Question

A conducting rod of length $L$ and mass $m$ is moving down a smooth inclined plane of inclination $\theta$ with constant speed $V$. A current $I$ is flowing in the conductor perpendicular to the paper and pointing inwards. A uniform magnetic field directed vertically upwards exists in the space. The magnnitude of magnetic field $B$ is:

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

Answer 1

The correct option is C $\frac{mg}{IL}\tan \theta$

Using the relation for magnetic force

${\overrightarrow{F}}_m=I(\overrightarrow{L}\times \overrightarrow{B})$

$F_m=ILBsin{90}^o=ILB$

The direction of magnetic force is shown in figure below:


Using the equations for equilibrium of conductor along the inclined plane.

$\because V=constant$

$\Rightarrow a=0$

$\therefore mgsin\theta =F_{m}cos\theta$

$mgsin\theta =BILcos\theta$

$B=\frac{mgsin\theta }{ILcos\theta }$

$\therefore B=\frac{mgtan\theta }{IL}$