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Mobile App Chapter 5: Q 5-5.47 P (page 259)
The magnetic field on the axis of a circular current loop (Eq. 5.41) 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 distanced apart (Fig. 5.59).
(a) Find the field (B) as a function of \(z\), and show that \(\frac{\partial \mathbf{B}}{\partial \mathbf{z}}\) 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 the midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing relatively uniform fields in the laboratory. Determine \(d\) such that
\(\partial^{2} B / \partial z^{2}=0\) at the midpoint, and find the resulting magnetic field at the center.
\(\frac{A I_{0}}{5 \sqrt{5} R}\)
Short Answer
(a) The magnetic field as a function of $z$ is
$\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{\left[R^{2}+\left(\frac{d}{2}+z\right)^{2}\right]^{3 / 2}}+\frac{1}{\left[R^{2}+\left(\frac{d}{2}-z\right)^{2}\right]^{3 / 2}}\right)$ and first derivative of this magnetic field is zero at the midway between both loops.
(b) The distance between loops for zero second derivative at midpoint is equal to radius of loop and the resulting magnetic field at the centre is $\frac{8 \mu_{0} l}{5 \sqrt{5} R}$.
Step by step solution
Determine the magnetic field as a function of z
Consider the figure for the field as:
The magnetic field due to the upper loop by using equation 5.41 is given as:
The net magnetic field due to both loops is given as:
$B=B_{1}+B_{2}$
$B=\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{[1}\right)+\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{\Gamma 3^{3 / 2}}\right)$
The magnetic field due to the upper loop by using equation 5.41 is given as:
$$
B_{2}=\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{\left[R^{2}+\left(\frac{d}{2}-z\right)^{2}\right]^{3 / 2}}\right)
$$
The net magnetic field due to both loops is given as:
$B=B_{1}+B_{2}$
$B=\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{[1}\right)+\frac{\mu_{0} I R^{2}}{2}\left(\frac{1}{\Gamma 3^{3 / 2}}\right)$
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