The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10 -4 Tesla. If the solenoid carries a current of 0.29 A , then the length of the solenoid is π cm .

Physics · Electromagnetic Induction

JEE Main 2025 — 29 January, Evening Shift — Question 37

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

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

Assuming long solenoid $\mathrm{B}=\mu_{0}\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i}$

$\ell=\frac{\mu_{0} \mathrm{Ni}}{\mathrm{B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m}$

$=8 \pi \mathrm{~cm}$

Answer key and solution verified before publishing.

Practise Electromagnetic Induction

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density