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Question

A circular coil of diameter 10 cm has 10 turns and carries a current of 5A. It is placed in a uniform field of 0.5 T with its plane parallel to the field. The torque on the coil in Nm, is

A
(6.25π)×102 N-m
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B
(6.25π)×103 N-m
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C
(6.25π)×104 N-m
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D
Zero
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Solution

The correct option is A (6.25π)×102 N-m

θ=90;sin90=1
T =Bi An
=0.5×5×πr2×10=25×π(5×102)2=625π×104=6.25π×102Nm.

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