# Torque on Magnetic Dipoles

So loops in a magnetic field spin until their dipole moment points in the direction of the magnetic field. What else spins until it points in the direction of the field? Why, compass needles, of course! Dipoles! In fact, back in Section 12.1, we defined the magnetic dipole moment of a bar magnet as pointing from its S pole to its N pole. The torque these dipoles feel is the same as on the loop:
$$\vec\tau=\vec\mu\times\vec B$$
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.
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.

Of course, we can use this equation for numerical results as well:
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.

We will see further evidence in later chapters that loops of current and magnetic dipoles are basically identical objects.