Question

The magnetic field at the center of a current carrying loop of radius $0.1\, m$ 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)

0.2 m

Answer 2

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