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

Find at the centre `B_(1)=(mu_(0)NI)/(2r)`
Field at distance h from the centre
`B=(mu_(0)nir^(2))/(2(r^(2)+h^(2))^(3//2))=B_(1)(1=(h^(2))/(r^(2)))^((-3)/(2)),`
`B_(1)=(1-(3)/(2)(h^(2))/(r^(2)))`