Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 5.50

The magnetic field on the axis of a circular current loop (Eq. 5.41)

\begin{aligned} B (z) & = \frac{μ_{0} I}{4 π} (\frac{\cos θ}{R^{2}}) 2 π R \\ (Eq. 5.41, as seen on example 5.6) & = \frac{μ_{0} I}{2} \frac{R^{2}}{(R^{2} + z^{2})^{3 / 2}} \end{aligned}

is far from uniform (it falls off sharply with increasing $z$ ). You can produce a more nearly uniform field by using two such loops a distance $d$ apart (Fig. 5.62).

(a) Find the field ( $B$ ) as a function of $z$ , and show that $\frac{\partial B}{\partial z}$ is zero at the point midway between them ( $z = 0$ ).

(b) If you pick $d$ just right, the second derivative of $B$ will also vanish at this midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing locally uniform fields in the laboratory. Determine $d$ such that $\frac{\partial^{2} B}{\partial z^{2}} = 0$ at the midpoint, and find the resulting magnetic field at the center. (Answer: $8 μ_{0} I / 5 \sqrt{5} R$ )

Part (a)

We know the field of each ring, so let's just add them up. Now, we have to be careful here: At the origin $z = 0$ , we are a distance $d / 2$ from each one of them, so we are going to add the field from Eq. 5.41 with $z \to d / 2 + z$ and with $z \to d / 2 - z$ :

B (z) = \frac{μ_{0} I}{2} \frac{R^{2}}{(R^{2} + (d / 2 + z)^{2})^{3 / 2}} + \frac{μ_{0} I}{2} \frac{R^{2}}{(R^{2} + (d / 2 - z)^{2})^{3 / 2}}

The derivative with respect to $z$ is:

\frac{\partial B}{\partial z} = \frac{3 I μ_{0} R^{2} (\frac{d}{2} - z)}{2 {({(\frac{d}{2} - z)}^{2} + R^{2})}^{5 / 2}} - \frac{3 I μ_{0} R^{2} (\frac{d}{2} + z)}{2 {({(\frac{d}{2} + z)}^{2} + R^{2})}^{5 / 2}}

Which when evaluated to $z = 0$ , vanishes as expected.

Part (b)

The second derivative is

\frac{\partial^{2} B}{\partial z^{2}} = \frac{15 I μ_{0} R^{2} {(\frac{d}{2} - z)}^{2}}{2 {({(\frac{d}{2} - z)}^{2} + R^{2})}^{7 / 2}} + \frac{15 I μ_{0} R^{2} {(\frac{d}{2} + z)}^{2}}{2 {({(\frac{d}{2} + z)}^{2} + R^{2})}^{7 / 2}} - \frac{3 I μ_{0} R^{2}}{2 {({(\frac{d}{2} - z)}^{2} + R^{2})}^{5 / 2}} - \frac{3 I μ_{0} R^{2}}{2 {({(\frac{d}{2} + z)}^{2} + R^{2})}^{5 / 2}}

Evaluate at $z = 0$ :

\frac{\partial^{2} B}{\partial z^{2}} |_{z = 0} = \frac{15 d^{2} I μ_{0} R^{2}}{4 {(\frac{d^{2}}{4} + R^{2})}^{7 / 2}} - \frac{3 I μ_{0} R^{2}}{{(\frac{d^{2}}{4} + R^{2})}^{5 / 2}}

Set it equal to zero, and we solve for $d$ :

\begin{aligned} \frac{15 d^{2} I μ_{0} R^{2}}{4 {(\frac{d^{2}}{4} + R^{2})}^{7 / 2}} & = \frac{3 I μ_{0} R^{2}}{{(\frac{d^{2}}{4} + R^{2})}^{5 / 2}} \\ \frac{15 d^{2}}{4 {(\frac{d^{2}}{4} + R^{2})}^{7 / 2}} & = \frac{3}{{(\frac{d^{2}}{4} + R^{2})}^{5 / 2}} \\ \frac{15 d^{2}}{4 \cdot 3} & = \frac{d^{2}}{4} + R^{2} \\ 15 d^{2} & = 3 d^{2} + 12 R^{2} \\ 12 d^{2} & = 12 R^{2} \\ d & = \pm R \end{aligned}

Where we, of course, pick

d = R

The resulting field at the center is given by setting $z \to 0$ and $d \to R$ in

\begin{matrix} (as discussed in part (a)) & B (z) = \frac{μ_{0} I}{2} \frac{R^{2}}{(R^{2} + (d / 2 + z)^{2})^{3 / 2}} + \frac{μ_{0} I}{2} \frac{R^{2}}{(R^{2} + (d / 2 - z)^{2})^{3 / 2}} \end{matrix}

This gives:

B_{d = R} (z = 0) = \frac{8 I μ_{0}}{5 \sqrt{5} \sqrt{R^{2}}}