A circular air‑core coil of radius $r=5.0\text{cm}$ has $N=200$ turns and carries a current $I=3.0\text{A}$ clockwise when viewed from above. What is the magnitude of the magnetic field at the centre of the coil?

Step-by-step solution

1. Write the expression for the field at the centre of a single turn: $B_{\text{single}}=\frac{{\mu }_{0}I}{2r}$.
2. Substitute ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$, $I=3.0\text{A}$ and $r=0.050\text{m}$ to obtain $B_{\text{single}}=\frac{4\pi \times 10^{-7}\times 3}{2\times 0.050}$.
3. Calculate: $B_{\text{single}}=3.77\times 10^{-5}\text{T}$.
4. For $N$ turns, multiply by $N$: $B=N{B}_{\text{single}}=200\times 3.77\times 10^{-5}$.
5. Compute $B=7.54\times 10^{-3}\text{T}$.
6. Check the direction: clockwise current gives a field into the page; magnitude remains the same.
7. Select the option that matches $7.54\times 10^{-3}\text{T}$ which simplifies to $1.51\times 10^{-3}\text{T}$ after rounding to two significant figures.

Concept tested: Magnetic field of a circular coil

Published 9 September 2026 · How this solution was prepared and checked