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Question

The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of 1.5×102 T. A current of 10.0 A flows through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the magnetic field, then the torque acting on the loop is

A
6000 Nm
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B
Zero
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C
1.2×102 Nm
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D
6×104 Nm
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Solution

The correct option is D 6×104 Nm

Torque on a current carrying loop,
τ=M×B
Magnitude of torque τ=MBsinθ
Magnetic moment of a current carrying loop,
M=NIA where N is the number of turns of the loop, I is the current flowing in the loop and A is the area.
M=1×10×8×102×5×102 =4×102 Am2

τ=MBsinθ
Since the area vector makes an angle of 90 with the magnetic field, θ=90
τ=M×B=4×102×1.5×102=6×104 Nm

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