Question

The distance at which the magnetic field on axis as compared to the magnetic field at the center of the coil carrying current i and radius r is 1/8 would be

Answer 1

Explanation:

The magnetic field at a distance x from a circular coil of radius r is :

$B=\dfrac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}}$

Magnetic field at the center of the coil is :

$B'=\dfrac{\mu_0I}{2r}$

According to question :

$B=\dfrac{1}{8}B'$

$\dfrac{\mu_0Ir^2}{2(r^2+x^2)^{3/2}}=\dfrac{1}{8}\times \dfrac{\mu_0I}{2r}$

$63r^2=x^2$

$x=r\sqrt{63}$

At $r\sqrt{63}$, the magnetic field on axis as compared to the magnetic field at the center of the coil carrying current i and radius r is 1/8.

Hence, this is the required solution.