A bar magnet with magnetic moment $m$ is placed in a uniform field $B$ making angle $\theta$ with the field. The torque on it is:

1. A $mB\cos\theta$
2. B $mB\sin\theta$ Correct answer
3. C $mB$
4. D $mB\tan\theta$

Solution

$\vec\tau = \vec m \times \vec B \Rightarrow \tau = mB\sin\theta$.

About this question

Subject: Physics · Chapter: Magnetism · Topic: Bar Magnet and Magnetic Field

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