$(a)\;\mu=\large\frac{e h n}{\pi m}\qquad(b)\;\mu=\large\frac{e h n}{2 \pi m}\qquad(c)\;\mu=\large\frac{e h n}{4 \pi m}\qquad(d)\;\mu=\large\frac{ 2e h n}{\pi m}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Answer : (b) $\;\mu=\large\frac{e h n}{ 4\pi m}$

Explanation :

The orbiting electron behaves as a current loop. The equivalent current

$i=\large\frac{net\;charge}{time\;of\;revolution }=\large\frac{e}{T}=e f_{rot}$

Now , magnetic dipole moment

$\mu=n i A$

n=no. of turns of loop =1 , A= area of loop = $\;n r^2$

$\mu=(1)\;(e f_{rot})\;(\pi r^2) \quad \;=> \mu=\pi e r^2 f_{rot}$

Putting the values of $\;f_{rot}\;$ and r we obtain ,

$\mu=\pi e \;(\large\frac{n^2 h^2 \varepsilon }{n m e^2 z})^{2}\;(\large\frac{m z^2 e^4}{4 \varepsilon^2 h^3 n^3})$

$\mu=\large\frac{e h n}{4 \pi m}$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...