Torque on a Circuit and Magnetic Dipoles

There are two ways to approach this. Both are valid, but the first approach is fairly tedious, while the second approach uses the magnetic dipole moment to reach the same result much faster!

Approach 1:

Let's find the torque by finding the contribution from each length of the wire loop, $\tau=\vec r \times \vec F$. We know there is no magnetic force contribution from the top or bottom sides of the rectangle because the field and current are parallel/antiparallel ($|\vec F|=IlB\sin\theta=0$ because $\theta=0^o$ and $\theta=180^o$).

For the left part of the wire loop, we can use the right-hand rule to see that the force points out of the page with magnitude $|\vec F|=IlB\sin\theta=I(y)B(1)=IyB$. For the right part of the loop, we can either use the right-hand rule and magnetic force formula to find a force that points into the page with magnitude $IyB$, or we can simply write this down, because we know that the total force must be zero for a closed loop of current in a uniform magnetic field.

Now we add together the torque from each side of the wire. Note that we are using the point at the center of the rectangle as our reference point to measure $\vec r$. The angle between each magnetic force and vector $\vec r$ is $90^o$, so the sine term becomes 1.

For the direction of the torque, we can see (based on the directions of the forces) that the loop rotates such that the left part comes out of the page and the right part goes in to the page. We could also say this is counter-clockwise with respect to the positive y-axis if we define the positive y-axis as going up the page.

Approach 2:

We know the magnetic dipole moment points out of the page using the right-hand rule (curl your fingers counter-clockwise and check where your thumb is pointing). The magnitude is $|\vec \mu|=IA$.

Using the formula for torque on a current loop (the angle between the field and dipole is $90^o$ and the sine term becomes 1).

The direction is such that the vector $\vec \mu$, which current points out of the page, wants to reach the direction of $\vec B$, which points to the right. This is equivalent to the direction found above.

That was much easier and got the same answer as above! This is why we introduced the magnetic dipole moment. It makes life easier!