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Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 24 January, Evening Shift — Question 64

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 \qquad .[0pt]

[Take mass of electron me=9×10−31 kgm_{e}=9 \times 10^{-31} \mathrm{~kg}, charge of electron ∣qe∣=1.6×10−19C\left|\mathrm{q}_{\mathrm{e}}\right|=1.6 \times 10^{-19} \mathrm{C},

μ0=4π×10−7 N A2,1 ns=10−9 s]\left.\mu_{0}=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^{2}}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right]

Answer: 250

Numerical answer — enter this value.

Step-by-step solution

Since time period of a revolving charge is 2πmqB\frac{2 \pi m}{\mathrm{qB}}

Where B=\mathrm{B}= magnetic field

due to a solenoid =μ0nI=\mu_{0} n I

∴T=2π mq(μ0nI)\therefore \mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{q}\left(\mu_{0} \mathrm{nI}\right)}

75×10−9=(2π)(9×10−31)1.6×10−19×4π×10−7×n×1.575 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times n \times 1.5}

N=250\mathrm{N}=250

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Motion of Charged Particles in magnetic Fields

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