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A closely wound solenoid of 2000 turns and area of cross-section 1.5×10−4 m21.5 \times 10^{-4} \text{ m}^2 carries a current of 2.0 A2.0 \text{ 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 5×10−2 T5 \times 10^{-2} \text{ T} making an angle of 30∘30^{\circ} with the axis of the solenoid. The torque on the solenoid will be

A

3×10−3 N m3 \times 10^{-3} \text{ N m}

B

1.5×10−3 N m1.5 \times 10^{-3} \text{ N m}

C

1.5×10−2 N m1.5 \times 10^{-2} \text{ N m}

D

3×10−2 N m3 \times 10^{-2} \text{ N m}

Step-by-Step Solution

  1. Magnetic Moment: The solenoid acts like a magnetic dipole. Its magnetic moment (mm) is given by the product of the number of turns (NN), the current (II), and the cross-sectional area (AA): m=NIAm = NIA . m=2000×2.0×1.5×10−4=0.6 A m2m = 2000 \times 2.0 \times 1.5 \times 10^{-4} = 0.6 \text{ A m}^2
  2. Torque Calculation: The torque (τ\tau) exerted on a magnetic dipole in a uniform magnetic field (BB) is given by the cross product τ⃗=m⃗×B⃗\vec{\tau} = \vec{m} \times \vec{B}, with magnitude τ=mBsin⁡θ\tau = mB \sin\theta, where θ\theta is the angle between the magnetic moment (axis of the solenoid) and the magnetic field .
  3. Substitution: Given θ=30∘\theta = 30^{\circ} and B=5×10−2 TB = 5 \times 10^{-2} \text{ T}: τ=0.6×(5×10−2)×sin⁡(30∘)\tau = 0.6 \times (5 \times 10^{-2}) \times \sin(30^{\circ}) τ=0.6×0.05×0.5\tau = 0.6 \times 0.05 \times 0.5 τ=1.5×10−2 N m\tau = 1.5 \times 10^{-2} \text{ N m}
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