A solenoid of 0.4 m length with 500 turns carries a current of 3 A. A coil of 10 turns and of radius 0.01 m carries a current of 0.4 A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is

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$6 π^{2} \times 10^{- 7} N m$

$12 π^{2} \times 10^{- 7} N m$

$9 π^{2} \times 10^{- 7} N m$

$3 π^{2} \times 10^{- 7} N m$

answer is A.

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Detailed Solution

$\begin{array}{l} Torque, τ = MBsin 90^{0} = MB \times 1 = MB \\ where, M = i_{2} AN and B = μ_{0} \frac{N_{1}}{l} i_{1} \\ τ = (i_{2} AN) . (μ_{0} \frac{N_{1}}{l} i_{1}) \\ = [0.4 \times π \times {(0.01)}^{2} \times 10] . [4 π \times 10^{- 7} \times \frac{500}{0.4} \times 3] \\ = 6 π^{2} \times 10^{- 7} Nm \end{array}$