A bar magnet with poles 25cm apart and p...

A bar magnet with poles `25cm` apart and pole strength `14.4 Am` rests with its center on a frictionless pivot. If it is held in equilibrium at `60^(0)` to a uniform magnetic field on induction `0.25 T` by applying a force `F` at right angles to its axis `10cm` from the pivolt, the value of `F` in newton is (nearly)

`3.9N`

`7.8N`

`15.6N`

`31.2N`

Text Solution

Verified by Experts

The correct Answer is:

`MB_(H)sin theta=Fr`

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