fingers in the direction of the current. And your thumb will be in the direction of the magnetic field. So, to determine the magnetic field, so you can see that the magnetic field is only inside of the solenoid. And then it's moving, it's uniform, inside the uniform. And anywhere else. outside, B is equal to zero. So we only care about the magnetic field that is set up inside of this solenoid or corel. So B inside is equal to N times mu zero times I. And then we know mu zero is a constant. So mu zero is equal to four times pi times ten to negative seven. So this is a constant. And then anywhere outside the solenoid or corel is equal to zero. Where small n, this is equal to big N divided by L. Where N is equal to pi times n, small n, is the number of turns per unit length, where n is the number of turns in the coil. So here, what it means that, how many turns do we have in this length of wire? So small n is the number of turns per unit length, so this comes out to be b is equals to n divided by l, okay, where l is the length of the cylinder, or the wire, times mu zero times i. So if we have 50 turns, so we would actually have 50 over here, and the length, if the length is 100 meters, so it will be 50 divided by 100. Now, again, I just went over that. By the right-hand rule for a solenoid, fingers of the right-hand curl around the solenoid in the direction of the current, and your thumb points in the direction of the magnetic field B. So remember that the magnetic field only exists inside the coil. so we have the magnetic field is moving in the positive x direction so this is my magnetic field and then this is the only place it is uniform it is uniform inside of the coil or the solenoid Okay, now to determine the magnetic force on a current carrying a wire, so let's say for example, if we take a wire that's carrying a current and then we place that in a uniform magnetic field and then we can see that the magnetic field is going into the page. It's going in. Alright, because remember that if you have a magnetic field X, this is going in into the page. And then if the magnetic field is a bunch of dots, we know that it is coming out of the page. And then one more thing. So the force acting on this wire, F is equals to ILB. Force that's acting on the wire itself, F is equals to ILB. Where L, this is called the length of the wire. I is the current that is set up in the wire. And then B, it is the magnetic field that the wires end. Okay? Now, so this is how we can determine the magnetic force that is acting on a wire. So this is only for a wire, magnetic force on a wire. Okay? That is carrying a current. Now, the force is always perpendicular to the conductor. in the field and then we'll see why. Now, for example, we saw that last time if we have a charged particle, if we have a charge, let's say for example, let's say a positive charge, right, with some velocity v moving in a uniform magnetic field, that the force that's acting on the charge F is equals to Q times VB right and then this is the magnitude of the charge times the magnitude of the velocity times B so this is the maximum force you can have that's acting on a charge moving in a informagnetic field. Now if the velocity is perpendicular, if the velocity is perpendicular to B, velocity is perpendicular to B, is that the angle theta between V and B is 90 degrees. So and then sine of 90 is equal to 1. So this is why we have F is equal to QVB and in reality this is the magnitude of the charge times the velocity times the magnetic field. This is for a particle moving in a uniform magnetic field. And then if the particle is at some angle to the magnetic field, so we know that F is equal to Q times sine of the angle between V and B. So this is if the particle is at some angle to the magnetic field. So if the velocity V is at some angle to B, then we can say that the force acting on the particle is equal to the charge times the velocity times the magnetic field and sign up the angle between V and B so this is the angle Phi between V and B now as a vector if the velocity is given as a vector and the magnetic field is given as a vector we can use the cross product where F is equals to Q times V cross B right and then we have to find V cross B in order to find once we find V cross B and then we can multiply it by Q we can find the force. Now, pretty much the same thing here. Now, if the magnetic field makes an angle phi with the wire, right, like we have here. So, we have the magnetic field B is going this way and the wire is going this way. And then here, this is the angle between the wire and the magnetic field. So, again, we can determine the force acting on the wire, F is equals to ILB times sine of phi, where in this case, phi is the angle between B and the current or the wire, right? Now, Now, we can also determine the force as a vector. For example, since we know that the length of the wire is a vector, and then the magnetic field is a vector, so we can find the cross product, L cross B. So if L is given as a vector and B is given as a vector, so we have to use the cross product. So in this case here, F is equal to I times L cross B. And in this case, where the vector L is always in the direction of the current, is always in the direction of the current. direction of the current I. So we can see that the vector L is the direction of the current. So now, the force, to determine the force we can use the cross product. So now if you imagine that you're holding a, sheet in the palm of your left hand and then if you place your right hand the fingers of your right hand in direction of the of the current or the vector L and then you try to sweep your fingers with a small angle towards B you can see that the force is is pushing up. So, this is using what we call the right hand rule for a current carrying a wire. So, is that the right hand of your, I mean the fingers of your right hand in the direction of the vector L and then the fingers are going to curl towards the magnetic field B through the smallest angle. And as you can see that the force is always perpendicular to both the wire and the magnetic field. Now, here's an example. So we have a current occurring a wire. So the current is going this way. That's I. And then remember that the vector L is always in the same direction as the current. And then we have B that's going up. Now, in this case, here, if you place your palm facing up of your right hand and then your fingers in the direction of the current and then you want to curl your fingers towards the magnetic field that is going up in the Y direction and as you can see that the magnetic force is coming out of the page in the Z direction. So, you can see that F is coming out of the page. And then this is in the positive K direction or in the Z direction. So this is the Z direction. It's coming out of the Z direction. And one more thing. So again, imagine that again you have a sheet placed in the palm of your left hand. And then... you have the wire coming a current that is I and then this is L is always in the same direction as the current and then we're using F is equals to I times L cross B right and then and then we have B go going in the negative x direction so B is going this way. Now we want to take your right hand and then you place your right hand, the fingers of your right hand in the direction of the current and then you want to sweep them towards the smallest angle towards B and as you can see your thumb of your right hand is pointing down and then this is the direction of the force F. So here the direction of the force F is in a negative j direction, negative y. Okay, now for this example, suppose that the short side of the wire measures 10 centimeters and the long side measures 20 centimeters. So what it means is that this side is 20 centimeters and this side is 10 centimeters, right? And the current in the wire is I, it's given, so I is 0.225 amps. And then my Magnetic force is given so F is given as 0.035 Newton and the They tell us what is the strength of the magnetic field here the magnetic field B is going in so it's going in to the page right now the thing is to determine the force acting on the wire so we need to find the force that's acting on every segment of the wire so now this is the portion of the wire that is in the uniform magnetic field So, here if we place, if we're looking at this part of the wire, if we're looking at this one, so you place the right hand, your fingers of your right hand in the direction of I. So, fingers in the direction of the current and then you're going to curl your fingers towards B. Now, you can see that the magnetic force is pushing to the left. So, in this case here, the magnetic force that's acting on this wire is pushing to the left. Okay. And then if we look at this segment, so the current is going down, so you want to place the right hand. of the fingers of your right hand at the direction of the current that's going down. And then you want to curl them towards B that's going in. And then you can see that force is pushing to the right. So here the magnetic force is pushing to the right. And then if we're looking at this segment, we can see that the current is going in this direction. The current is going in this direction. The current is going down, flowing this way. So here, for this piece of the wire for that segment, if you place the fingers of your right hand, hand in the direction of the current and then we call them towards B that's going and you can see that the force that's acting on this segment is going down so this is the force here so we can see that these two forces this one and that one are equal and opposite so they cancel each other out so the net force on of these two are cancel each other out it's added to zero force that we have acting on this segment under so the only force that we have acting on the wire it's only on this segment so what we have is that F is equals to ILB and then since we can see that the the wire is perpendicular to the magnetic field so this is basically the sine of 90 right so we know that sine of 90 is equals to 1 so if we're looking for B B is equals to what F divided by I times L and then what we get this comes out to be 0.035 Newton divided by 0.25 amps right and l for this segment is 10 centimeters convert that to meters comes out to be 0.1 meters and then what we get is that the magnetic field that is going into the page is equals to 0.15 tesla for this example what we have is a rectangular loop that is set up by 0.6 and by 0.5 meters so 0.6 meter is the long side and 0.5 is the shorter side and and then we see that the magnetic field B is going into the page. And then what's given is that the current that is set up, I is equal to three amps, and then the magnetic field that is set up at the wire's end is 2.4 Tesla. And then the mass of the wire, the mass of the loop, I should say, the mass of the loop is equal to 0.5 kilogram. So now we want to find the magnitude and direction of the magnetic force acting on the loop. Again, if we're looking at the direction of the current, current so we know that on this side the current is going this way the current is coming down this way traveling this way into this loop and up on this side now this is the portion of the of the loop that is in the magnetic field so this is what we care about so now if we're looking at this section of the loop we can see that the current is going down so if we place our fingers of the right hand in direction of the current and then we curl them towards b we can see that the force that's acting on this segment is pushing this way right FB. And then if we look on this segment, on this segment, again, you place your fingers of your right hand in direction of the current, and then you curl them so it's B. You can see that the force is going to the left. So these two forces are equal and opposite because it's the same segment, same magnetic field, same current, so they can't see each other out. So the only force that's acting on the loop, so if we're looking at this segment of the loop for this segment of the loop, so again, you place your right hand in direction of the current. so you curl them towards B, B is going in you can see that the magnetic force is pushing up. So for this segment the magnetic force is pushing up. So what we have is that the magnetic force FB is equal to ILB times sine of phi. And then since the wire is perpendicular to B So we know that this is 1 because this is 90 degrees. So what we have the force magnetic force F is equal to 3 amps right that's the current times L which is 0.5 and then times B which is 2.4. So what we get is that the force that's acting on the wire, on the loop I should say, is 3.6 Newton and then it is going, it's pushing up on the wire, on the loop. Now, and then pushing up, or it is the same thing, or we could say this is in the plus J direction. Okay. Now, what is the tension in the string supporting the loop? Now, to find the tension, so we have to draw a free body diagram. So, if we're looking at the loop, we know that there's gravity and G is pushing down. And then we know that there's a tension in the loop, there's a tension T going up, right? And then we also know that that there is a magnetic force due to the current that is passing through the loop is pushing on the loop this way so fb is pushing on a loop this way again if we take up as positive and down as negative and then since the loop is at equilibrium is that the tension in the loop plus the magnetic force that it's n minus mg has to equals to a net force of zero okay so again if we solve for t so what we have t would equals to bring mg on a are they assigned it becomes plus this becomes mg minus the force the magnetic force so by solving for T so mg will be the mass of the wire so that would be 0.05 kilogram times 9.81 okay and then minus the magnetic force which is which we just calculated that that's 3.6 Newton and then this is give us a tension of 1.3 Newton so this is the tension in the loop that is set up by the wire. So again, we have to find the direction of the magnetic force for this segment and we can see that the magnetic force is pushing up on the segment of the loop according to the right hand rule for current carrying wire. Alright, for this example, we see two parallel wires. We have wire 1 and wire 2 and point A are all on the same plane. Wire 1 is parallel to wire 2 and separated by a distance of 67. centimeters and point A is a distance of 35 centimeter away from wire 2. So as we can see in the diagram so we have this current I current I of 3 amps is going this way and I of five amps is going that way so uh question number question a is asking find a magnetic force pushing on a 50 centimeter segment of wire two okay so um now we know that to find the force acting on wire two so f has the equals to i2 the current of wire two times the length times b1 right Because what happens is that the magnetic field that is produced by the current on wire 1 is acting on wire 2. This is why it's called B1. And then it's acting on wire 2, and then it's acting on the current, which is, if we call this one I2, and we call this one I1. And then it's the same force that wire 2 would actually exert on wire 1, but these two forces are equal and opposite. So, first, before... we can find the force that's acting on wire 2 we need to find a magnetic field that is produced by current I1 so in this case this is B1 is equals to mu 0 times I1 over 2 pi times R right which is the distance from wire 1 to the position of wire 2 so and then we know mu 0 mu 0 is a constant this is equals to 4 pi times 10 to negative 7 right So we know this is equal to 2 times 10 to negative 7, right, times I1 divided by R. right so so now if we solve for that so b1 is equals to 2 times 10 to negative 7 times i1 which is 3 amps so that would be 3 amps divided by the distance from wire 1 to wire 2 so that would be 0.6 meters and then what we get is that this is comes out to be 1 times 10 to negative 6 Tesla that's my That's my magnetic field. Now, to find the direction of the magnetic field that is produced by current I1 at the position of wire 2, again we used to use the magnetic field. we need to use the right hand rule and that is that if we place our thumb in direction of the current and then we curl it around until our fingers is pointing at the position of wire two we can see that the magnetic field is coming out so b here so b1 is coming out of the page This is B1. It's coming out of the page. It's coming in a Z direction. So this is in the positive K direction. Now to find the force on a 50 segment of 1. wire 2 so if we have two long strip parallel wires so now f is equals to 5 amps which is i2 times 50 segments so that'll be 0.5 right i'm 0.5 and then times b which is 1 times 10 to negative 6 tesla okay so now what we get is that the force that we get is 2.5 times 10 to the negative 6 newton so this is the force at the position of wire two and then if we want to find the the direction of that force again we can use right hand rule so to find the to find the direction again we know that the length of this wire L is in the same direction as the current so this is the direction of L this is the vector L now if we want to do L cross B so if we do L cross B. So fingers in the direction of the length or the current. And then we want to curl them towards B which is coming out. We can see that the thumb is pointing down. So the direction of the current, I mean of the force is going down or it is in a negative J direction. So this is the direction of the force, a negative J. Now for part B, we want to find the net magnetic field magnitude and direction at point A. Now, so we know that that wire 1 will produce a magnetic field at point A. And we also know that wire 2 will produce a magnetic field at point A. So now again, if we place our thumb in the direction of the current I1 and then we curl them until the fingers get to the position A, we can see that B1 is coming out. So what we have is that B1 A is equal to 2 times 10 to negative 7, 2 times 10 to negative 7 times I1 which is 3 amps times 3 amps. over the distance between wire 1 and position A. So this is 60 centimeters plus 35 to give me 0.95. So that comes out to be 0.95 meters. And then what we get is that this comes out to be 6.32. 6.32 times 10 to the negative 7 Tesla and then this is coming out in a positive K direction, right? Now we want to do the same thing for wire 2 now if we place our thumb in direction of the current I2. So, and then if we curl our fingers all the way to the position of point A, we can see that our finger is actually going to be pointing in at the position of point A. So, for B2A, right, for B2A, we're going to be pointing in at the position of a right let's come over here so for b2a right so again this is something this is 2 times 10 to negative 7 right uh times i which is 5 amps and then over the distance between wire 2 and point a which is 0.35 and then what we get is that this comes out to be 2.86 times 10 to the negative seven and then this is tesla i mean to the negative six this comes to the negative six to the negative six tesla right now this is coming out so b2a over here is going in okay so these two magnetic field that lie exactly you know along the same axis So one is going in and the other one is coming out. So they're both in the Z direction. So to find the magnitude is looking at the difference, right? So if we're adding these two, so B at point A should equal to B1A plus B2A. Okay? Now if we... Come over here on this page. So BA is equals to 6.32 times 10 to negative 7. And then this is plus or minus 2.86 times 10 to negative 6. And then when we add this to negative. 2.22 times 10 to a negative 6 Tesla and that is going into the page Or something like saying this is in the minus k direction. So if we put minus k, we don't need to put the negative sign here. We can actually put negative k over here. So this would be in a negative k direction. Now, for the last question, now if we place a proton charge at point A, a proton is released from rest at point A, so if we have a proton charge here, a positive. charge and moving to the right, so it's moving to the right this way, it's moving to the right, right? With velocity of this magnitude 2 times 10 to the 6 meters per second, find the force magnitude and reaction acting on the proton. So what it means that we know that the force acting on the proton, F is equals to what? F is equals to QVB and this is the magnitude, right? So we know that this equals to 1.6 times 10 to the negative 19 coulombs times the velocity of the proton, that would be 2 times 10 to the 6 meters per second times the magnitude of the magnetic field at point A, which is 2.22 times 10 to the negative 19 coulombs times 10 to a negative 6 tesla and then what we get is that the force that's acting on the proton is equals to 7.12 times 10 to negative 19 newton okay now if we want to find the direction of that force right if we want to find the direction of that force again we have to use the right hand rule uh if we go back so we know that the velocity of the proton is moving to the right, V is moving this way, and then the resultant magnetic field is going in, right? So, if we do V cross B, right? V cross B, we can see that the force is pushing up. So the force here is pushing up in a positive direction, which is plus J direction is pushing up. So what we have is that we have the proton moving this way. The magnetic field is going into the page. And then you take the fingers of your right hand, you place them in the direction of the velocity, and then you curl them towards B, and your thumb puts in the direction of the force, which is in the way up, so which is plus J direction.

Now, now we have the direction of the B field between the two wires points out for current in opposite directions. Now, if we assume that the current that is moving in this wire is going this way, right? I is going this way and then the current that's going in this wire is going this way.

Now if you place your thumb in the direction of the current for this wire, for this wire, you can see that your fingers are going to curl and then it's going to curl all the way way in the middle between the two wires your fingers are going up so the magnetic field is going up. And then if we do the same thing for this wire, and then you place your thumb in the direction of the current and then you can see that your fingers are going up again in the direction between the wires. So here the resultant magnetic field is going in the up direction. Now to determine the magnetic field strength, we are going to use the magnetic for a straight long wire carrying a current.

So, B is equal to some constant mu zero times the current over 2 pi r. And then the reason for that is because if we have a wire, we know that the magnetic field lines are moving in a circle or concentric circles around the wire. So, the radius is the radius at any distance away from the wire.

So, r is the radius distance in meters, i is the current and amps in the wire, and mu zero times the current. it is a constant. This is equal to 4 pi times 10 to the negative 7 and this is tesla times meters per amps. And then this is called a magnetic constant.

Now, if we replace this constant within this equation, what we get is that the pi cases out with the pi. 4 divided by 2 will get rid of this 2, and then I have 2 over here. So what we are left with is that the magnetic field for a long straight wire is simply equal to 2 times 10 to the negative 7 times I over the distance away from the wire to where you want to find the magnetic field strength.

Now, if we have a ring, okay, now for this type of a ring, so you want to place your fingers in the direction of the current. So, in this case, if you place your fingers in the direction of the current, your thumb will point in the direction of the magnetic field. case here your fingers in direction of the current and your thumb is actually in a direction of magnetic field so this is the magnetic field direction for a ring okay okay first by the right hand rule fingers in direction of current and thumb points in the direction of the magnetic field. Now for a ring, we only care about the magnetic field at the center of the ring because this is the only place it is uniform. Okay, so we only care about the center of the ring.

So the B field at the center of the ring is B center is equals to mu zero times I over 2R and R would be the radius of the ring. Okay, and then if we have N loops, if we have more than one loops, N number of loops, then it simply becomes because the b at the center is equal to n times u0 times i over 2r, where n is the number of turns, and r is the radius of the loop. Now, for this one we have a magnetic field due to a solenoid. Now, once we set up the current in a solenoid, and the current in this case, you can see that the current is going this way.

It's going down this way around the ring of the solenoid. Now, again, if you place your right hand, your fingers of your right hand in the direction of the current, so in this case, fingers... and direction of current, okay?

For this, so fingers in the direction of the current, and then your thumb will be pointing in the direction of the magnetic field. So, in this case, you can see that the magnetic field, B, will be in this direction, along the positive X direction. Okay.

Your fingers place in the direction of the current. Fingers in the direction of the current. And your thumb will be in the direction of the magnetic field. So, to determine the magnetic field, so you can see that the magnetic field is only inside of the solenoid.

And then it's moving, it's uniform, inside the uniform. And anywhere else. outside, B is equal to zero.

So we only care about the magnetic field that is set up inside of this solenoid or corel. So B inside is equal to N times mu zero times I. And then we know mu zero is a constant.

So mu zero is equal to four times pi times ten to negative seven. So this is a constant. And then anywhere outside the solenoid or corel is equal to zero.

Where small n, this is equal to big N divided by L. Where N is equal to pi times n, small n, is the number of turns per unit length, where n is the number of turns in the coil. So here, what it means that, how many turns do we have in this length of wire? So small n is the number of turns per unit length, so this comes out to be b is equals to n divided by l, okay, where l is the length of the cylinder, or the wire, times mu zero times i. So if we have 50 turns, so we would actually have 50 over here, and the length, if the length is 100 meters, so it will be 50 divided by 100. Now, again, I just went over that.

By the right-hand rule for a solenoid, fingers of the right-hand curl around the solenoid in the direction of the current, and your thumb points in the direction of the magnetic field B. So remember that the magnetic field only exists inside the coil. so we have the magnetic field is moving in the positive x direction so this is my magnetic field and then this is the only place it is uniform it is uniform inside of the coil or the solenoid Okay, now to determine the magnetic force on a current carrying a wire, so let's say for example, if we take a wire that's carrying a current and then we place that in a uniform magnetic field and then we can see that the magnetic field is going into the page.

It's going in. Alright, because remember that if you have a magnetic field X, this is going in into the page. And then if the magnetic field is a bunch of dots, we know that it is coming out of the page.

And then one more thing. So the force acting on this wire, F is equals to ILB. Force that's acting on the wire itself, F is equals to ILB. Where L, this is called the length of the wire.

I is the current that is set up in the wire. And then B, it is the magnetic field that the wires end. Okay? Now, so this is how we can determine the magnetic force that is acting on a wire.

So this is only for a wire, magnetic force on a wire. Okay? That is carrying a current. Now, the force is always perpendicular to the conductor.

in the field and then we'll see why. Now, for example, we saw that last time if we have a charged particle, if we have a charge, let's say for example, let's say a positive charge, right, with some velocity v moving in a uniform magnetic field, that the force that's acting on the charge F is equals to Q times VB right and then this is the magnitude of the charge times the magnitude of the velocity times B so this is the maximum force you can have that's acting on a charge moving in a informagnetic field. Now if the velocity is perpendicular, if the velocity is perpendicular to B, velocity is perpendicular to B, is that the angle theta between V and B is 90 degrees. So and then sine of 90 is equal to 1. So this is why we have F is equal to QVB and in reality this is the magnitude of the charge times the velocity times the magnetic field. This is for a particle moving in a uniform magnetic field.

And then if the particle is at some angle to the magnetic field, so we know that F is equal to Q times sine of the angle between V and B. So this is if the particle is at some angle to the magnetic field. So if the velocity V is at some angle to B, then we can say that the force acting on the particle is equal to the charge times the velocity times the magnetic field and sign up the angle between V and B so this is the angle Phi between V and B now as a vector if the velocity is given as a vector and the magnetic field is given as a vector we can use the cross product where F is equals to Q times V cross B right and then we have to find V cross B in order to find once we find V cross B and then we can multiply it by Q we can find the force.

Now, pretty much the same thing here. Now, if the magnetic field makes an angle phi with the wire, right, like we have here. So, we have the magnetic field B is going this way and the wire is going this way. And then here, this is the angle between the wire and the magnetic field. So, again, we can determine the force acting on the wire, F is equals to ILB times sine of phi, where in this case, phi is the angle between B and the current or the wire, right?

Now, Now, we can also determine the force as a vector. For example, since we know that the length of the wire is a vector, and then the magnetic field is a vector, so we can find the cross product, L cross B. So if L is given as a vector and B is given as a vector, so we have to use the cross product.

So in this case here, F is equal to I times L cross B. And in this case, where the vector L is always in the direction of the current, is always in the direction of the current. direction of the current I. So we can see that the vector L is the direction of the current.

So now, the force, to determine the force we can use the cross product. So now if you imagine that you're holding a, sheet in the palm of your left hand and then if you place your right hand the fingers of your right hand in direction of the of the current or the vector L and then you try to sweep your fingers with a small angle towards B you can see that the force is is pushing up. So, this is using what we call the right hand rule for a current carrying a wire. So, is that the right hand of your, I mean the fingers of your right hand in the direction of the vector L and then the fingers are going to curl towards the magnetic field B through the smallest angle. And as you can see that the force is always perpendicular to both the wire and the magnetic field.

Now, here's an example. So we have a current occurring a wire. So the current is going this way. That's I. And then remember that the vector L is always in the same direction as the current. And then we have B that's going up.

Now, in this case, here, if you place your palm facing up of your right hand and then your fingers in the direction of the current and then you want to curl your fingers towards the magnetic field that is going up in the Y direction and as you can see that the magnetic force is coming out of the page in the Z direction. So, you can see that F is coming out of the page. And then this is in the positive K direction or in the Z direction. So this is the Z direction.

It's coming out of the Z direction. And one more thing. So again, imagine that again you have a sheet placed in the palm of your left hand. And then...

you have the wire coming a current that is I and then this is L is always in the same direction as the current and then we're using F is equals to I times L cross B right and then and then we have B go going in the negative x direction so B is going this way. Now we want to take your right hand and then you place your right hand, the fingers of your right hand in the direction of the current and then you want to sweep them towards the smallest angle towards B and as you can see your thumb of your right hand is pointing down and then this is the direction of the force F. So here the direction of the force F is in a negative j direction, negative y.

Okay, now for this example, suppose that the short side of the wire measures 10 centimeters and the long side measures 20 centimeters. So what it means is that this side is 20 centimeters and this side is 10 centimeters, right? And the current in the wire is I, it's given, so I is 0.225 amps. And then my Magnetic force is given so F is given as 0.035 Newton and the They tell us what is the strength of the magnetic field here the magnetic field B is going in so it's going in to the page right now the thing is to determine the force acting on the wire so we need to find the force that's acting on every segment of the wire so now this is the portion of the wire that is in the uniform magnetic field So, here if we place, if we're looking at this part of the wire, if we're looking at this one, so you place the right hand, your fingers of your right hand in the direction of I.

So, fingers in the direction of the current and then you're going to curl your fingers towards B. Now, you can see that the magnetic force is pushing to the left. So, in this case here, the magnetic force that's acting on this wire is pushing to the left. Okay. And then if we look at this segment, so the current is going down, so you want to place the right hand.

of the fingers of your right hand at the direction of the current that's going down. And then you want to curl them towards B that's going in. And then you can see that force is pushing to the right.

So here the magnetic force is pushing to the right. And then if we're looking at this segment, we can see that the current is going in this direction. The current is going in this direction. The current is going down, flowing this way. So here, for this piece of the wire for that segment, if you place the fingers of your right hand, hand in the direction of the current and then we call them towards B that's going and you can see that the force that's acting on this segment is going down so this is the force here so we can see that these two forces this one and that one are equal and opposite so they cancel each other out so the net force on of these two are cancel each other out it's added to zero force that we have acting on this segment under so the only force that we have acting on the wire it's only on this segment so what we have is that F is equals to ILB and then since we can see that the the wire is perpendicular to the magnetic field so this is basically the sine of 90 right so we know that sine of 90 is equals to 1 so if we're looking for B B is equals to what F divided by I times L and then what we get this comes out to be 0.035 Newton divided by 0.25 amps right and l for this segment is 10 centimeters convert that to meters comes out to be 0.1 meters and then what we get is that the magnetic field that is going into the page is equals to 0.15 tesla for this example what we have is a rectangular loop that is set up by 0.6 and by 0.5 meters so 0.6 meter is the long side and 0.5 is the shorter side and and then we see that the magnetic field B is going into the page.

And then what's given is that the current that is set up, I is equal to three amps, and then the magnetic field that is set up at the wire's end is 2.4 Tesla. And then the mass of the wire, the mass of the loop, I should say, the mass of the loop is equal to 0.5 kilogram. So now we want to find the magnitude and direction of the magnetic force acting on the loop.

Again, if we're looking at the direction of the current, current so we know that on this side the current is going this way the current is coming down this way traveling this way into this loop and up on this side now this is the portion of the of the loop that is in the magnetic field so this is what we care about so now if we're looking at this section of the loop we can see that the current is going down so if we place our fingers of the right hand in direction of the current and then we curl them towards b we can see that the force that's acting on this segment is pushing this way right FB. And then if we look on this segment, on this segment, again, you place your fingers of your right hand in direction of the current, and then you curl them so it's B. You can see that the force is going to the left. So these two forces are equal and opposite because it's the same segment, same magnetic field, same current, so they can't see each other out. So the only force that's acting on the loop, so if we're looking at this segment of the loop for this segment of the loop, so again, you place your right hand in direction of the current.

so you curl them towards B, B is going in you can see that the magnetic force is pushing up. So for this segment the magnetic force is pushing up. So what we have is that the magnetic force FB is equal to ILB times sine of phi.

And then since the wire is perpendicular to B So we know that this is 1 because this is 90 degrees. So what we have the force magnetic force F is equal to 3 amps right that's the current times L which is 0.5 and then times B which is 2.4. So what we get is that the force that's acting on the wire, on the loop I should say, is 3.6 Newton and then it is going, it's pushing up on the wire, on the loop. Now, and then pushing up, or it is the same thing, or we could say this is in the plus J direction. Okay.

Now, what is the tension in the string supporting the loop? Now, to find the tension, so we have to draw a free body diagram. So, if we're looking at the loop, we know that there's gravity and G is pushing down. And then we know that there's a tension in the loop, there's a tension T going up, right? And then we also know that that there is a magnetic force due to the current that is passing through the loop is pushing on the loop this way so fb is pushing on a loop this way again if we take up as positive and down as negative and then since the loop is at equilibrium is that the tension in the loop plus the magnetic force that it's n minus mg has to equals to a net force of zero okay so again if we solve for t so what we have t would equals to bring mg on a are they assigned it becomes plus this becomes mg minus the force the magnetic force so by solving for T so mg will be the mass of the wire so that would be 0.05 kilogram times 9.81 okay and then minus the magnetic force which is which we just calculated that that's 3.6 Newton and then this is give us a tension of 1.3 Newton so this is the tension in the loop that is set up by the wire.

So again, we have to find the direction of the magnetic force for this segment and we can see that the magnetic force is pushing up on the segment of the loop according to the right hand rule for current carrying wire. Alright, for this example, we see two parallel wires. We have wire 1 and wire 2 and point A are all on the same plane.

Wire 1 is parallel to wire 2 and separated by a distance of 67. centimeters and point A is a distance of 35 centimeter away from wire 2. So as we can see in the diagram so we have this current I current I of 3 amps is going this way and I of five amps is going that way so uh question number question a is asking find a magnetic force pushing on a 50 centimeter segment of wire two okay so um now we know that to find the force acting on wire two so f has the equals to i2 the current of wire two times the length times b1 right Because what happens is that the magnetic field that is produced by the current on wire 1 is acting on wire 2. This is why it's called B1. And then it's acting on wire 2, and then it's acting on the current, which is, if we call this one I2, and we call this one I1. And then it's the same force that wire 2 would actually exert on wire 1, but these two forces are equal and opposite.

So, first, before... we can find the force that's acting on wire 2 we need to find a magnetic field that is produced by current I1 so in this case this is B1 is equals to mu 0 times I1 over 2 pi times R right which is the distance from wire 1 to the position of wire 2 so and then we know mu 0 mu 0 is a constant this is equals to 4 pi times 10 to negative 7 right So we know this is equal to 2 times 10 to negative 7, right, times I1 divided by R. right so so now if we solve for that so b1 is equals to 2 times 10 to negative 7 times i1 which is 3 amps so that would be 3 amps divided by the distance from wire 1 to wire 2 so that would be 0.6 meters and then what we get is that this is comes out to be 1 times 10 to negative 6 Tesla that's my That's my magnetic field.

Now, to find the direction of the magnetic field that is produced by current I1 at the position of wire 2, again we used to use the magnetic field. we need to use the right hand rule and that is that if we place our thumb in direction of the current and then we curl it around until our fingers is pointing at the position of wire two we can see that the magnetic field is coming out so b here so b1 is coming out of the page This is B1. It's coming out of the page.

It's coming in a Z direction. So this is in the positive K direction. Now to find the force on a 50 segment of 1. wire 2 so if we have two long strip parallel wires so now f is equals to 5 amps which is i2 times 50 segments so that'll be 0.5 right i'm 0.5 and then times b which is 1 times 10 to negative 6 tesla okay so now what we get is that the force that we get is 2.5 times 10 to the negative 6 newton so this is the force at the position of wire two and then if we want to find the the direction of that force again we can use right hand rule so to find the to find the direction again we know that the length of this wire L is in the same direction as the current so this is the direction of L this is the vector L now if we want to do L cross B so if we do L cross B.

So fingers in the direction of the length or the current. And then we want to curl them towards B which is coming out. We can see that the thumb is pointing down.

So the direction of the current, I mean of the force is going down or it is in a negative J direction. So this is the direction of the force, a negative J. Now for part B, we want to find the net magnetic field magnitude and direction at point A. Now, so we know that that wire 1 will produce a magnetic field at point A.

And we also know that wire 2 will produce a magnetic field at point A. So now again, if we place our thumb in the direction of the current I1 and then we curl them until the fingers get to the position A, we can see that B1 is coming out. So what we have is that B1 A is equal to 2 times 10 to negative 7, 2 times 10 to negative 7 times I1 which is 3 amps times 3 amps. over the distance between wire 1 and position A. So this is 60 centimeters plus 35 to give me 0.95.

So that comes out to be 0.95 meters. And then what we get is that this comes out to be 6.32. 6.32 times 10 to the negative 7 Tesla and then this is coming out in a positive K direction, right? Now we want to do the same thing for wire 2 now if we place our thumb in direction of the current I2.

So, and then if we curl our fingers all the way to the position of point A, we can see that our finger is actually going to be pointing in at the position of point A. So, for B2A, right, for B2A, we're going to be pointing in at the position of a right let's come over here so for b2a right so again this is something this is 2 times 10 to negative 7 right uh times i which is 5 amps and then over the distance between wire 2 and point a which is 0.35 and then what we get is that this comes out to be 2.86 times 10 to the negative seven and then this is tesla i mean to the negative six this comes to the negative six to the negative six tesla right now this is coming out so b2a over here is going in okay so these two magnetic field that lie exactly you know along the same axis So one is going in and the other one is coming out. So they're both in the Z direction. So to find the magnitude is looking at the difference, right?

So if we're adding these two, so B at point A should equal to B1A plus B2A. Okay? Now if we... Come over here on this page.

So BA is equals to 6.32 times 10 to negative 7. And then this is plus or minus 2.86 times 10 to negative 6. And then when we add this to negative. 2.22 times 10 to a negative 6 Tesla and that is going into the page Or something like saying this is in the minus k direction. So if we put minus k, we don't need to put the negative sign here.

We can actually put negative k over here. So this would be in a negative k direction. Now, for the last question, now if we place a proton charge at point A, a proton is released from rest at point A, so if we have a proton charge here, a positive.

charge and moving to the right, so it's moving to the right this way, it's moving to the right, right? With velocity of this magnitude 2 times 10 to the 6 meters per second, find the force magnitude and reaction acting on the proton. So what it means that we know that the force acting on the proton, F is equals to what?

F is equals to QVB and this is the magnitude, right? So we know that this equals to 1.6 times 10 to the negative 19 coulombs times the velocity of the proton, that would be 2 times 10 to the 6 meters per second times the magnitude of the magnetic field at point A, which is 2.22 times 10 to the negative 19 coulombs times 10 to a negative 6 tesla and then what we get is that the force that's acting on the proton is equals to 7.12 times 10 to negative 19 newton okay now if we want to find the direction of that force right if we want to find the direction of that force again we have to use the right hand rule uh if we go back so we know that the velocity of the proton is moving to the right, V is moving this way, and then the resultant magnetic field is going in, right? So, if we do V cross B, right? V cross B, we can see that the force is pushing up. So the force here is pushing up in a positive direction, which is plus J direction is pushing up.

So what we have is that we have the proton moving this way. The magnetic field is going into the page. And then you take the fingers of your right hand, you place them in the direction of the velocity, and then you curl them towards B, and your thumb puts in the direction of the force, which is in the way up, so which is plus J direction.

Transcript for:Understanding Magnetic Forces and Fields

Transcript for:
Understanding Magnetic Forces and Fields