$$\vec\tau=\vec\mu\times\vec B$$

where \(\vec\mu\) is the magnetic dipole moment of
the dipole, which we mentioned back in Section 12.1.
Notice that when \(\vec\mu\) and \(\vec B\) point in the same
direction, their cross-product is zero, and there is no torque.
These pictures show a magnetic dipole in a magnetic
field that points upward. Find the direction of the torque
in each case, and explain what the dipole does.

Of course, we can use this equation for numerical results as well:
Both vectors lie on the plane of the screen, so their
cross-product either points into or out of the screen. The
right-hand rule gives us
$$\vec\tau=\vec\mu\times\vec
B=\nwarrow\times\uparrow=\into$$
Using the torque right-hand rule, we point our right thumb in the
direction of \(\vec\tau\), and our fingers tell us that the
dipole will rotate *clockwise* (to eventually line up
with the magnetic field).

Using the right-hand rule, we see that
$$\vec\tau=\vec\mu\times\vec
B=\swarrow\times\uparrow=\into$$
so again the dipole will rotate *clockwise*.

In this case,
$$\vec\tau=\vec\mu\times\vec B = \nearrow\times\uparrow =
\outof$$
which means the dipole will rotate *counterclockwise*:
still, the direction it needs to go to align itself with the field.

Because the dipole in the magnetic field point in opposite
directions,
$$\vec\tau=\vec\mu\times\vec B =
\downarrow\times\uparrow=0$$
The dipole feels no torque. It's like a pencil balanced on
one end: it *wants* to fall over, but it can't decide
which direction to go, and so is in a state of *unstable
equilibrium*. Give it the slightest bump, and it will
flip to point up.

A typical bar magnet has a magnetic dipole moment of
\(\mu=0.5\u{Am^2}\). The Earth's magnetic field in a particular
room is \(50\u{\mu T}\) in magnitude, and points north and downward,
dipping \(50\degrees\) below the horizontal. What is the magnitude of
the torque our bar magnet feels, if it points due east?

The dipole and the magnetic field are perpendicular to each other (EXPLAIN!),
and so
$$\begin{eqnarray}
\abs{\vec\tau}&=&\abs{\vec\mu\times\vec B}\\
&=&\abs{\vec\mu}\abs{\vec B}\sin90\degrees\\
&=&(0.5\u{Am^2})(50\ten{-6})=\boxed{2.5\ten{-5}\u{Nm}\\
\end{eqnarray}$$

If the magnet is 5 cm long, how much force would I have to apply to
the end of the magnet to balance this torque? Assume the magnet
pivots at its geometric center.

If I apply a force \(F\) perpendicularly at one end of the bar magnet, the lever arm
is \(2.5\u{cm}=0.025\u{m}\) long, and the torque is
$$\tau=rF=(0.025\u{m})F=2.5\ten{-5}\u{Nm} \implies
F=\frac{2.5\ten{-5}}{2.5\ten{-2}\u{m}}=1\u{mN}$$
Since it takes \(10\u{mN}\) to pick up a paper clip, this is a very
small force, easily provided by friction. If you *want* the dipole to spin, you
need to remove as much friction as possible, by balancing the magnet
on a pin or suspending it from a string.

If you had a circular loop of wire with a radius of \(5\u{cm}\), how
much current would have to flow through it to have the same dipole
moment?

The dipole moment of a current loop is \(\mu=IA\), so the current
through this loop should be
$$I={\mu\over A}={0.5\u{Am^2}\over \pi (0.05\u{m}^2)}=\boxed{63.7\u{A}}$$
which is a pretty hefty current.