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 $A{m}^2$)

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

Answer 1

The correct option is D $1.26\times {10}^{-23}$

The magnetic moment of an electron revolving in a circular path is $iA=\frac{q}{T}A$

$=q\nu \pi r^2$

$=1.6\times {10}^{-19}\times {10}^{16}\times 3.14\times (0.05\times {10}^{-9}{)}^2$

$=1.26\times {10}^{-23}A{m}^2$