Question

Work done in rotating a bar magnet from 0 to 120°

Answer 1

Answer:

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Answer 2

Given :

Initial angle of bar magnet , $\theta_i=0^o$ .

Final angle of bar magnet , $\theta_f=120^o$.

To find :

Work done in rotating a bar magnet from 0° to 120° .

Solution :

Let , us assume that the magnetic moment of bar magnet be M .

Also , let magnetic field be B .

We know , work done is in rotating a bar magnet from $\theta_i$ to $\theta_f$ is :

$W=U_f-U_i$             ....... ( 1 )

Here , U is potential energy and given by :

$U=-M.B\\\\U=-MBcos\ \theta$

Putting value of U in equation 1 .

We get :

$W=MBcos\ \theta_i-MBcos\ \theta_f\\\\W=MB-MB(cos\ 120^o)\\\\W=\dfrac{3MB}{2}$

Therefore , Work done in rotating a bar magnet from 0° to 120° is $\dfrac{3MB}{2}$ .

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