Question

If two charged particles each of charge $q$ and mass $m$ are connected to the ends of a rigid massless rod of length $l$ and is rotated about an axis passing through the center and perpendicular to length. Then the ratio of magnetic moment to angular momentum is-

• A. $\frac{q}{m}$
• B. $\frac{2q}{m}$
• C. $\frac{q}{3m}$
• D. $\frac{q}{2m}$

Answer 1

The correct option is D $\frac{q}{2m}$


(i) The magnetic moment of the two charge system is,

$M=NiA$

$\Rightarrow M=\frac{2q}{T}\times \pi {\left(\frac{l}{2}\right)}^2$

$\Rightarrow M=\frac{\pi q{l}^2}{2T}........\left(1\right)$

(ii) Angular moment of the two charge system is,

$L=2\left(m{r}^2\omega \right)$

$L=2m(\frac{l^2}{4}\times \frac{2\pi }{T})=\frac{m\pi l^2}{T}........\left(1\right)$

From (1) and (2),

$\frac{M}{L}=\frac{\left(\frac{\pi q{l}^2}{2T}\right)}{\left(\frac{m\pi l^2}{T}\right)}=\frac{q}{2m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.