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

Answer 1

Answer: (B)

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

or $F_m\cos \theta =mg\sin \theta$
or $ilB\cos \theta =mg\sin \theta$
$\therefore B=\left(\frac{mg}{il}\right)\tan \theta$