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A horizontal rod of mass 10g and length ...

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

A

1.73 tesla

B

`(1)/(1.73)` tesla

C

1 tesla

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C
At equilibrium - `mg sin 60^(@) - I//B cos 60^(@)`
`implies B = (0.01 xx 10 xx sqrt3)/(0.1 xx 1.732)`
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