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NEET 2010 (AIPMT, Mains) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from NEET 2010 (AIPMT, Mains)

Q82NEET 2010 · AIPMT · MAINSMedium

Current Loop as a Magnetic Dipole

A closely wound solenoid of 2000 turns and area of cross-section 1.5 ×\times 10−-4 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 ×\times 10 −-2 tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be

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Answer: (c)

  1. Torque on a coil: τ=NIABsin⁡θ\tau = NIAB\sin\theta, with θ\theta the angle between the axis and B⃗\vec B.
  2. τ=2000×2×1.5×10−4×5×10−2×sin⁡30∘\tau = 2000 \times 2 \times 1.5\times10^{-4} \times 5\times10^{-2} \times \sin 30^\circ
  3. τ=0.6×0.05×0.5=1.5×10−2 N m\tau = 0.6 \times 0.05 \times 0.5 = 1.5\times10^{-2}\text{ N m}
  4. So the answer is (c) 1.5×10−21.5\times10^{-2} N m.
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