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MHT CET 2023 (14 May, Shift 1) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from MHT CET 2023 (14 May, Shift 1)

Q45MHT CET 2023 · 14 MAY · SHIFT 1Difficult

Current Loop as a Magnetic Dipole

A solenoid of length 0.4 m0.4\text{ m} and having 500 turns of wire carries a current 3 A3\text{ A}. A thin coil having 10 turns of wire and radius 0.1 m0.1\text{ m} carries current 0.4 A0.4\text{ A}. the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid is (μ0=4π×10−7(\mu_0 = 4\pi \times 10^{-7} SI units, π2=10) (sin⁡90∘=1)\pi^{2} = 10)\ (\sin 90^\circ = 1)

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Answer: (c)

  1. Field inside the solenoid: B=μ0NLI=4π×10−7×5000.4×3=1.5π×10−3B = \mu_0\frac{N}{L}I = 4\pi\times10^{-7} \times \frac{500}{0.4} \times 3 = 1.5\pi\times10^{-3} T.
  2. Coil moment: m=NIπr2=10×0.4×π(0.1)2=0.04π A m2m = NI\pi r^2 = 10 \times 0.4 \times \pi(0.1)^2 = 0.04\pi\text{ A m}^2.
  3. Axes perpendicular: τ=mB=0.04π×1.5π×10−3=0.06π2×10−3\tau = mB = 0.04\pi \times 1.5\pi\times10^{-3} = 0.06\pi^2\times10^{-3}
  4. With π2=10\pi^2 = 10: τ=6×10−4\tau = 6\times10^{-4} N m.
  5. So the answer is (c).
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