Helmholtz Coils (Theory)

The Helmholtz coils are two large circular coils with N turns of radius R that can give a uniform magnetic field. Please find where on the x axis is the magnetic field uniform if the two coils are separated by a distance R.

The magnetic field of a coil having $N$ turns (packed together) and radius $R$, along its $z$ axis (perpendicular to the plane of current loop) is

$B_z=μNI/2*R^2/(R^2+z^2 )^(3/2)$

This equation can be found by integration starting from the Biot-Savart law. If you have two identical parallel coils of radius R at distance R one from another the total field on the common z axis at a distance z from the first coil is

$B_z=B_z1+B_z2=(μNIR^2)/2 [1/(z^2+R^2)^{3/2} +1/((R-z)^2+R^2)^{3/2}$

$B_z=(μNIR^2)/2 [1/(z^2+R^2)^{3/2} +1/(2R^2+z^2-2Rz)^{3/2}$

The condition of uniform magnetic field along z is ($B$ does not depend on $z$ at least up until the 3rd order of $z$)

$dB/dz=0$ and  $(d^2 B)/(dz^2)=0$

$dB/dz=(πNIR^2)/2 [(-3/2) (2z) (z^2+R^2)^{-5/2}-(3/2) (2R^2+z^2-2Rz)^{-5/2} (2z-2R)]$

This is true for $z=R/2$ (one can simply replace above this value of z and verify):

$-(3/2)*R*(R^2/4+R^2)^{-5/2}-(3/2) (2R^2+R^2/4-R^2 )(R-2R)=0$

For the second derivative we get

$(d^2 B)/(dz^2 )=-(3μINR^2)/2 [(z^2+R^2)^{-5/2}-5z^2 (z^2+R^2 )^{-7/2}-$

$-(2R^2+z^2-2Rz)^{-5/2}+5(z-R)^2 (2R^2+z^2-2Rz)^{-7/2}]$

Again this is true expression is 0 for $z=R/2$:

$(R^2/4+R^2 )^{-5/2}-(5 R^2/4) (R^2/4+R^2 )^{-7/2}-$

$-(2R^2+R^2/4-R^2 )^{-5/2}+5(R/2-R)^2 (….)^{-7/2}=0$

So the distance where the field is uniform is half distance between coils and you need to show that the first and second derivatives of the field (with respect to z) are zero at this distance.

Reference

Helmholtz Coils (Embry Riddle Aeronauutical University)