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A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the angle between the magnetic field and the plane of the coil?
3 years ago

Kevin Nash
332 Points
Sol. n = 50, r = 0.02 m

A = π × (0.02)^2, B = 0.02 T
i = 5 A, μ = niA = 50 × 5 × π × 4 × 10^–4
τ is max. when θ = 90°
τ = μ × B = μB Sin 90° = μB = 50× 5 × 3.14 × 4 × 10^–4 × 2 × 10^–1 = 6.28 × 10^–2 N-M
Given τ = (1/2) τ base max
⇒ Sin θ = (1/2)
or, θ = 30° = Angle between area vector & magnetic field.
⇒ Angle between magnetic field and the plane of the coil = 90° – 30° = 60°
3 years ago
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