A 250 -Turn rectangular coil of area 2.1 cm × 1.25 cm carries a current of 85 μA and is subjected to a magnetic field of 0.85 T . Work done for rotati - Physics MCQ NEET (UG) 2017 2026 | ExamTest.live

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A 250 -Turn rectangular coil of area 2.1 cm \times 1.25 cm carries a current of 85 μ A and is subjected to a magnetic field of 0.85 T . Work done for rotating the coil by 18 0^{\circ} from its stable equilibrium position is:

Solution & Step-by-step Explanation

Work done W = Δ U = - MB (cos θ_{2} - cos θ_{1}) .Stable equilibrium θ_{1} = 0^{\circ} .Final position θ_{2} = 18 0^{\circ} . W = - MB (cos 18 0^{\circ} - cos 0^{\circ}) = - MB (- 1 - 1) = 2 MB . M = N I A = 250 \times (85 \times 1 0^{- 6}) \times (2.1 \times 1.25 \times 1 0^{- 4}) $ W = 2 \times [250 \times 85 \times 1 0^{- 6} \times 2.625 \times 1 0^{- 4}] \times 0.85 \approx 9.1 \times 1 0^{- 6} J = 9.1 μ J $

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A 250 -Turn rectangular coil of area 2.1 cm \times 1.25 cm carries a current of 85 μ A and is subjected to a magnetic field of 0.85 T . Work done for rotating the coil by 18 0^{\circ} from its stable equilibrium position is:

4.55 μ J

2.3 μ J

1.15 μ J

9.1 μ J

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