A horizontal rod of mass `10g` and length `10cm` is placed on a smooth plane inclined at an angle of `60^@` with the horizontal with the length of the rod parallel to the edge of the inclined plane. A uniform magnetic field induction B is applied vertically downwards. If the current through the rod is `1*73ampere`, the value of B for which the rod remains stationary on the inclined plane is

Options

A. 1.73 tesla
B. `(1)/(1.73)` tesla
C. 1 tesla
D. None of the above

Answer

C

Explanation

At equilibrium - `mg sin 60^(@) - I//B cos 60^(@)`
`implies B = (0.01 xx 10 xx sqrt3)/(0.1 xx 1.732)`