Why is $m_{\ell}$ called the magnetic quantum number? What is its association with magnets? I am going over my quantum lecture notes and I can't seem to link the quantum number $m_{\ell}$ with any magnetic property. It just seems to specify the shape of an orbital with a particular principal quantum number. Is there any reason for it being labelled as magnetic?
 A: An angular momentum of an electron about a nucleus of $\bf L$ implies a magnetic moment $-\mu_B \bf L$. See here.
A: The magnetic quantum number $m_\ell$ determines the energy shift
when the atom is in an external magnetic field.
Quoted from Magnetic quantum number - Effect in magnetic fields:

The quantum number $m$ refers, loosely, to the direction of the angular
   momentum vector. The magnetic quantum number $m$ only affects the electron's
   energy if it is in a magnetic field because in the absence of one,
   all spherical harmonics corresponding to the different arbitrary values
   of $m$ are equivalent. The magnetic quantum number determines the energy
   shift of an atomic orbital due to an external magnetic field (the Zeeman
   effect) — hence the name magnetic quantum number. 

A: The  wikipedia page linked in my2cts's answer says:

The revolution of an electron around an axis through another object, such as the nucleus, gives rise to the orbital magnetic dipole moment. 

I would say that such  a statement is not fully  consistent with a truly quantum-mechanical description (did they wrote revolution?), or, at the least it would require a few additional words for its justification. Here they are.
States with non-zero magnetic quantum number correspond to complex wavefunctions such that the quantum probability current density 
$$
{\bf j}({\bf r})=\frac{1}{2m}\left(  \Psi^* {\bf \hat p} \Psi - \Psi {\bf \hat p} \Psi^* \right),
$$
is different from zero as well as  the resulting electric current density and magnetic moment
$$
{\bf m}=\frac{e}{2}\int {\bf r}\times {\bf j}({\bf r}) d^{3}{\bf r},
$$
where $e$ is the electron charge.
By performing the integral in the case of the hydrogen atom wavefunctions, one can check the proportionality between magnetic moment and angular momentum via the Bohr's magneton ($\mu_B$).
