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Question

Physics Question on Moving charges and magnetism

If a magnetic dipole of moment MM situated in the direction of a magnetic field BB is rotated by 180180^{\circ}, then the amount of work done is

A

MB

B

2 MB

C

MB2\frac{MB}{\sqrt{2}}

D

0

Answer

2 MB

Explanation

Solution

If the magnet be rotated from an initial orientation θ1=0\theta_{1}=0^{\circ} to final orientation θ2=180\theta_{2}=180^{\circ} the total work done
W=θ1θ2MBsindθW =\int\limits_{\theta_{1}}^{\theta_{2}} M B \sin d \theta
=MB[cosθ]θ1θ2=M B[-\cos \theta]_{\theta_{1}}^{\theta_{2}}
=MB(cosθ1cosθ2)=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)
where B=B= magnetic field induction
M=M = magnetic moment
=MB(+cos0cos180)=M B\left(+\cos 0^{\circ}-\cos 180^{\circ}\right)
=MB[1+1]=2MB=M B[1+1]=2 \,M B