Why compass needle doesn't precess about but aligns by magnetic field?

Question

A magnetic dipole in magnetic field, precess about the axis of magnetic field. So why compass needle doesn't precess but lines up with magnetic field?

Answer 1

A good compass contains a liquid that dampens the motion of the needle. An ideal frictionless needle would constantly oscillate around the direction of the magnetic field and if the needle was free to move in three dimensions, then it would indeed show precession.

Answer 2

Precession occurs if the object has angular momentum. The torque then locates the angular momentum vector causing precession. A compass needle has no angular momentum, so there is no precession

Answer 3

Jerrold Franklin's answer has the right idea. Unless the compass needle has an appreciable spin angular momentum, it will not precess about the direction of the magnetic field. It may still oscillate if the damping is sufficiently low, but this would not normally be referred to as precession.

A stationary compass needle actually does have some angular momentum just by virtue of being magnetized, since its dipole moment is associated with spin angular momenta of electrons. However this angular momentum is very small.

To get some numbers, suppose we have a cylindrical magnet of radius $r$, volume $V$, mass density $\rho$ and an average magnetic field $B$ inside it. The dipole moment can be estimated as $$ \mu = \frac{BV}{\mu_0}.$$ The gyromagnetic ratio (ratio of magnetic moment to angular momentum) of ferromagnetic materials is about $e/m_e$ (the gyromagnetic ratio of electrons), where $e$ is the elementary charge and $m_e$ is the electron mass. We can write $$\mu = \frac{e}{m_e}L=\frac{e}{m_e}I\omega_\text{eq}$$ where $L$ is the angular momentum, $I=\frac12Mr^2=\frac12\rho Vr^2$ is the moment of inertia and $\omega_\text{eq}$ is the equivalent angular speed that would yield the same angular momentum as that due to magnetization alone. We have $$\omega_\text{eq} = \frac{2m_eB}{e\mu_0\rho r^2}.$$ For a neodymium magnet which might have $B\approx 1\text{ T}$, taking $\rho = 7.5\text{ g/cm}^3$ and $r = 1\text{ mm}$, we get $\omega_\text{eq} \approx 1.2\times10^{-3}\text{ rad/s}$, corresponding to a spin period of about 1.4 hours. This is a tiny angular speed (and angular momentum) as far as macroscopic objects and torques are concerned.

To see a clear precession without too much nutation, the "rotational kinetic energy" due to angular momentum should be large compared to the potential energy of the magnetic dipole due to the surrounding magnetic field $B_0$. This means $$\frac{L^2/2I}{\mu B_0}=\frac{m_e\omega_\text{eq}}{2eB_0} \gg 1.$$ With the value of $\omega_\text{eq}$ calculated above, this requires $$B_0 \ll 3\times10^{-15}\text{ T}.$$ The Earth's magnetic field is about $5\times10^{-5}\text{ T}$.

Answer 4

Stating the obvious here, in case anyone misses it. The compass needle is not a freely suspended magnetic dipole. It is mechanically confined to rotating in a plane. If you look at the compass needle from the side, it is suspended from the top of the hollow center cap through a pin, making it a pendulum effectively. This makes the needle align to the parallel component of the earth's magnetic field. The perpendicular component is negligible compared to the torque due to the weight of this pendulum.