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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2025 · 29 January · Evening Shift · Physics · Magnetic Effects of Current

The magnetic field inside a 200 turns solenoid of radius 10 cm is...

The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9×10−4Tesla2.9 \times 10^{-4} \mathrm{Tesla}. If the solenoid carries a current of 0.29 A , then the length of the solenoid is ________ πcm\pi \mathrm{cm}.

Correct answer

0

Step-by-step solution

Assuming long solenoid B=μ0( Nℓ)iℓ=μ0Ni B=(4π×10−7)(200)(0.29)2.9×10−4 m=8π cm\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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