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Current Loop as a Magnetic Dipole question from NEET 2026

Q12NEET 2026Medium

Current Loop as a Magnetic Dipole

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14×10−3 T3.14 \times 10^{-3}\text{ T} at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively:

(Take μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7}\text{ T m/A})

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Answer: (d)

  1. B=μ0NI2R⇒I=2RBμ0N=2×0.05×3.14×10−34π×10−7×100B = \frac{\mu_0 NI}{2R} \Rightarrow I = \frac{2RB}{\mu_0 N} = \frac{2\times0.05\times3.14\times10^{-3}}{4\pi\times10^{-7}\times100}
  2. I=3.14×10−41.256×10−4=2.5I = \frac{3.14\times10^{-4}}{1.256\times10^{-4}} = 2.5 A
  3. M=NIπR2=100×2.5×3.14×(0.05)2≈1.96≈2M = NI\pi R^2 = 100\times2.5\times3.14\times(0.05)^2 \approx 1.96 \approx 2 A m2^2
  4. So the answer is (d) 2.52.5 A, 22 A m2^2.
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