Magnetic field of two Helmholtz coils

Configuration of two Helmholtz coils to produce a uniform magnetic field

To setup a Helmholtz coil two similar coils with radius R are placed in the same distance R. When the coils are so connected that the current through the coils flows in the same direction, the Helmholtz coils produce a region with a nearly uniform magnetic field.

The magnetic field at center of the coils with N wire windings is proportional to current through coils:

${B=\mu_0\cdot \frac{8\cdot I\cdot \text N}{\sqrt{125}\cdot R}}$ $I$ = coil current, $\mu_0$ = vacuum permeability, N = windings, $R$ = radius and distance of coils

The magnetic field of the Helmholtz coil used on the next pages, depending on the coil current $I$, is:$$\bbox[5px,border:2px solid red]{B\approx 7.48\cdot 10^{-4}\frac{\text T}{\text A}\cdot I}$$$I$ = coil current, $\mu_0 = 4\pi \cdot 10^{-7} \frac{\text N}{{\text{A}}^2}$, N = 124 windings, $R$ = 14.9 cm

This derivation is beyond school level!

This equation is indicated by Biot-Savart law. For the magnetic field in horizontal direction and a coil with only one winding applies

$\vec B(x)=\frac{\mu_0\cdot I}{2}\cdot \frac{R^2}{\left(R^2+x^2\right)^{\frac{3}{2}}}\cdot \vec e_x$ The magnetic field at center of two Helmholtz coils is the superposition of two circular currents. For symmetry reasons it becomes:

$B=B\left(\frac{R}{2}\right)+B\left(-\frac{R}{2}\right)=2\cdot B\left(\frac{R}{2}\right)$ $=\mu_0\cdot \frac{I\cdot R^2}{\left(R^2+\frac{R^2}{4}\right)^{\frac{3}{2}}}$ $=\mu_0\cdot \frac{8\cdot I}{\sqrt{125}\cdot R}$