Question:
A magnet of magnetic moment $20\, CGS$ units is freely suspended in a uniform magnetic field of intensity $0.3\, CGS$ units. The amount of work done in deflecting it by an angle of $30^\circ$ in $CGS$ units is
A magnet of magnetic moment $20\, CGS$ units is freely suspended in a uniform magnetic field of intensity $0.3\, CGS$ units. The amount of work done in deflecting it by an angle of $30^\circ$ in $CGS$ units is
Updated On: Aug 4, 2024
- $6$
- $3 \, \bar{3}$
- $3(2 - \bar{3})$
- $3$
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The Correct Option is C
Solution and Explanation
Work done, $ W =-MB_{\mu} (1 - \cos \, \theta)$
$ = 20 \times 0.3 (1- \cos \, 30^{\circ})$
$ = 6 \, 1 - \frac{\bar{3}}{2} = 3 \, 2 - \bar{3} $
$ = 20 \times 0.3 (1- \cos \, 30^{\circ})$
$ = 6 \, 1 - \frac{\bar{3}}{2} = 3 \, 2 - \bar{3} $
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Concepts Used:
Electric Dipole
An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.
p=q×2a
where,
p denotes the electric dipole moment, pointing from the negative charge to the positive charge.
Force Applied on Electric Dipole
