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Torque on the solenoid is given by τ = MB sin θ, where θ is the angle between the magnetic field and the axis of solenoid.
- M = niA
- τ = niA B sin 30°
- τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × 1/2
- τ = 1.5 × 10⁻² N-m
Correct answer: τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × 1/2
Solution
The torque on a solenoid is given by τ = MB sin θ, where M = nIA. Substituting the given values (n = 2000, I = 2 A, A = 1.5 × 10⁻⁴ m², B = 5 × 10⁻² T, and sin 30° = 1/2), we get τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × 1/2.
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