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_("centre"))/(B_("axis")) = (1 + (h^2)/(R^2))^(3/2) implies B_("axis")= B _("centre") (1 + (h^2)/(R^2))^((-3)/(2))`
`implies B_("axis") = B_("centre") [ 1- (3h^2)/(2R^2)] = B_("centre") - (3h^2)/(2R^(2)) . B_("centre")`
`therefore B_("axis") lt B_("centre")` by fraction `(3h^2)/(2R^2)`