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 the paper inwards. A vertically upward magnetic field $\overrightarrow{B}$ exists in space. Then, magnitude of magnetic field $\overrightarrow{B}$ 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

The correct option is B $\frac{mg}{Il}\tan \theta$

Let $F_m$ be the force on the current carrying conductor due to magnetic field.
Let us understand the direction of this force with the help of cartesian coordinate system

$mg$ also acts on the rod in downward direction.
The components of $mg$ and $F_m$ along the inclined plane can be shown as below:
As the rod is moving with constant velocity along the inclined plane.
We can say, $mgsin\theta =F_{m}cos\theta$
$⟹mgsin\theta =BILcos\theta$
$\therefore B=\frac{mgtan\theta }{IL}$