Question

A particle of charge $2\text{C}$ moves along a circle of radius $4\text{m}.$ The speed of the particle is constant and equal to $2\text{m/s}$. The magnetic moment associated with the motion of the particle is

​​​​​​​$[$0.77 Mark$]$

• A. $6{\text{Am}}^2$
• B. $2{\text{Am}}^2$
• C. $10{\text{Am}}^2$
• D. $8{\text{Am}}^2$

Answer 1

The correct option is D $8{\text{Am}}^2$

Magnetic moment, $\mu =IA$

For a charged particle moving in a circular path,

$I=\frac{Q}{T}$

Where, $T$ - Time period of revolution

Also, $T=\frac{2\pi }{\omega },A=\pi R^2$

So, $\mu =\left(\frac{\omega Q}{2\pi }\right)\pi R^2$

$=\frac{\omega Q{R}^2}{2}$

$=\frac{QvR}{2}$ $[\because \omega =\frac{v}{R}]$

$=\frac{2\times 2\times 4}{2}$

$=8{\text{Am}}^2$