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. `0*5T`
B. `1*0T`
C. `1*5T`
D. `1*7T`

Answer

B

Explanation

the forces acting on the rod are

(i) mg acting vertically downwards
(ii) BI l acting horizontally. The rod will be in equilibrium if the resolved components of the forces parallel to the inclined plane are equal and opposite i.e.,
`mgsin 60^@=BIl cos 60^@`
or `B=(mg)/(Il)tan 60^@=(10^-2xx10xxsqrt3)/(1*73xx0*1)=1T`