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JEE Main 2025 (24 Jan, Shift 2) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2025 (24 Jan, Shift 2)

Q18JEE Main 2025 · 24 JAN · SHIFT 2Medium

Ampere's Law, Cylinders, Solenoid & Toroid

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is _________.

[Take mass of electronme=9×10−31 kgNA2,1 ns=10−9 s]\mathrm{m_e} = 9 \times 10^{-31}\text{ kg} \quad \frac{\mathrm{N}}{\mathrm{A}^2}, 1\text{ ns} = 10^{-9}\text{ s}], charge of electron ∣qe∣=1.6×10−19 C,μ0=4π×10−7|\mathrm{q_e}| = 1.6 \times 10^{-19}\text{ C}, \mu_0 = 4\pi \times 10^{-7}

Numerical answer type. Work it out, then open the answer.

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Answer: 250

  1. Period of circular motion T=2πmeB⇒B=2πmeTT = \frac{2\pi m}{eB} \Rightarrow B = \frac{2\pi m}{eT}.
  2. B=2π×9×10−311.6×10−19×75×10−9=1.5π×10−4B = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 75 \times 10^{-9}} = 1.5\pi \times 10^{-4} T
  3. B=μ0nI⇒n=1.5π×10−44π×10−7×1.5B = \mu_0 n I \Rightarrow n = \frac{1.5\pi \times 10^{-4}}{4\pi \times 10^{-7} \times 1.5}
  4. n=250n = 250 turns per metre.
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