A circular coil of radius `R` carries a current `i`. The magnetic field at its centre is `B`. The distance from the centre on the axis of the coil where the magnetic field will be `B//8` is

Options

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

Answer

B

Explanation

According to the question, magnetic field at the centre of circular coil is given by
`B_("centre")=(mu_(0)Ni)/(2R)` . . .(i)
`B_("axis")=(mu_(0))/(4pi)(2piNiR^(2))/(4pi(x^(2)+R^(2))^(3//2))`
`(B)/(8)=(mu_(0)NiR^(2))/(2(x^(2)+R^(2))^(3//2))`
From Eq. (i), we get
`(mu_(0)Ni)/(8xx2R)=(mu_(0)NiR^(2))/(2(x^(2)+R^(2))^(3//2))`
`(1)/(8R)=(1)/((x^(2)+R^(2))^(3//2))`
`(1)/(8R^(3))=(1)/((x^(2)+R^(2))^(3//2))`
`8R^(3)=(x^(2)+R^(2))^(3//2)` (on taking cube root)
`2R=(x^(2)+R^(2))^(1//2)` (on taking square root)
`4R^(2)=(x^(2)+R^(2))`
`4R^(2)-R^(2)=x^(2)`
`3R^(2)=x^(2)`
`x=Rsqrt(3)`