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JEE Main 2023 (31 Jan, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2023 (31 Jan, Shift 1)

Q29JEE Main 2023 · 31 JAN · SHIFT 1Medium

Ampere's Law, Cylinders, Solenoid & Toroid

A rod with circular cross-section area 2 cm22\text{ cm}^2 and length 40 cm40\text{ cm} is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A0.4\text{ A} flows in the wire windings, the total magnetic flux produced inside windings is 4π×10−6 Wb4\pi \times 10^{-6}\text{ Wb}. The relative permeability of the rod is

(Given: Permeability of vacuum μ0=4π×10−7 NA−2\mu_0 = 4\pi \times 10^{-7}\text{ NA}^{-2})

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Answer: (a)

  1. n=4000.4=1000n = \frac{400}{0.4} = 1000 turns/m, so μ0nI=4π×10−7×1000×0.4=1.6π×10−4\mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 0.4 = 1.6\pi \times 10^{-4} T.
  2. Total flux through all windings: Φ=NBA⇒B=4π×10−6400×2×10−4=5π×10−5\Phi = N B A \Rightarrow B = \frac{4\pi \times 10^{-6}}{400 \times 2 \times 10^{-4}} = 5\pi \times 10^{-5} T.
  3. μr=Bμ0nI=5π×10−51.6π×10−4=516\mu_r = \frac{B}{\mu_0 n I} = \frac{5\pi \times 10^{-5}}{1.6\pi \times 10^{-4}} = \frac{5}{16}
  4. So the answer is (a) 516\frac{5}{16}.
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