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Question

A rectangular coil of wire of 500 turns of area 10×5cm2 carries a current of 2A in a magnetic field of induction 2×103T . If the plane of the coil is parallel to the field. The torque on the coil is (inNm):
  1. 0.01
  2. 0.001
  3. 0.1
  4. 1

A
0.01
B
0.1
C
0.001
D
1
Solution
Verified by Toppr

The torque on the rectangular coil due to presence of magnetic field is given,
τ=NIABsinθ
where number of turns N=500,
Current in the coil I=2A,
Area of the coil A=(10×5)104m2,
Magnetic field B=2×103T,
Angle between area and magnetic field vector is θ
As area vector is always normal to plane and given the plane is parallel to field, so the angle between area and field is 90o.
So, τ=500×2×(10×5)×104×(2×103)sin90=0.01Nm

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