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JEE Main 2024 (6 Apr, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2024 (6 Apr, Shift 2)

Q38JEE Main 2024 · 6 APR · SHIFT 2Medium

Current Loop as a Magnetic Dipole

A coil having 100 turns, area of 5×10−3 m25 \times 10^{-3}\text{ m}^{2}, carrying current of 1 mA1\text{ mA} is placed in uniform magnetic field of 0.20 T0.20\text{ T} such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90∘90^\circ is _________ μJ\mu\text{J}.

Numerical answer type. Work it out, then open the answer.

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Answer: 100

  1. Plane perpendicular to B⃗\vec B means m⃗∥B⃗\vec m \parallel \vec B (θ=0\theta = 0).
  2. W=mB(cos⁡0∘−cos⁡90∘)=NIABW = mB(\cos0^\circ - \cos90^\circ) = NIAB
  3. W=100×10−3×5×10−3×0.2=10−4 JW = 100 \times 10^{-3} \times 5\times10^{-3} \times 0.2 = 10^{-4}\text{ J}
  4. So W=100 μW = 100\ \muJ.
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