8.3 Current carrying wire in an external magnetic field
Alright, now let’s consider the magnetic force generated by an external magnetic field on a current carrying wire through an example. In this case, let’s assume that we have a wire which has a straight part and a semi-circular part, and another straight part, something like this. Let’s assume that the radius of the semi-circular part is big R and also the length of the straight part is R and the length of this straight part is 2R.
It is carrying a current, let’s say from left to right, i, and we place this current carrying wire inside of a magnetic field region, which is pointing out of plane. So, B is pointing out of plane, and let’s assume that it is uniform and we’d like to determine the net force that this external magnetic field generates on this current carrying wire.
Now if you recall, the magnetic force generated by an external magnetic field on an incremental segment of a current carrying wire is equal to, dF is equal to i dl cross B. So, this is the magnetic field exerted along an incremental segment of a current carrying wire. Now, we’re going to, therefore, calculate the magnetic force generated along the incremental segments of this whole wire and once we determine these incremental forces, then we will add them vectorially to be able to get the total force. Our purpose is to calculate the net force at this point, at the center of the semi-circular region.
So, the question is B at point O, is the question mark. Alright, well, if you look at this straight segment first and we choose an incremental current element with a length of dl, so let’s say this is our incremental element, dl, and the current is flowing to the right, and the magnetic field is out of plane. Therefor the direction of the force that generates on this current carrying incremental segment, from i dl cross B, applying right-hand rule, keeping the right-hand fingers in the direction of first vector, which is to the right, and then curling them towards the second vector and that is B pointing out of plane, i dl cross B, while keeping the thumb in open position, is going to generate a force, let’s call this one as F1, in downward direction. And this is for the segment 1.
In a similar way, if we look at the other straight segment, here also due to the same reason i dl is pointing to the right, B is out of plane and applying right-hand rule, i dl cross B is going to generate the net force, let’s call this segment as segment 3, F3 in downward direction. The net forces are going to be in downward direction because if we go ahead and calculate the force on the next i dl, then the next i dl, they all will be pointing in downward direction. Therefore, the total force will be sum of all those forces, which will generate total force of F1, pointing in downward direction. And same thing is going to be true, of course, for the force F3.
Now, let’s look at this semi-circular region. In the semi-circular region, let’s choose our incremental current element, i dl, something like this. And here i dl is pointing in a tangent, or very small segment along the semi-circular part in an arbitrary location. B is out of plane and therefore the force on the segment, dF, incremental force, will be perpendicular to this, and as well as perpendicular to the external magnetic field, and therefore it is going to be pointing in radial direction, towards the center of the semi-circular region. Let’s call this force as dF.
When we look at this region of the wire, it is symmetric relative to this vertical axis. Means that we can always find a symmetrical i dl, just right across from this unit somewhere around here. And therefore if you look at the magnetic force on that segment by applying right-hand rule, that too is going to be pointing radially inward direction, towards the center of the semi-circular region. So, since both these i dl‘s will have the same amount of current, same length, therefore, and they’re the same distance away from this central point, therefore the magnitude of these forces will be the same.
And for every i dl, we will have a symmetrical one, relative to this axis on the other side of the semi-circular region. So, the total force will be the vector sum of all these forces. If we just consider this pair, and if we resolve them into two components, namely a vertical and a a horizontal component — this is the horizontal component and this one will be the vertical component. Similarly, for this regional, or for this dF, so we will have a horizontal component like this and the vertical component, pointing in downward direction.
And here, we can easily see that the horizontal components will have the same magnitude, but they’re going to be pointing in opposite directions. Therefore, when we add them vectorially, we will realize that the horizontal components will cancel. The vertical components are going to be pointing downward direction, so for every pair of this incremental force dF, the horizontal components will cancel, but the vertical components will add. So at this location, the net force due on this segment, the semi-circular segment, and let’s call that one as segment 2, the net force is going to be pointing in downward direction.
So, without doing any calculation, we have realized that all the forces, acting each one of these segments will be pointing in downward direction. Therefor the total force, Ftotal, is going to be direct sum of these forces, the sum of their magnitudes because they directly point in the same direction, which is downward direction.
Well, let’s just go ahead and calculate the magnitude of these forces for the segment 1. F1 is going to be equal to integral of all the incremental forces acting on these incremental segments, and that’s going to be equal to i dl magnitude, B magnitude times sine of the angle between these two vectors. We can even see that the angle between i dl and the magnetic field i dl is on the plane magnetic field is out of plane, therefore that angle is 90 degrees, so we will have sine of 90 degrees over here.
Sine of 90 degrees is just 1, and here current is constant. We can take it outside of the integral, and the magnetic field is uniform and constant. Therefore we can take the magnetic field also outside of this integral. Then, we end up with i times B times integral of dl. And integral of dl for the first segment is, we’re taking this integral, of course, over the first segment, we’re adding all these magnitude of displacement vectors dl‘s along the segment, and when we do that we will end up with the total length of that segment. And that is basically equal to R. So, the magnitude of the net force on the first segment, due to this external magnetic field on this current carrying conductor will be equal to i times B times R.
Now, similarly, for segment 3, the other segment, F3 will be equal to integral of dF again and that will be equal to i times B times, again there the angle between i dl and B is 90 degrees, so it’s exactly identical to the first segment. Now, that integral has taken over the third segment and integral of dl over that third segment will be equal to the length of that segment, which is 2R. Therefore, here we will have 2iB times R.
Alright, now we’re going to try to determine the force F2, which is the net force on the semicircular region. For that region, for the segment two, we have shown that dFhorizontals cancel, due to the symmetry, then, the total force F2 becomes the integral of dFverticals. So, we need to express dFvertical and then integrate them over all this semicircular region and in order to express the vertical component, we will again try to take the advantage of these right triangles forming once we resolve the force into its components.
In order to do that, therefore, we will define an angle. Let’s go ahead and define this angle as