Question

Which expression gives the magnitude of the small magnetic field contribution \(dB\) from a wire segment \(dl\), according to the Biot–Savart law?

Options

• \(dB=\frac{\mu_0}{4\pi}\frac{I r^2}{dl}\)

• \(dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}\)

• \(dB=\frac{4\pi}{\mu_0}\frac{I\,dl}{r^2}\)

• \(dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{r}\)

Answer 1

The Biot–Savart law gives \(dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{r^2}\). The contribution is proportional to the current and segment length, and inversely proportional to the square of the distance \(r\).