Question

A square wire loop 10 cm on each side carries a clockwise current of 8 A . Find the magnitude and direction of the magnetic field at its centre due to the four 1.2 mm wire segments at the midpoint of each side.

Short answer

The total magnetic field is$1.54\times {10}^{-6}T$ .

Step-by-step solution

Formula to calculate the magnetic field

The magnetic field due to a segment of length and carrying a current of is given by Biot and Savart law

$dB=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{ldlsin\varphi }{r^2}$

Where the angle$\mathit{\varphi }$ is the angle between the position unit vector and the length of the line segment.

Calculate the magnetic field

Since we want to find the magnetic field at the center, then$\varphi =90°$ and the distance is half the length of the side, that is, r = 5 cm . The magnetic field due to the four segments is into the page and will add.

The magnetic field can be calculated as:

$\begin{array}{r}d{B}_{\text{tot.}}=\frac{4{\mu }_0}{4\pi }\frac{Idl\sin \varphi }{r^2}\\ B_{\text{tot.}}=\frac{4\pi \times {10}^{-7}}{\pi }\times \frac{1.2\times {10}^{-3}\times 8\times \sin {90}^{\circ }}{{\left(5\times {10}^{-2}\right)}^2}\\ B_{\text{tot.}}=1.54\times {10}^{-6}\mathrm{T}\end{array}$

So, the total magnetic field is$1.54\times {10}^{-6}T$ into the page.