A solenoid of length $L=0.50\text{m}$ has $N=500$ tightly wound turns and carries a current $I=2.0\text{A}$. Find the magnitude of the magnetic field at any point inside the solenoid.

Step-by-step solution

1. Calculate the turn density $n=\frac{N}{L}$ (turns per metre).
2. Substitute $n$ into the solenoid field formula $B={\mu }_{0}nI$, where ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$.
3. Compute $n=\frac{500}{0.50}=1.0\times 10^3\text{turns/m}$.
4. Insert the numbers: $B=(4\pi \times 10^{-7})(1.0\times 10^3)(2.0)\text{T}$.
5. Evaluate $B=8\pi \times 10^{-4}\text{T}=2.51\times 10^{-3}\text{T}$.

Concept tested: Magnetic field inside a solenoid

Published 11 September 2026 · How this solution was prepared and checked