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JEE Main 2021 (17 Mar, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2021 (17 Mar, Shift 1)

Q40JEE Main 2021 · 17 MAR · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is: (permeability of free space = 4π×\pi \times 10−-7 H/m)

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Answer: (a)

  1. With a core, B=μ0μrnIB = \mu_0 \mu_r n I.
  2. B=4π×10−7×500×1000×5=4π×10−7×2.5×106B = 4\pi \times 10^{-7} \times 500 \times 1000 \times 5 = 4\pi \times 10^{-7} \times 2.5 \times 10^{6}.
  3. B=πB = \pi T.
  4. So the answer is (a) π\pi T.
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