Question

A bar magnet of magnetic moment M is hung by
a thin cotton thread in a uniform magnetic field B.

Work done by the external agent to rotate the bar
magnet from stable equilibrium position to 120°
with the direction of magnetic field is (consider
change in angular speed is zero) [NCERT Pg. 178]​

Answer 1

Explanation:

Work done by the magnetic field is W=−→M. →B=−MBcosθ. Therefore, we have to do a work W=→M. →B=MBcosθ.

Answer 2

Answer:

Required work done = $-\frac{MB}{2}$

Explanation:

Work done by a magnetic field to rotate a magnet by anle $\theta$ is given by,

$W= \vec{M} .\vec{B}=MBcos(\theta)$

Given: magnetic moment = $M$

Angle = $\theta = 120^o$

Magnetic field = $B$

Work done = $M.B.cos(120^o)=M.B.\frac{-1}{2} = -\frac{MB}{2}$

Work done by the external agent to rotate the bar magnet from stable equilibrium position to 120° = $-\frac{MB}{2}$

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