The field due to a wire of n turns and radius r which carries a current I is measure on the axis of the coil at a small distance h form 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

(a) The magnetic field on the axis of a coil carrying current I.
having n turns, radius r and at a distance h from the centre
of the coil, is given by
`B=(mu_0)/(4pi)xx(2piNIr^2)/((r^2+h^2)^(3//2))....(i)`
The field at the centre is given by
`B_C=(mu_0)/(4pi)xx(2piNI)/r....(ii)`
`:. B/(B_C)=(r^3)/((r^2+h^2)^(3//2))=(r^3)/(r^3[1+(h^2)/(r^2)]^(3//2))=[1-3/2 (h^2)/(r^2)]`
We have to find: `f=(B_C-B)/(B_C)=1-B/(B_C)=3/2 (h^3)/(r^2)`