Question

A closely wound, circular coil with radius 2.40 cm has 800 turns. The distance x from the centre of the coil, along the axis, at which the magnetic field is half of its value at the centre, is $184\times {10}^{-x}m$. Find the value of x :

Answer 1

Magnetic field along the axis of a circular coil at a distance $x$ from its center is

$B=\frac{{\mu }_{0}Ni{R}^2}{2(R^2+x^2{)}^{3/2}}$

$⟹\frac{B^{\prime }}{B}=\frac{1}{2}=\frac{(R^2+x^2{)}^{3/2}}{R^3}$

$⟹x=184\times {10}^{-4}m$

$⟹x=4$