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

Answer: 20

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
$25\times 5={\left(1+\frac{x^2}{{\left(0.1\right)}^2}\right)}^3$
$\sqrt[3]{125}=1+\frac{x^2}{{\left(0.1\right)}^2}$
$\Rightarrow 0.01+x^2=0.05$
$\Rightarrow x^2=0.05-0.01$
$\Rightarrow x^2=0.04$
$\Rightarrow x=0.2m=20cm$