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"/"sqrt3`
B. `Rsqrt3`
C. `2R"/"sqrt3`
D. `(2R"/"sqrt3)R`

Answer

B

Explanation

`B_(ax"i"s)=(mu_0)/(4pi)xx(2piIR^(2))/((R^(2)+x^(2))^(3"/"2)),B_(centre)=(mu_(0))/(4pi)xx(2piL)/(R)`
In the given problem ,
`(mu_(0))/(4pi)xx(2piIR^(2))/((R^(2)+x^(2))^(3"/"2))=(1)/(8)[mu_(o)/(4pi)xx(2piI)/(R)]or(R^(2)+x^(2))^(3"/"2)=8R^(3)`
By solving , we get `x=sqrt3R`