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. A vertically upwards magnetic field $\vec{ B }$ exists in space. Then magnitude of magnetic field $\vec{ B }$ is

• A. $\frac{m g}{i l} \sin \theta$
• B. $ \frac{mg}{il}\tan \theta $
• C. $ \frac{mg\cos \theta }{il} $
• D. $ \frac{mg}{il\,\sin \,\theta } $

Answer 1

Answer: (B)

Magnetic force $\vec{ F }_{m}=i l B$ acts in the direction as shown in given figure.
Rod will move downwards with constant velocity, if net force on it is zero.

or $ F_{m} \cos \theta=m g \sin \theta $
or $ i l B \cos \theta=m g \sin \theta $
$\therefore B=\left(\frac{m g}{i l}\right) \tan \theta$