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Question

A galvanometer coil 5 cm× 2 cm with 200 turns is suspended vertically in a field of 5×102T.The suspension fibre needs a torque of 0.125×107Nm to twist it through one radian. If i is the strength of the current required to be maintained in the coil when we require a deflection of 6o then find x such that x=i×108.

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Solution

i=(kNBA)ϕ

=(0.125×107)(6)(π/180)200×5×102×5×2×104

=1.3×107A

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