JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 25)
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 $\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}$, charge of electron $\left|\mathrm{q}_{\mathrm{e}}\right|=1.6 \times 10^{-19} \mathrm{C}$, $$ \left.\mu_0=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^2}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right] $$
Explanation
Since time period of a revolving charge is $\frac{2 \pi \mathrm{~m}}{\mathrm{qB}}$
Where $\mathrm{B}=$ magnetic field
due to a solenoid $=\mu_0 \mathrm{nI}$
$$\begin{aligned} & \therefore \mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{q}\left(\mu_0 \mathrm{nI}\right)} \\ & 75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times \mathrm{n} \times 1.5} \\ & \mathrm{~N}=250 \end{aligned}$$
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