$A$ proton has spin and magnetic moment just like an electron. Why then is its effect neglected in the magnetism of materials?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The magnetic dipole moment of a particle is inversely proportional to its mass,given by the relation $M = \frac{eh}{4\pi m}$.
For a proton,the magnetic dipole moment is $M_{p} = \frac{eh}{4\pi m_{p}}$.
For an electron,the magnetic dipole moment is $M_{e} = \frac{eh}{4\pi m_{e}}$.
Taking the ratio,we get $\frac{M_{p}}{M_{e}} = \frac{m_{e}}{m_{p}}$.
Since the mass of a proton $m_{p}$ is approximately $1837$ times the mass of an electron $m_{e}$ $(m_{p} \approx 1837 m_{e})$,we have $\frac{M_{p}}{M_{e}} = \frac{m_{e}}{1837 m_{e}} = \frac{1}{1837}$.
Thus,$M_{p} = \frac{M_{e}}{1837}$.
This shows that the magnetic moment of a proton is about $1837$ times smaller than that of an electron. Therefore,the magnetic effect of a proton is negligible compared to that of an electron in the magnetism of materials.

Explore More

Similar Questions

Which of the following curves represents the variation of magnetic moment $(M)$ with temperature $(T)$ for a ferromagnetic material?

Explain paramagnetism and paramagnetic substances.

If a magnetic substance is kept in a magnetic field,then which of the following substances is thrown out?

Curie temperature is the temperature above which

The Curie temperatures of cobalt and iron are $1400 \,K$ and $1000 \,K$ respectively. At $T=1600 \,K$, the ratio of the magnetic susceptibility of cobalt to that of iron is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo