Uniform Magnetic Fields

Taha Azzaoui - 2018.07.07


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.

Figure 1: The magnetic field along the axis of a single current loop of radius R.

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 dl⃗ and dr⃗ 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 p1, there must exist another point p2 on the opposite side of the loop such that magnetic field at p1 is equal in magnitude to that at p2 (since they’re equidistant from the center) but opposite in direction (since dl⃗ is tangent to the loop). Therefore, we are only interested in the z-component of the field.
$$dB_z = dBcos\theta = dB\frac{R}{r} = \frac{\mu_0}{4 \pi} \frac{IR}{r^3} dl \tag{4}$$

Figure 2: Planar components of vecB cancel due to the symmetry of the loop.

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