The electron in hydrogen atom moves around the proton with a speed of 2.2 x 106 m/s in a circular orbit of radius 5.3 x 10-11m. calculate?
1. The equivalent current?
2. Equivalent dipole moment?
3. The magnetic field at the site of proton.

Dear Student,
Given,
Speed of the electron, v = 2.2 x 10⁶ m/s 
Radius of the circular orbit, R = 5.3 x 10^-11m
Charge of the electron, e = 1.6 x 10^-19C

1.
Number of revolutions made by the electron in 1s, n = $\frac{v}{2\pi R}=\frac{2.2\times {10}^6}{2\times 3.14\times 5.3\times {10}^{-11}}=6.61\times {10}^{15} revolutions$
Equivalent current, $I = ne=6.61\times {10}^{15}\times 1.6\times {10}^{-19}=1.06 mA$

2.
Equivalent dipole moment, $p=e\times R=1.6\times {10}^{-19}\times 5.3\times {10}^{-11}=8.48\times {10}^{-30} C-m$

3.
The magnetic field at the site of proton.:
$B=\frac{{\mu }_{o}I}{2R}=\frac{4\pi x {10}^{-7}\times 1.06\times {10}^{-3}}{2\times 5.3\times {10}^{-11}}=1.256\times 10=12.56 T$
Regards