JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 25)
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 _______ $\pi \mathrm{cm}$.
Answer
8
Explanation
We know, magnetic field due to solenoid, $$B = {\mu _0}nI$$
where, I = current, $$n = {N \over L}$$ = no. of turns per unit length
$$ \Rightarrow B = {{{\mu _0}NI} \over L}$$
$$ \Rightarrow L = {{{\mu _0}NI} \over B} = {{4\pi \times {{10}^{ - 7}} \times 200 \times 0.29} \over {2.9 \times {{10}^{ - 4}}}}$$
$$ \Rightarrow L = 8\pi \times {10^{ - 2}}\,m$$
$$ \Rightarrow L = 8\pi \,cm$$
Hence, answer = 8.
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