A solenoid has $400$ turns and a length of $0.40\text{m}$. It carries a current of $2\text{A}$. What is the magnetic field inside the solenoid? (Take ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$.)

Step-by-step solution

1. Calculate the number of turns per unit length: $n=\frac{N}{L}=\frac{400}{0.40}=1000\text{turns/m}$.
2. Use the formula for magnetic field inside a solenoid: $B={\mu }_{0}nI$.
3. Substitute the values: $B=4\pi \times 10^{-7}\times 1000\times 2$.
4. Simplify: $B=8\pi \times 10^{-4}\text{T}=2.5\times 10^{-3}\text{T}$ (since $\pi \approx 3.14$).
5. Select the matching option.

Concept tested: Magnetic field inside a solenoid

Published 23 August 2026 · How this solution was prepared and checked

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