Q 12-04-064JEE MainJEE Main 2025 (29 Jan, Shift 2)Easy
The magnetic field inside a 200-turn solenoid of radius $10\ \text{cm}$ is $2.9\times10^{-4}\ \text{T}$. If the solenoid carries a current of $0.29\ \text{A}$, then the length of the solenoid is ______ $\pi$ cm.
Numerical value type. Enter your answer.
Answer: 8
$B = \dfrac{\mu_0NI}{L} \Rightarrow L = \dfrac{\mu_0NI}{B}$:
$$L = \frac{4\pi\times10^{-7}\times200\times0.29}{2.9\times10^{-4}} = 4\pi\times10^{-7}\times2\times10^{5} = 0.08\pi\ \text{m} = 8\pi\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics