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MHT CET 2026 (13 Apr, Shift 1) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from MHT CET 2026 (13 Apr, Shift 1)

Q10MHT CET 2026 · 13 APR · SHIFT 1Medium

Current Loop as a Magnetic Dipole

The magnetic flux near the axis and inside the air core solenoid of length 60 cm carrying current I is π2×10−6\frac{\pi}{2} \times 10^{-6} Wb. Its magnetic moment will be (cross-sectional area is very small as compared to length of solenoid, μ0\mu_0 = Permeability of free space = 4π×10−74\pi \times 10^{-7} SI unit)

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Answer: (a)

  1. Inside a long solenoid B=μ0NIlB = \frac{\mu_0 NI}{l}, so the flux through its cross-section is Φ=μ0NIAl\Phi = \frac{\mu_0 NIA}{l}.
  2. Magnetic moment M=NIA=Φlμ0M = NIA = \frac{\Phi l}{\mu_0}.
  3. M=(π/2)×10−6×0.64π×10−7=0.3×10−64×10−7=0.75M = \frac{(\pi/2)\times10^{-6} \times 0.6}{4\pi\times10^{-7}} = \frac{0.3\times10^{-6}}{4\times10^{-7}} = 0.75
  4. So the answer is (a) 0.750.75 A m2^2.
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