A straight wire carries a current of $4.0\text{A}$ and a point $8.0\text{cm}$ from the wire lies in its perpendicular bisector. What is the magnitude of the magnetic field at that point?

Step-by-step solution

1. Step 1: Identify the formula for the magnetic field due to an infinitely long straight wire: $B=\frac{m{u}_{0}I}{2pir}$, where $I$ is the current and $r$ is the radial distance from the wire.
2. Step 2: Substitute the given values: $I=4.0\text{A}$, $r=8.0\text{cm}=0.080\text{m}$, and ${\mu }_0=4\pi \times 10^{-7}{\text{N,A}}^{-2}$.
3. Step 3: Compute the numerator: ${\mu }_{0}I=4\pi \times 10^{-7}\times 4.0=1.6\pi \times 10^{-6},{\text{N,A}}^{-1}$.
4. Step 4: Compute the denominator: $2\pi r=2\pi \times 0.080=0.160\pi$.
5. Step 5: Divide: $B=\frac{1.6\pi \times 10^{-6}}{0.160\pi }=10^{-5}\times 10;=;1.27\times 10^{-4},\text{T}$.
6. Step 6: State the result with appropriate significant figures: $1.27\times 10^{-4},\text{T}$.

Concept tested: Magnetic field of straight wire

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked