The field normal to the plane of a wire of `n` turns and radis `r` which carriers `i` is measured on the axis of the coil at a small distance `h` from the centre of the coil. This is smaller than the field at the centre by the fraction.

Options

A. `3/2.(h^(2))/(r^(2))`
B. `2/3.(h^(2))/(r^(2))`
C. `3/2.(r^(2))/(h^(2))`
D. `2/3.(r^(2))/(h^(2))`

Answer

A

Explanation

`B=(mu_(0)nir^(2))/(2(r^(2)+h^(2))^(3//2))=(mu_(0)ni)/(2r(1+(h^(2))/(r^(2)))^(3/2))`
`=(mu_(0)ni)/(2r)[1+(h^(2))/(r^(2))]^(3/2)=B_(c)(1-(3h^(2))/(2r^(2)))`
`B/(B_(c))-1=-(3h^(2))/(2r^(2))`