JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Easy
- Numerical
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 \(\left|\mathrm{q}_{\mathrm{e}}\right|=1.6 \times 10^{-19} \mathrm{C}\),
\(\left.\mu_0=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^2}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 250
Step-by-step Solution
Detailed explanation
Since time period of a revolving charge is \(\frac{2 \pi m}{q B}\)
Where \(\mathrm{B}=\) magnetic field
due to a solenoid \(=\mu_0 \mathrm{nI}\)
\(\begin{aligned}
& \therefore \mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{q}\left(\mu_0 \mathrm{nI}\right)} \\
& 75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times \mathrm{n} \times 1.5} \\
& \mathrm{~N}=250
\end{aligned}\)
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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