Question

If an electron is revolving in a circular orbit of radius $0.5A^{\circ }$ with a velocity of $2.2\times {10}^6$ m/s. The magnetic dipole moment of the revolving electron is

• A. $8.8\times {10}^{-24}A{-m}^2$
• B. $8.8\times {10}^{-23}A-m^2$
• C. $8.8\times {10}^{-22}A-m^2$
• D. $8.8\times {10}^{-21}A-m^2$

Answer 1

Answer: (A)

$8.8\times {10}^{-24}A{-m}^2$

$\text{Given}:$r=$0.5\times {10}^{-10}m$, v=$2.2\times {10}^6$m/s

$\text{Solution}:$

The magnetic dipole moment is given by,

M=I n A

Also,

I=$\frac{q}{A}$=$\frac{qvr}{2}$=$\frac{1.6\times {10}^{-19}\times 2.2\times {10}^6\times 0.5\times {10}^{-10}}{2}$=$8.8\times {10}^{-24}A{m}^2$

$\text{Hence A is the correct option}$