A long solenoid has $1200$ turns per metre and carries a current of $2.5\text{A}$. Find the magnitude of the magnetic field inside the solenoid (in tesla).

Step-by-step solution

1. Use Ampère's circuital law for a long solenoid: $B={\mu }_{0}nI$, where $n$ is the number of turns per unit length.
2. Substitute ${\mu }_0=4\pi \times 10^{-7}{\text{T m A}}^{-1}$, $n=1200{\text{m}}^{-1}$ and $I=2.5\text{A}$.
3. Calculate $B=4\pi \times 10^{-7}\times 1200\times 2.5=4\pi \times 10^{-7}\times 3000=12000\pi \times 10^{-7}\text{T}$.
4. Simplify: $12000\pi \times 10^{-7}=1.2\pi \times 10^{-3}\text{T}$.
5. Evaluate $1.2\pi \approx 3.77$, so $B\approx 3.8\times 10^{-3}\text{T}$.
6. Thus the magnetic field inside the solenoid is $3.8\times 10^{-3}\text{T}$.

Concept tested: Magnetic field inside a solenoid

Published 6 September 2026 · How this solution was prepared and checked