Question

The magnetic field at the center 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

• A. 0.2 m
• B. 0.1 m
• C. 0.05 m
• D. 0.25 m

Answer 1

Answer: (A)

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