# Uniform Magnetic Fields

# Taha Azzaoui - 2018.07.07

### Introduction

A Helmholtz coil is a circular apparatus used to produce an approximately uniform magnetic field. For this reason, it is ideal for studying the behavior of objects under the influence of a uniform magnetic field. In what follows, we will make use of symmetry to derive the expression for the magnetic field between two Helmholtz coils.

Recall that magnetic fields ascribe to every point in space a vector that describes the magnetic influence of an
electric current on an object at that point. The value of the magnetic field at an infinitesimally small point in
space is defined according to the Biot-Savart
law

$$d\vec{B} = \frac{\mu_0}{4 \pi} \frac{Id\vec{l} \times \hat{r}}{r^2}
\tag{1}$$

In the case of the loop in Figure 1, the vectors *d**l⃗* and
*d**r⃗* are perpendicular, therefore we have

$$d\vec{B} = \frac{\mu_0}{4 \pi} \frac{I}{r^2} dl \tag{2}$$

We can then integrate over the entire loop to find the cummulative field

$$\vec{B} = \frac{I\mu_0}{4 \pi} \oint \frac{1}{r^2} dl \tag{3}$$

Note that by the symmetry of the loop, we can ignore the planar components of the magnetic field. To see this,
recall that we are looking for the magnetic field through the exact center of loop. This means that, for every
point on the loop *p*_{1}, there must exist another point
*p*_{2} on the opposite side of the loop such that magnetic field
at *p*_{1} is equal in magnitude to that at *p*_{2} (since they’re equidistant from the center) but opposite in direction
(since *d**l⃗* is tangent to the loop). Therefore, we are only
interested in the z-component of the field.

$$dB_z = dBcos\theta = dB\frac{R}{r} = \frac{\mu_0}{4 \pi} \frac{IR}{r^3} dl
\tag{4}$$

Finally, we can integrate this value along the loop. Note that by the Pythagorean theorem, we have $r = \sqrt{R^2 + z^2}$.

$$\begin{eqnarray} B &=& \int cos\theta dB\nonumber \\ &=& \int_0^{2\pi R}
\frac{\mu_0}{4 \pi} \frac{IR}{r^3} dl \nonumber \\ &=& \frac{IR\mu_0}{4 \pi (R^2 + z^2)^{3/2}} \int_0^{2\pi R}
dl \nonumber \\ &=& \frac{IR^2\mu_0}{2 (R^2 + z^2)^{3/2}} \nonumber \end{eqnarray}$$

Our resulting equation for the magnetic field along the axis of a charged loop is

$$\vec{B} = \frac{IR^2\mu_0}{2 (R^2 + z^2)^{3/2}} \hat{k} \tag{5}$$

In the case of two Helmholtz coils with *N* turns each, separated by a
distance *h*, with one coil centered at the origin, we have

$$\vec{B}_H = \frac{NIR^2\mu_0}{2 (R^2 + z^2)^{3//2}}\hat{k} + \frac{NIR^2\mu_0}{2 (R^2
+ (h-z)^2)^{3//2}} \hat{k} \tag{6}$$

We can find a relative optimum in the magnetic field by differentiating $B$ and setting the result to 0.

$$\frac{\partial \vec{B}}{\partial z} = 0 \implies z = \frac{h}{2}
\tag{7}$$

Using (7), we can calculate the corresponding spacing between the coils

$$\frac{\partial \vec{B}\|_{z = h/2}}{\partial h} = 0 \implies h = R
\tag{8}$$

Substituting (7) and (8) into (6), we can obtain the magnetic field under the optimal values of *z* and *h*. Finally, we can substitute (7) and
(8) into (6) to obtain the expression of the magnetic field under the optimal values of *z* and *h*.

$$B_H = \frac{8\mu_0 NI}{\sqrt{125}R} \hat{k} \tag{9}$$