Torque required to deflect a bar magnet in a uniform magnetic field through an angle of 300 is 25 N - m. Then the work done by an external agent to deflect the magnet through an additional angle of 300 will be
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a
50 J
b
503 J
c
253 −1J
d
503 −2J
answer is C.
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Detailed Solution
(3)τ=MBsin α ⇒ 25=MB sin 300⇒ MB=50 N−mW=−MB cos 600− −MB cos300=MB 32−12⇒ W=503−12J = 253−1J
Torque required to deflect a bar magnet in a uniform magnetic field through an angle of 300 is 25 N - m. Then the work done by an external agent to deflect the magnet through an additional angle of 300 will be