Question

The magnetic induction at the centre of a current carrying circular coil of radius $10cm$ is $5\sqrt{5}$ times the magnetic induction at a point on its axis. The distance of the point from the centre of the coil in cm is

• A. $5$
• B. $10$
• C. $20$
• D. $25$

Answer 1

The correct option is C $20$

The magnetic field at the center of a current carrying circular loop of radius R is: $B_{center}=\frac{{\mu }_{0}i}{2R}$
The magnetic field on the axis at a distance x from the center of a current carrying circular loop of radius R is: $B_{axis}=\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

Given $B_{center}=5\sqrt{5}{B}_{axis}$

$\therefore \frac{{\mu }_{0}i}{2R}=5\sqrt{5}\frac{{\mu }_{0}i{R}^2}{2(R^2+x^2{)}^{3/2}}$

$\Rightarrow \frac{(R^2+x^2{)}^3}{R^6}=(5\sqrt{5}{)}^2=125$
$\Rightarrow {\left(\frac{R^2+x^2}{R^2}\right)}^3=125$
$\Rightarrow \frac{R^2+x^2}{R^2}=5$
$\Rightarrow 1+\frac{x^2}{R^2}=5$
$\frac{x^2}{R^2}=4$
$x=2R=2\times 10=20cm$ since given $R=10cm$