QJEEM25Ph4ac6: JEE Main 2025 Physics - Class 12 JEEM25Ph4ac6 Solution: The magnetic field inside a 200 turns solenoid of radius 10 cm…

Question

Verified Solution • 2025–26 Session moderate

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}

.

Accepted Answer

We know, magnetic field due to solenoid, $B = μ_{0} n I$

where, I = current, $n = \frac{N}{L}$ = no. of turns per unit length

$\Rightarrow B = \frac{μ_{0} N I}{L}$

$\Rightarrow L = \frac{μ_{0} N I}{B} = \frac{4 π \times 10^{- 7} \times 200 \times 0.29}{2.9 \times 10^{- 4}}$

$\Rightarrow L = 8 π \times 10^{- 2} m$

$\Rightarrow L = 8 π c m$

Hence, answer = 8.

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