A magnetic field is established by a circular current i of radius a. Find the magnetic field gradient (i.e. the derivative of the magnetic field induction vector) in the direction of the circular current's axis at a point whose distance from the centre of the turn is x.

Answer

`(d B)/(dx)=-(6mu_(0)p_(m)x)/(4pi r^5)," where "r=sqrt(a^2+x^2)`.

Explanation

The magnetic field induction at the point of interest is, `B=(2mu_(0) p_(m))/(4pir^2)=(mu_(0)p_(m))/(2pi) (a^(2)+x^2)^(-3/2)`
The gradient is `(dB)/(dx) =(mu_(0)p_(m))/(2pi)=d/(dx) (a^2+x^2)^(-3/2)`
`=-3/2 (mu_(0)p_m)/(2pi) (a^2+x^2)^(-5//2), 2x=-(3mu_(0) p_(m)x)/(2pir^3)`