A circular coil of area 0.01 m^2 and 40 turns is rotated about its vertical diameter with an angular speed of 50 rads^-1 in a uniform horizontal magnetic field 0.05 T. If the average power loss due t…

Question

A circular coil of area 0.01 m^2 and 40 turns is rotated about its vertical diameter with an angular speed of 50 rads^-1 in a uniform horizontal magnetic field 0.05 T. If the average power loss due to joule heating is 25 mW, then the closed loop resistance of the coil is

Accepted Answer

Correct answer: D. (d) We have e= d dt= NBA d dt ( sin ⁡ ωt )= NBAωsin ⁡ t e_ max= NBA e_ rms= NBAω 2 . P_ avg= e_ mss^2R ⇒R= e_ mms^2P_ avg= N^2B^2A^2 ^22 P_ avg = 40^2 0.05^2 0.01^2 50^22 25 10^-3=20

$A circular coil of area 0.01 m^{2} and 40 turns is rotated about its vertical diameter with an angular speed of 50 rad s^{- 1} in a uniform horizontal magnetic field 0.05 T . If the average power loss due to joule heating is 25 mW, then the closed loop resistance of the coil is$

Solution

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