JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Easy
- Numerical
The magnetic field inside a 200 turns solenoid of radius 10 cm is \(2.9 \times 10^{-4} \mathrm{Tesla}\). If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ \(\pi \mathrm{cm}\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 8
Step-by-step Solution
Detailed explanation
Assuming long solenoid
\(\begin{aligned}
& \mathrm{B}=\mu_0\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \\
& \ell=\frac{\mu_0 \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \\
& =8 \pi \mathrm{~cm}
\end{aligned}\)
\(\begin{aligned}
& \mathrm{B}=\mu_0\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \\
& \ell=\frac{\mu_0 \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \\
& =8 \pi \mathrm{~cm}
\end{aligned}\)
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