Question

A square loop of side $a=6cm$ carries a current $I=1A$. 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

Answer: 9

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
$r=\sqrt{x^2+{\left(\frac{a}{2}\right)}^2}=4cm$
Now, consider only one side $AB$ 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 $AB$, 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 $AB$ and $CD$ 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 $2B_0\cos \theta$ (along the axis).
Similarly, resultant of magnetic inductions produced by currents in remaining two opposite sides $BC$ and $AD$ will also be equal to $2B_0\cos \theta$ (along the axis in same direction).
Hence, resultant magnetic induction,
$B=4B_0\cos \theta$
$B=4\times \left(3\times 10^{-6}\right)\frac{(a/2)}{r}$
$=9\times 10^{-6}T=9\mu T$