Question

A copper rod of length $L$ and mass $m$ carrying current $I$ flowing inwards and perpendicular to the plane(as shown in the figure) , is placed on a fixed smooth inclined plane of inclination $\theta$ . If a vertically upward magnetic field $\overrightarrow{B}$ exists in this region, find the magnitude of the magnetic field $\overrightarrow{B}$ for which the copper rod slides down the inclined plane with a constant speed.

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

Answer 1

The correct option is C

$\frac{\text{mg}}{\text{IL}}\text{tan}\theta$

As the rod is sliding down with constant speed
Net force along incline plane should be Zero.
$\Rightarrow F_{net}=0$
$\Rightarrow BIL\text{cos}\theta -mg\text{sin}\theta =0$
$\Rightarrow BIL\text{cos}\theta =mg\text{sin}\theta$
$B=\frac{mg}{IL}\frac{\sin \theta }{\cos \theta }=\frac{mg}{IL}\tan \theta$