A long solenoid has 800 turns per meter and carries a current of $I=1.5\text{A}$. What is the magnetic field inside the solenoid?

Step-by-step solution

1. The magnetic field inside an ideal long solenoid is given by $B={\mu }_{0}nI$, where $n$ is the number of turns per unit length and ${\mu }_0=4\pi \times 10^{-7}{\text{T m A}}^{-1}$.
2. Substitute $n=800{\text{turns m}}^{-1}$ and $I=1.5\text{A}$: $B=(4\pi \times 10^{-7})(800)(1.5)$.
3. Calculate: $4\pi \times 10^{-7}\times 800=1.005\times 10^{-3}$, then multiply by $1.5$ to get $1.5075\times 10^{-3}\text{T}$? Wait, re‑evaluate: $4\pi \times 10^{-7}=1.2566\times 10^{-6}$; $1.2566\times 10^{-6}\times 800=1.0053\times 10^{-3}$; $1.0053\times 10^{-3}\times 1.5=1.508\times 10^{-3}\text{T}$, i.e., $1.51\times 10^{-3}\text{T}$.
4. Thus $B\approx 1.51\times 10^{-3}\text{T}$, i.e., $1.51\text{m}T$.
5. Since the answer format expects tesla, write $1.51\times 10^{-3}\text{T}$.
6. (If the exam expects $1.51\text{T}$, the given data would be different; here we keep the computed value.)

Concept tested: Magnetic field inside a solenoid

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