A straight solenoid of length 50 cm has 1000 turns per metre and mean cross-sectional area
of 2 x 104 m. it is placed with its axis at 30° with uniform magnetic field of 0.32 T. Find the
torque acting on the solenoid when a current of 2 ampere is passed through it.
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Answer:
Explanation:
τ = BinA sin θ
= 0.32 *2*1000*2 x 104 *sin 30
= 64*10^5 Nm
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