Question

A conductor of length $l$ and mass $m$ carrying $I$ is placed on a smooth incline plane making an angle $\theta$ with horizontal.A magnetic field $B$ is directed vertically upwards. Then for equilibrium of the conductor tan $\theta$ is given by

• A. $\frac{BIl}{mg}$
• B. $\frac{mg}{BIl}$
• C. $\frac{mg}{2BIl}$
• D. $\frac{2mg}{BIl}$

Answer 1

The correct option is A $\frac{BIl}{mg}$

A conductor of length L and mass m. Carrying current 1 is placed on a smooth inclined plane making an angle $\theta$ with horizontal, a magnetic field B.

Ref. image

Applying fleming's left hand rule the magnetic force exerted on the current carrying wire is F is horizontal direction is given by.

$F=BIL$

Now the component division of forces mg and F suggests that,

$mg\sin \theta =BIL\cos \theta$

$\tan \theta =\frac{BIL}{mg}$