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NEET 2020 (Phase 1) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from NEET 2020 (Phase 1)

Q48NEET 2020 · PHASE 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is:

(μ0=4π×10−7 TmA−1)\left(\mu_0 = 4\pi \times 10^{-7}\, TmA^{-1}\right)

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Answer: (d)

  1. n=NL=1000.5=200n = \dfrac{N}{L} = \dfrac{100}{0.5} = 200 turns/m.
  2. B=μ0nI=4π×10−7×200×2.5=2π×10−4B = \mu_0 n I = 4\pi \times 10^{-7} \times 200 \times 2.5 = 2\pi \times 10^{-4} T.
  3. B=6.28×10−4B = 6.28 \times 10^{-4} T.
  4. So the answer is (d).
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