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Question

Physics Question on Electric Dipole

A magnet of magnetic moment 20CGS20\, CGS units is freely suspended in a uniform magnetic field of intensity 0.3CGS0.3\, CGS units. The amount of work done in deflecting it by an angle of 3030^\circ in CGSCGS units is

A

66

B

33ˉ3 \, \bar{3}

C

3(23ˉ)3(2 - \bar{3})

D

33

Answer

3(23ˉ)3(2 - \bar{3})

Explanation

Solution

Work done, W=MBμ(1cosθ) W =-MB_{\mu} (1 - \cos \, \theta)
=20×0.3(1cos30)= 20 \times 0.3 (1- \cos \, 30^{\circ})
=613ˉ2=323ˉ= 6 \, 1 - \frac{\bar{3}}{2} = 3 \, 2 - \bar{3}