Question

The magnetic field at the centre of a current carrying loop of radius $0.1m$ is $5\sqrt{5}$ times that at a point along its axis. The distance of this point from the centre of the loop is(in $cm$ )

Answer 1

We know that,
$\frac{B_{\text{centre}}}{B_{\text{axis}}}=\left(1 + \frac{x^{2}}{r^{2}}\right)^{\frac{3}{2}}$
Given that, $B_{\text{centre}}=5\sqrt{5} \, B_{\text{axis}}$
$\frac{B_{\text{centre}}}{B_{\text{axis}}}=5\sqrt{5}$
$\therefore \, \, 5\sqrt{5}=\left(1 + \frac{x^{2}}{\left(0.1\right)^{2}}\right)^{3 / 2}$
On squaring both sides, we get
$25\times 5=\left(1 + \frac{x^{2}}{\left(0.1\right)^{2}}\right)^{3}$
$\sqrt[3]{125}=1+\frac{x^{2}}{\left(0.1\right)^{2}}$
$\Rightarrow \, 0.01+x^{2}=0.05$
$\Rightarrow \, \, x^{2}=0.05-0.01$
$\Rightarrow \, \, x^{2}=0.04$
$\Rightarrow \, \, x=0.2\,m=20\,cm$