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If a long solenoid of 600 turns /m and 10^–4m² carries current of 10A. It is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30⁰ with the direction of applied field ?

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$6.5 \times 10^{- 2} J$

$4.5 \times 10^{- 2} J$

$4.8 \times 10^{- 2} J$

$7.5 \times 10^{- 2} J$

answer is D.

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Detailed Solution

$τ = niAB \sin θ$

$(\begin{array}{l} ∵ n = \frac{N}{L} = 600 \\ A = 10^{- 4}, i = 10 A, B = 0 .25 T \end{array})$

$τ = 600 \times 10^{- 4} \times 0.25 \times 10 \times \frac{1}{2}$

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