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A solenoid of self-inductance 1.2 H is in series with a tangent galvanometer of reduction factor 0.9A. They are connected to a battery and the tangent galvanometer shows a deflection of $53^{0}$ . The energy stored in the magnetic field of the solenoid is
(tan $53^{0}$ = 4/3)

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0.864 J

0.72 J

0.173 J

1.44 J

answer is A.

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Detailed Solution

$i = K \tan θ = 0.9 \times \frac{4}{3} = 1.2 A, U = \frac{1}{2} L i^{2} = \frac{1}{2} \times 1.2 \times {1.2}^{2} = 0.864$

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