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 \(A=HW\).

- 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.

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.