A coil of 50 turns is situated in a magnetic field b = 0.25weber/m 2 as shown in figure. A current of 2A is flowing in the coil. Torque acting on the coil will be
Text Solution
Verified by ExpertsThe correct answer is:
B
Since plane of the coil is parallel to magnetic field. So θ θ = 90 o
Hence τ τ = NBiA sin 90 o = NBiA
= 50 × × 0.25 × × 2 × × (12 × × 10 –2 × × 10 × × 10 –2 ) = 0.3 N.
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