A square loop of edge a constructed from a uniform wire is shown in...

Let the resistance of each side of the square loop be $r$.

The current $i$ enters at $A and exits at$B$. It divides into two parallel paths: the direct path$AB$and the path$ADCB$.<br><br>The resistance of path$AB$is$r$, and the resistance of path$ADCB$is$3r$.<br><br>The current$i_1$through$AB$is$i_1 = i \left( \dfrac{3r}{r + 3r} \right) = \dfrac{3i}{4}$.<br><br>The current$i_2$through$ADCB$is$i_2 = i \left( \dfrac{r}{r + 3r} \right) = \dfrac{i}{4}$.<br><br>The magnetic field at the centre of the square due to a single wire of length$a$carrying current$I$is$B = \dfrac{\mu_0 I}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \dfrac{\sqrt{2} \mu_0 I}{\pi a}$.<br><br>The magnetic field at the centre due to the segment$AB$is$B_1 = \dfrac{\sqrt{2} \mu_0}{\pi a} \ \left(\$\dfrac{3i}{4}$\right)$. According to the right-hand thumb rule, this field is directed outwards from the plane of the loop.<br><br>The segments$AD$,$DC$, and$CB$each carry current$i_2 = \$\dfrac{i}{4} in the same sense, so their magnetic fields add up and are directed inwards.<br><br>The total magnetic field due to the path$ADCB$is$B_2 = 3 \times \$\dfrac{\sqrt{2} \mu_0}{\pi a}$\ \left(\dfrac{i}{4}\right) = \dfrac{\sqrt{2} \mu_0}{\pi a} \left(\dfrac{3i}{4}\right)$.<br><br>Since$B_1$and$B_2$ are equal in magnitude and opposite in direction, the net magnetic field at the centre of the loop is zero.