Question

A conducting rod of length $l$ and mass $m$ is moving down a smooth inclined plane of inclination $\theta$ with constant velocity $V$. A current $i$ is flowing in the conductor in a direction perpendicular to paper inwards. The magnitude of field $\overrightarrow{B}$ is :
(Given acceleration due to gravity $=g$)

• A. $\frac{mg}{il}\sin \theta$
• B. $\frac{mg}{il}\tan \theta$
• C. $\frac{mg\cos \theta }{il}$
• D. $\frac{mg}{il\sin \theta }$

Answer 1

Answer: (B)

$\frac{mg}{il}\tan \theta$

Net force is given by,

$F=Mg\sin \theta -Bil\cos \theta$

Since particle is moving down with constant magnetic field,

$F=0$

$mg\sin \theta -Bil\cos \theta =0$

$B=\frac{mg}{il}\tan \theta$