An air‑core circular coil of radius $0.10\text{m}$ carries a current $I=20\text{A}$ in the clockwise direction 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 formula for the magnetic field at the centre of a circular loop: $B=\frac{{\mu }_{0}I}{2R}$.
2. Substitute ${\mu }_0=4\pi \times 10^{-7}{\text{T m A}}^{-1}$, $I=20\text{A}$ and $R=0.10\text{m}$.
3. Calculate $B=\frac{4\pi \times 10^{-7}\times 20}{2\times 0.10}=\frac{80\pi \times 10^{-7}}{0.20}=4\pi \times 10^{-5}\text{T}$.
4. Evaluate $4\pi \times 10^{-5}\approx 1.26\times 10^{-4}\text{T}$, which is closest to $1.0\times 10^{-4}\text{T}$.
5. Thus the correct option is the first one.

Concept tested: Magnetic field of a current loop

Published 5 September 2026 · How this solution was prepared and checked