Magnetic dipole moment because of spinning electron We always consider only the spin of electron. But protons are spinning charges as well. So what about the magnetic dipole moment caused by them?
 A: All else being equal the magnetic moment of a particle is inversely proportional to its mass. Since the mass of the proton is much greater than the mass of the electron the magnetic moment of the proton is much less than the magnetic moment of the electron. This means that when both electrons and protons are present (e.g. in neutral matter) the field from the electrons dominates and to a good approximation we can simply ignore the protons.
However there are lots of situations where the proton magnetic moment is very important. For example nuclear magnetic resonance measures the interaction of the proton magnetic moment with an external magnetic field. Another example is the famous hydrogen 21 cm line that is produced by the interaction of the electron magnetic dipole with the proton magnetic dipole.
A: A magnetic moment is a vector quantity, and the direction of the proton's magnetic moment is defined by its spin. The torque on the proton resulting from an external magnetic field is towards aligning the proton's spin vector in the same direction as the magnetic field vector.
The value for the magnetic moment of the proton is $μ_p = 2.7928473508(85) μ_N$.In these values, $μ_N$ is the nuclear magneton, a physical constant and standard unit for the magnetic moments of nuclear components.
The nuclear magneton is the spin magnetic moment of a Dirac particle, a charged, spin $1/2$ elementary particle, with a proton's mass mp. In SI units, the nuclear magneton is
$$\mu_N=\frac{e\hbar}{2m_p}$$
When a proton is put into a magnetic field produced by an external source, it is subject to a torque tending to orient its magnetic moment parallel to the field (hence its spin also parallel to the field).

~References

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*Proton magnetic moment

*Proton

*The magnetic moment of the proton, I. The value in nuclear magnetons
