The field normal to the plane of a wire of `n` turns and radius `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

The magnetic field on the axis of a coil carrying current `I`, having `n` turns, radius `r` and at a distance from the centre of the coil, is given by :
`B=(mu_(0))/(4pi)xx(2piN.Ir^(2))/((r^(2)+h^(2))^(3//2))........(1)`
The field at the centre is given by :
`B_(C)=(mu_(0))/(4pi)xx(2piN.I)/r`
`:. B/(B_(C))=(r^(2))/((r^(2)+h^(2))^(3//2))=r^(3) (r^(3))/([1+(h^(2))/(h^(2))]^(3//2))`
`=1/([1+3/2 (h^(2))/(h^(2))])`
or `B[1+3/2 (h^(2))/(h^(2))]=B_(c)`
`:. ((B_(c)-B))/B=3/2 (h^(2))/(r^(2))`