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Standard XII

Physics

More about Magnetic Dipoles

A circular co...

Question

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of $0.1 w e b e r / m^{2}$ . The amount of work done in rotating it through $180^{\circ}$ from its equilibrium position will be

0.1 J

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0.2 J

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0.5 J

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0.8 J

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Solution

The correct option is C 0.5 J
Work done in rotating a coil through an angle $θ$ from its equilibrium position is $W = M B (1 - c o s θ)$ where $θ = 180^{\circ}$ and $M = 50 \times 2 \times π (4 \times 10^{- 2})^{2} = 50.24 \times 10^{- 2} A - m^{2}$ . Hence W = 0.5 J

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A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of 0.1 weber/m2. The amount of work done in rotating it through 180∘ from its equilibrium position will be

The correct option is C 0.5 JWork done in rotating a coil through an angle θ from its equilibrium position is W=MB(1−cosθ) where θ=180∘ and M=50×2×π(4×10−2)2=50.24×10−2A−m2. Hence W = 0.5 J

A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of $0.1 w e b e r / m^{2}$ . The amount of work done in rotating it through $180^{\circ}$ from its equilibrium position will be

The correct option is C 0.5 J
Work done in rotating a coil through an angle $θ$ from its equilibrium position is $W = M B (1 - c o s θ)$ where $θ = 180^{\circ}$ and $M = 50 \times 2 \times π (4 \times 10^{- 2})^{2} = 50.24 \times 10^{- 2} A - m^{2}$ . Hence W = 0.5 J