8.4 Torque on a current loop from Office of Academic Technologies on Vimeo.

**8.4 Torque on a current loop**

All right. Let’s consider a system which consists of two bar magnets. They’re placed such that one of them’s north pole is located at this point and the other one’s south pole is located across from the north pole of the other magnet, like this. We know that such an orientation will generate magnetic field lines, as we all know, which will, they will originate from the north pole and enter into the south pole, filling the region between the poles of these two magnets this way.

At this point, let’s consider a rectangular current loop and place inside of this magnetic field region at an arbitrary position like this. And then, let current *i* flow through this rectangular loop. Let’s also give some dimensions to this loop and say that this side has the length of *a*, and the long side has the length of *b*. Well, once we place this current-carrying loop inside of this external magnetic field, naturally, the magnetic field generates magnetic force on the segments of this current loop, and that force is going to be equal to integral of *i d***l** cross the external magnetic field, **B**.

To be able to see the directions of these forces clearly, let’s look at this picture from the front field, and if you do that, this will be the orientation of the loop, and in the upper part of the loop, which is this part, current is coming out of plane, and it will be flowing along the side segment, or the front segment of the loop, and going into the plane at this point. And of course, at the back of the plane, it will be flowing along the back segment of the rectangular loop in upward direction.

Magnetic field is pointing to the right and at the location of the upper segment, also, it is pointing this way, and if you look at the direction of the force generated by this external magnetic field on the upper segment, and let’s number that segment as one, and the lower segment as two, this segment at three, and the other one as fourth segment. So, on the upper segment, along segment one, *i d***l** is coming out, **B** is to the right. Holding the right-hand fingers in the direction of *i d***l** vector, which is out of plane, and curling them towards **B**, keeping the right-hand thumb in open, or up position, we will see that the force acting on the segment will be pointing in upward direction.

And in three-dimensional picture, it will be something like this. Of course, *i d***l** cross **B** will give us the direction of the force acting on an incremental segment, that is *d***F**, but all those *d***F**‘s along the straight-line segment will be acting in upward direction, the net force will be the sum of all those forces, and we’re going to end up with a net force of **F** pointing in upward direction.

And if you consider the symmetrical segment, which is segment two right across from the first one, and there, the current is going into the plane, again, the magnetic field is pointing to the right, and *i d***F** cross **B** in this case, is going to give us a force, net force, pointing in downward direction. These forces, since they are generated by the same magnetic field acting along a segment which, segments that they have the same length, *a*, therefore magnitude of these two forces will be the same, and in the upper segment, the force acting in upward direction, and for the lower segment, it will be acting in downward direction.

If we look at the forces acting along the side segments over here, segment three and four, for segment three, *i d***l** is going to be pointing along the wire, down direction, and **B** is to the right, *i d***l** cross **B**, therefore, will generate a force coming out of plane. Let’s call this force as **F′**. And if you look at the segment four, which is behind, and the current is flowing in upward direction, in that case, that force is going to be pointing into the plane.

All right. Now, we see how the forces are oriented in three-dimensional picture. For this segment, **F′** is going to be pointing like this, and for the other segment, it’s going to be pointing like this.

When we look at these forces, we observe that the **F′** are acting along the same line, and whereas the vertical forces, **F**‘s, are not acting along the same line. Since **F′ **will have the same magnitude, because again, these forces are generated by the same magnetic field on the same length of current, or wires carrying the same amount of current, therefore they will have the same magnitude and they will orient in opposite directions. And as a result of this, the net force along these side segments will cancel, but the forces acting along segment one and segment two, despite the fact that they have the same magnitude, since they do not act along the same line, they will naturally cause a rotation.

And the loop is going to rotate about an axis passing through its center, about this axis, axis of rotation. And we know that the cause of rotation is torque and let’s try to determine the direction and the magnitude of this torque, generated by these pair of forces **F** acting on segment one and segment two of this current loop.

Well, as you recall, torque was equal to position vector crossed with the force vector. Position vector was a vector drawn from the axis of rotation to the application point of the force. Therefore, for this case, it is going to be pointing in this direction and if you extend the direction of this position vector, or in other words, if you just carry such that the tails of **r** and **F** coincides, then the angle measured in that position will give us the angle between these two vectors.

Similarly, for the other force vector, **F**, the position vector relative to this axis of rotation is going to be drawn from that axis to the application point of the force. And again, the angle between that position vector and the force vector will be this angle, and that too will be equal to