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JEE Main 2020 (4 Sep, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2020 (4 Sep, Shift 2)

Q67JEE Main 2020 · 4 SEP · SHIFT 2Difficult

Current Loop as a Magnetic Dipole

A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying current to produce a magnetic moment of 20 Am2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical,it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be:

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Answer: (c)

  1. Coil plane vertical means m⃗\vec m is horizontal, so initially θ=90∘\theta = 90^\circ with the vertical B⃗\vec B; after turning 60∘60^\circ, θ=30∘\theta = 30^\circ.
  2. Loss in PE =mB(cos⁡30∘−cos⁡90∘)=20×4×32=403= mB(\cos30^\circ - \cos90^\circ) = 20 \times 4 \times \frac{\sqrt3}{2} = 40\sqrt3 J.
  3. 12Iω2=403⇒ω2=8030.8=1003\frac12 I\omega^2 = 40\sqrt3 \Rightarrow \omega^2 = \frac{80\sqrt3}{0.8} = 100\sqrt3
  4. ω=10(3)1/4\omega = 10(3)^{1/4} rad/s
  5. So the answer is (c).
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