The magnetic field normal to the plane of a coil of N turns and radius r which carries a current I is measured on the axis of the coil at a distance h from the centre of the coil. This is smaller than the field at the centre by ther 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

The magnetic field normal to the plane of a coil of N turns and radius r having current I on the axis at a distance h from the centre is given by
`B _(axis ) = (mu_(0) N I r^(2))/( 2(r^(2) + h^(2))^((3)/(2))` ,
`=(mu_(0) N i r^(2))/(2(r^(2))^(3//2) (1+ (n^(2))/(r^(2)))^((3)/(2)))= (mu_(0)Ni r^(2) [1+ (n^(2))/(r^(2))]^((3)/(2)))/(2r^(3))`
`B_(centre ) = (mu_(0) Ni )/( 2r )`
From Eqs . (i ) and (ii) we get
`B_(axis ) = B_(centre) (1 - (3h^(2))/(2h^(2)))`
From Eq. (iii) it is clear that the magnetic field on the axis is smaller than the field at the centure by a fraction `(3h^(2))/( 2r^(2))`