A rectangular coil of wire of 100 turns and 10 x 15cmsize carrying a current of 2Amp. is in a
magnetic field of induction 2 x 10 wb/m. If the normal drawn to the plane of the coil makes an
angle 30° with the field, then the torque on the coil is
1) 3x10-N-
m 2 ) 3x10-N-
m 3 ) 313 x 10-N-m 4) 313x10-N-m
Answers
Answered by
14
*I think, it should be like this*
Given: N = 100 turns , Area of rectangular coil = l x b = 10 x 15 = 150cm²
I = 2A
B = 2 x 10T
Ф = 30°
Solution: Torque = NIBAsinФ
= 100 x 2 x 10 x 150 x sin 30°
= 300000 x 1/2
= 150000Nm
Answered by
5
Answer:
3×10^-5
Explanation:
Torque=NIAB sin theta
=100×2×(10×15×10^-6)×2×10^-3×1/2
=2×150×10^-7
=300×10^-7N/m
=3×10^-5N/m
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