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MHT CET 2024 (3 May, Shift 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2024 (3 May, Shift 1)

Q26MHT CET 2024 · 3 MAY · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A current of 5 A flows through a toroid having a core of mean radius 20 cm. If 4000 turns of the conducting wire are wound on the core, then the magnetic field inside the core of the toroid is [permeability of free space =4π×10−7= 4\pi \times 10^{-7} SI units]

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Answer: (d)

  1. B=μ0NI2πrB = \dfrac{\mu_0 N I}{2\pi r}.
  2. B=4π×10−7×4000×52π×0.2=2×10−7×2×1040.2B = \dfrac{4\pi \times 10^{-7} \times 4000 \times 5}{2\pi \times 0.2} = \dfrac{2 \times 10^{-7} \times 2 \times 10^{4}}{0.2}.
  3. B=2×10−2B = 2 \times 10^{-2} Wb/m2^2.
  4. So the answer is (d).
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