A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be `(1//8)^(th)` of its value at the centre of the coil is,

Options

A. `R//sqrt(3)`
B. `Rsqrt(3)`
C. `2Rsqrt(3)`
D. `(2sqrt(3))R`

Answer

B

Explanation

`B_(axis)=(mu_(0))/(4pi)xx(2piIR^(2))/((R^(2)+x^(2)))`
At centre, `B_(centre)=(mu_(0))/(4pi)xx(2piI)/R`
In this given problem.
`(mu_(0))/(4pi)xx(2piIR^(2))/((R^(2)+x^(2)))=1/8[(mu_(0))/(4pi)xx(2piI)/R]`
or `(R^(2)+x^(2))^(3//2)=8R^(3)`
Solving, we get `=x=Rsqrt(3)`