Question

If the intensity of magnetic field at a point on the axis of a current carrying coil of radius $R$ is half of that at the centre of the coil, then the distance of that point from the centre of the coil will be

Answer 1

The magnetic field at the centre of the circular loop is,

$B_O=\frac{{\mu }_{0}I}{2R}....\left(1\right)$

The magnetic field on the axis of the circular current carrying coil is,

$B_P=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}$

According to the given information,

$B_P=\frac{B_O}{2}$

$\Rightarrow \frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{\frac{3}{2}}}=\frac{{\mu }_{0}I}{4R}$

$\Rightarrow 2R^3=(R^2+x^2{)}^{\frac{3}{2}}$

$\Rightarrow R^2\times 2^{\left(\frac{2}{3}\right)}=R^2+x^2$

$\Rightarrow x^2=R^2(2^{\left(\frac{2}{3}\right)}-1)$

$x=R\sqrt{4^{\left(\frac{1}{3}\right)}-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question ? To understand the concept of magnetic field on the axis of circular current carrying coil.