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Question:

A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10⁻⁸ m² carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10⁻⁶ T, making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be?

3 × 10⁻⁷ N-m

1.5 × 10⁻⁷ N-m

1.5 × 10⁻⁶ N-m

3 × 10⁻⁶ N-m

Solution:

Given :N=2000turnsA=1.5 × 10⁻⁸ m²I=2.0AB=5 × 10⁻⁶ Tθ=30°Torque τ=NI(A × B)=NIABsinθ∴τ=(2000)(2.0)(1.5 × 10⁻⁸)(5 × 10⁻⁶)(0.5)(∵sin30°=0.5)⇒τ=1.5 × 10⁻⁶ N-m