Question

An electron in a circular orbit of radius .$05$ nm performs $10^{16}$ revolutions per second. The magnetic moment due to this rotation of electron is (in $Am^2$)

• A. $2.16 × 10^{-23}$
• B. $3.21 × 10^{-22}$
• C. $3.21 × 10^{-24}$
• D. $1.26 × 10^{-23}$

Answer 1

Answer: (D)

Given, $r=0.05 \,nm =0.05 \times 10^{-9} m$
$\left(\because 1\, nm =10^{-9} m \right)$
$n=10^{16}$ revolutions
$e=1.6 \times 10^{-19}\, C$
We know that
The magnetic moment $M=A i$
$M=\pi r^{2} \times n e$
$M=3.14 \times\left(0.05 \times 10^{-9}\right)^{2} \times 10^{16} \times 1.6 \times 10^{-19}$
$M=0.1256 \times 10^{-18} \times 10^{16} \times 10^{-19}$
or $M=1.26 \times 10^{-23} \,Am ^{2}$