# Torque on a Current Loop

The figure shows a rectangular loop of wire which has current flowing through it counterclockwise. If the loop is placed in a magnetic field that points upward, we showed (Example 13.4.2) that its sides feel the forces shown. The net force on the loop is zero. However, if the field is pushing inward at the top and outward at the bottom, this will tend to make the loop spin around a horizontal axis. That is, the loop is experiencing a net torque. (Click the word Torque below if you need a refresher on the topic.)
Find the net torque $$\vec\tau$$ on the loop above.
The torque on the top wire is equal to $$\vec\tau=\vec r\times\vec F$$, where $$\vec F=IWB\into$$ is the force, and $$\vec r=H/2\uparrow$$ is the lever arm: the vector that runs perpendicularly from the axis to the point where the force is applied. Thus the torque on the top wire is $$\vec\tau={H\over2}\uparrow\times IWB\into = \frc2IHWB(\uparrow\times\,\into)=\frc2IHWB\leftarrow$$ Similarly, the torque on the bottom wire is $$\vec\tau={H\over2}\downarrow\times IWB\outof = \frc2IHWB(\downarrow\times\,\outof)=\frc2IHWB\leftarrow$$ and so the net torque is the sum of these: $$\boxed{\vec\tau=IHWB\leftarrow}$$ What does it mean for the torque to point to the left? Well, the net torque vector lies along the axis of rotation (so, the horizontal line), and a right-hand rule tells you the direction of rotation: point your right thumb to the left, and your fingers tell you the direction the loop will spin.

Interactive 13.4.1

If left alone, the loop will spin until it is horizontal, and the magnetic field is pointing up through it. At that point, the loop looks like result (b) from Example 13.3.2: the magnetic forces pull the wire outwards, but do not cause the loop to spin. Notice how the loop seems to turn to "face" the direction of the magnetic field? We can quantify that a little bit.

Let's define a new vector, the magnetic dipole moment $$\vec \mu$$, in the following way:

• it is a vector which points perpendicularly to the plane of the loop. There are two ways the vector could point, so we use a loop-normal right-hand rule (Section 1.5): curl the fingers of your right hand in the direction of the current (counterclockwise, here), and your thumb points in the direction of $$\vec\mu$$. In this picture, $$\vec\mu$$ points out of the page.
• the magnitude of the magnetic dipole moment is $$\abs{\vec\mu}=IA$$ where $$I$$ is the current running through the loop, and $$A$$ is the area. This loop is a rectangle, so $$A=HW$$.
Now given this definition, we notice the following things:
• The loop turns until $$\vec\mu$$ points in the same direction as $$\vec B$$.
• The initial torque on the loop is $$\vec\tau=I(HW)B\leftarrow$$, and $$IHW=IA=\mu$$, so we can write it as $$\abs{\vec\tau}=\abs{\vec\mu}\abs{\vec B}$$ Furthermore, notice that $$\vec\mu$$ points out of the page, $$\vec B$$ points up, and $$\outof\times\uparrow=\leftarrow$$, the direction of the torque. So we could write $$\vec\tau=\vec\mu\times\vec B$$
• Is the same true in case (b)? In that case, $$\vec\mu$$ and $$\vec B$$ point in the same direction, so $$\vec\tau=\vec\mu\times\vec B=0$$ which is true.
I wouldn't call that a proof, but it turns out to be true in general.
A flat loop with area $$A$$ and carrying current $$I$$, placed in a uniform magnetic field $$\vec B$$, feels a torque $$\vec\tau=\vec\mu\times\vec B$$ where the magnetic dipole moment $$\vec\mu$$ has magnitude $$IA$$ and points perpendicularly to the loop, according to a loop-normal right-hand rule.