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Question: The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform ...

The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is

(dipole moment of this dipole = 0.5 A m2)

A

0.07 J

B

0.08 J

C

0.09 J

D

0.1 J

Answer

0.09 J

Explanation

Solution

: Since the most stable position is at θ=0\theta = 0and the most unstable position is at θ=180∘\theta = 180 ^ { \circ } then the work done is given by

W=∫θ=0θ=180∘τ(θ)dθ\mathrm { W } = \int _ { \theta = 0 } ^ { \theta = 180 ^ { \circ } } \tau ( \theta ) \mathrm { d } \theta

=∫0180∘mBsin⁡θdθ=−mB[cos⁡θ]0180∘= \int _ { 0 } ^ { 180 ^ { \circ } } \mathrm { mB } \sin \theta \mathrm { d } \theta = - \mathrm { mB } [ \cos \theta ] _ { 0 } ^ { 180 ^ { \circ } }

=−mB[cos⁡180∘−cos⁡0]=−mB[−1−1]= - \mathrm { mB } \left[ \cos 180 ^ { \circ } - \cos 0 \right] = - \mathrm { mB } [ - 1 - 1 ]

=−mB[−2]=2mB= - \mathrm { mB } [ - 2 ] = 2 \mathrm { mB }

Here, m=0.5Am2\mathrm { m } = 0.5 \mathrm { Am } ^ { 2 } and B=0.09 T\mathrm { B } = 0.09 \mathrm {~T}

W=2×0.50×0.09=0.09 J\mathrm { W } = 2 \times 0.50 \times 0.09 = 0.09 \mathrm {~J}