PQ is an infinite current carrying conductor. AB and CD are smooth conducting rods on which a conductor EF moves with constant velocity V as shown. The force needed to maintain constant speed of EF is
A
1VR[μ0Iv2πln(ba)]2
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B
[μ0Iv2πln(ba)]−21VR
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C
[μ0Iv2πln(ba)]2VR
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D
VR1VR[μ0Iv2πln(ba)]−2
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Solution
The correct option is A1VR[μ0Iv2πln(ba)]2 A, Induced emf ∫baBvdx=∫baμ0I2πxvdx ⇒ Induced emf =μ0Iv2πln(ba) ⇒ Power dissipated =E2R Also, power = F.V ⇒F=E2VR ⇒F=1VR[μ0Iv2πln(ba)]2.