Question

A square loop of side $a=6 \,cm$ carries a current $I=1 \,A$. Calculate magnetic induction $B$ (in $\mu T$ ) at point $P$, lying on the axis of loop and at a distance $x=\sqrt{7}\, cm$ from the center of loop.

Answer 1

Axis of the loop means a line passing through center of loop and normal to its plane. Since distance of the point $P$ is $x$ from center of loop and side of square loop is $a$ as shown in Fig. (a).

Therefore, perpendicular distance of $P$ from each side of
$r=\sqrt{x^{2}+\left(\frac{a}{2}\right)^{2}}=4 \,cm$
Now, consider only one side $A B$ of the loop as shown in Fig. (b).
$\tan \alpha=\tan \beta=\frac{(a / 2)}{r}=\frac{3}{4} $
$\alpha=\beta=\tan ^{-1}\left(\frac{3}{4}\right)=37^{\circ}$
Magnitude of magnetic induction at $P$, due to current in this side $A B$, is
$B_{0=} \frac{\mu_{0} I}{4 \pi r}(\sin \alpha+\sin \beta)=3 \times 10^{-6} T$
Now, consider magnetic inductions, produced by currents in two opposite sides $A B$ and $C D$ as shown in Fig. (c).

Components of these magnetic inductions, parallel to plane of loop neutralise each other. Hence, resultant of these two magnetic inductions is $2 B_{0} \cos \theta$ (along the axis).
Similarly, resultant of magnetic inductions produced by currents in remaining two opposite sides $B C$ and $A D$ will also be equal to $2 B_{0} \cos \theta$ (along the axis in same direction).
Hence, resultant magnetic induction,
$B=4 B_{0} \cos \theta$
$B=4 \times\left(3 \times 10^{-6}\right) \frac{(a / 2)}{r}$
$=9 \times 10^{-6} T =9\, \mu T$