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

1

Explanation

`(B_("axis"))/(B_("centre"))=(r^(3))/(R^(2)+X^(2))^(3//2)`
`B_("axis")=B_("centre")xx(r^(3))/(r^(3)(1+(x^(2))/(r^(2)))^(3//2))`
`=B_(e)(1+(x^(2))/(r^(2)))^(-3//2)` , x=h
`B_("axis") (1-(3h^(2))/(2r^(2)))B_(c)`
`rArr (Delta B)/(B_(c))=(B_("axis"))/(B_(c))=(3)/(2) (h^(2))/(r^(2))`