Question

A square loop of side $a=6cm$ 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 centre of loop and normal to its plane.

Let distance of the point $P$ is $x$ from centre of loop and side of square loop is $a$.
Therefore, perpendicular distance of $P$ from each side of the loop is:
$r=\sqrt{x^2+{\left(\frac{a}{2}\right)}^2}=4cm$
$\tan \alpha =\tan \beta =\frac{a/2}{r}=\frac{3}{4}$
$\alpha =\beta ={\tan }^{-1}\frac{3}{4}={37}^o$
Magnitude of magnetic induction at $P$, due to current in this side $AB$, is
$B_0=\frac{{\mu }_0}{4\pi r}(\sin \alpha +\sin \beta )=3\times {10}^{-6}T$

Now, consider magnetic inductions, produced by currents in two opposite sides $AB$ and $CD$. 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,
$B=4B_0\cos \theta$
$B=4\times (3\times {10}^{-6})\frac{\left(a/2\right)}{r}=9\times {10}^{-6}T=9\mu T$