Since time period of a revolving charge is qB2πm Where B= magnetic field due to a solenoid =μ0nI $\begin{aligned}
& \therefore \mathrm{T}=\frac{2 \pi \mathrm{m}}{\mathrm{q}\left(\mu_0 \mathrm{nI}\right)} \
& 75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times \mathrm{n} \times 1.5} \
& \mathrm{N}=250
\end{aligned}$
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…
JEE Main 2025 — Physics Electromagnetism
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 electron me=9×10−31 kg, charge of electron ∣qe∣=1.6×10−19C, μ0=4π×10−7 A2 N,1 ns=10−9 s]
Held on 24 Jan 2025 · Verified 6 Jul 2026.
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