(a)

Show that the interaction magnetic energy \begin{align*} W_{12} = \int \bs{J} \cdot \bs{A}_2 d^3x = I_1 F_2 \end{align*}

(where \(F_2\) is the magnetic flux from \(I_2\) linking the rectangular circuit carrying \(I_1\)), is \begin{align*} W_{12} = \frac{\mu _0 I_1 I_2 a}{4 \pi } \ln \left [ \frac{4d^2 + b^2 + 4db \cos \alpha }{4d^2 + b^2 - 4db \cos \alpha } \right ] \end{align*}

Solution. \begin{align*} W_{12} = I_1 \oint d\bs{l}_1 \cdot \bs{A}_2 \end{align*}

Take the loop to be in the x-y plane, centered at the origin, with the wire above it as shown:

So \(\bs{A}_2\) is given by \begin{align*} \bs{A}_2 = - \frac{I_2 \hat{\bs{y}}}{4 \pi } \ln \left [{\left ( x - d \cos \alpha \right )}^2{\left ( x - d \sin \alpha \right )}^2 \right ] \end{align*}

And only the sides of length \(a\) will contribute: \begin{align*} \Rightarrow W_{12} &= I_1 \oint d\bs{l}_1 \cdot \bs{A}_2 \\ &= \frac{\mu _0 I_1 I_2 a}{4 \pi } \ln \left [ \frac{{(-\frac{b}{2} - d \cos \alpha )}^2 -{(d \sin \alpha )}^2}{{(\frac{b}{2} - d \cos \alpha )}^2 +{(d \sin \alpha )}^2} \right ] \\ &= \frac{\mu _0 I_1 I_2 a}{4 \pi } \ln \left [ \frac{4d^2 + b^2 + 4db \cos \alpha }{4d^2 + b^2 - 4db \cos \alpha } \right ] \end{align*}

(b)

Calculate the force between the loop and the wire for fixed currents.

Solution. \begin{align*} \bs{B} = \bs{\nabla } \times \bs{A} \end{align*}

Only the force on the sides of length a will be nonzero by symmetry \begin{align*} B_x(x, 0) &= - \frac{\partial A_y}{\partial z} \biggr \vert _{z=0} \\ &= \frac{-d \sin \alpha }{{(x - d \cos \alpha )}^2 +{(d \sin \alpha )}^2} \\ B_z(x, 0) &= \frac{\partial A_y}{\partial x} \biggr \vert _{z=0} \\ &= \frac{x-d \sin \alpha }{{(x - d \cos \alpha )}^2 +{(d \sin \alpha )}^2} \end{align*}

\begin{align*} F_x &= I_1 \left [ B_z\left (\frac{b}{2}, 0\right ) - B_z\left (-\frac{b}{2}, 0\right ) \right ] \\ &= \frac{2 \mu _0 I_1 I_2 a b}{\pi } \frac{4d^2 \cos (2\alpha ) - b^2}{b^4 - 8d^2\cos (2\alpha )b^2 + 16d^4} \\ F_z &= - I_2 \left [ B_x\left (\frac{b}{2}, 0\right ) - B_x\left (-\frac{b}{2}, 0\right ) \right ] \\ &= \frac{8 \mu _0 I_1 I_2 a b}{\pi } \frac{d^2 \sin (2\alpha )}{b^4 - 8d^2\cos (2\alpha )b^2 + 16d^4} \end{align*}

(c)

Repeat the calculation for a circular loop of radius \(a\), whose plane is parallel to the wire and makes an angle \(\alpha \) with respect to the plane containing the center of the loop and the wire. Show that the interaction energy is \begin{align*} W_{12} = \mu _0 I_1 I_2 d \cdot \Re \left \{ e^{i\alpha } - \sqrt{e^{2i\alpha } - a^2 / d^2} \right \} \end{align*}

Find the force.

Solution. Converting to cylindrical coordinates, \(x = a\cos \phi \) and \(dl_y = d\phi \; a \cos \phi \): \begin{align*} W_{12} = I_1 \int d\phi \; a \cos \phi A_y(a \cos \phi , 0) \end{align*}

Expand \(A_y\) in terms of \(1/d\): \begin{align*} W_{12} = \mu _0 I_1I_2 \left ( \frac{a^2 \cos{\alpha }}{2d} + \frac{a^4 \cos (3\alpha )}{8d^3} + \cdots \right ) \end{align*}

\begin{align*} \frac{1 - \sqrt{1-z^2}}{z} = \frac{z}{2} + \frac{z^3}{8} + \cdots \end{align*}

therefore \begin{align*} W_{12} &= \mu _0 a I_1I_2 \Re \left ( \frac{1 - \sqrt{1 -{(\frac{a}{d}e^{i\alpha })}^2}}{\frac{a}{d}e^{i\alpha }} \right ) \\ &= \mu _0 d I_1I_2 \Re \left ( e^{-i\alpha } - \sqrt{e^{-2i\alpha } - \frac{a^2}{d^2}} \right ) \\ &= \mu _0 d I_1I_2 \Re \left ( e^{i\alpha } - \sqrt{e^{2i\alpha } - \frac{a^2}{d^2}} \right ) \end{align*}

The last step is obtained from taking the real part.
For the force, a similar process is applied to the following integral: \begin{align*} \bs{F} &= \hat{\bs{x}} I_1 \int d\phi \; a \cos \phi B_z(a \cos \phi , 0) - \hat{\bs{z}}I_1 \int d\phi \; a \cos \phi B_x(a \cos \phi , 0) \end{align*}

yielding \begin{align*} F_x &= \mu _0 I_1 I_2 \Re \left [ \frac{1}{\sqrt{1 - \left ( \frac{a}{d}E^{i\alpha } \right )}} \right ] \\ F_z &= \mu _0 I_1 I_2 \Im \left [ \frac{1}{\sqrt{1 - \left ( \frac{a}{d}E^{i\alpha } \right )}} \right ] \end{align*}

(d)

For both loops, show that when \(d \gg a,b\) the interaction energy reduces to \(W_{12} \approx \bs{m} \cdot \bs{B}\), where \(\bs{m}\) is the magnetic moment of the loop. Explain the sign.

Solution. \begin{align} W_{12} &= \frac{\mu _0 I_1 I_2 a}{4 \pi } \ln \left [ \frac{4d^2 + b^2 + 4db \cos \alpha }{4d^2 + b^2 - 4db \cos \alpha } \right ] \\ &\approx \frac{\mu _0I_1I_2a}{4\pi } \frac{2b\cos (\alpha )}{d} \\ &= (I_1ab) \frac{\mu _0 I_2 \cos \alpha }{2 \pi d} \\ &= m_1 B_{2z} \end{align}

Similarly for the second case: \begin{align*} W_{12} &\approx \mu _0 I_1 I_2 a \frac{a \cos \alpha }{2d} \\ &= (I_1 \pi a^2) \frac{\mu _0 I_2 \cos \alpha }{2 \pi d} \\ &= m_1 B_{2z} \end{align*}

The sign is positive because the magnetic field and the magnetic moment oppose each other.

Show that the mutual inductance of two circular coaxial loops in a homogeneous medium of permeabililty \(\mu \) is \begin{align*} M_{12} = \mu \sqrt{ab} \left [ \left ( \frac{2}{k} - k \right ) K(k) - \frac{2}{k} E(k) \right ] \end{align*}

where \begin{align*} k^2 = \frac{4ab}{(a + b^2) + d^2} \end{align*}

and \(a, b\) are the radii of the loops, \(d\) is the distance between their centers, and \(K\) and \(E\) are the complete elliptic integrals.

Find the limiting value when \(d \ll a,b\) and \(a \simeq b\)

A circular loop of mean radius \(a\) is made of wire having a circular cross section of radius \(b\), with \(b \ll a\). The sketch shows the relevant dimensions and coordinates for this problem.

(a)

Using (5.37), the expression for the vector potential of a filamentary circular loop, and appropriate approximations for the elliptic integrals, show that the vector potential at the point \(P\) near the wire is approximately \begin{align*} A_\phi = (\mu _0 I / 2\pi ) [\ln (8a / \rho ) - 2] \end{align*}

where \(\rho \) is the transverse coordinate shown in the figure and corrections are of order \((\rho /a) \cos \phi \) and \({(\rho / a)}^2\)

Solution. \begin{align*} A_\phi (r, \theta ) = \frac{\mu _0}{4\pi } \frac{4Ia}{\sqrt{a^2 + r^2 + 2ar \sin \theta }} \left [ \frac{(2-k^2)K(k) - 2E(k)}{k^2} \right ] \end{align*}

where k is defined as \begin{align*} k^2 = \frac{4ar \sin \theta }{a^2 + r^2 + 2ar\sin \theta } \end{align*}

\begin{align*} r &= a + \rho \sin \phi \\ \Rightarrow a^2 + r^2 + 2ar \sin \theta = a^2 &+{(a + \rho \sin \phi )}^2 + 2a(a + \rho \sin \phi )\sin \theta \end{align*}

For \(P\) close to the wire, \(\sin \theta \approx 1\) \begin{align*} \Rightarrow a^2 + r^2 + 2ar \sin \theta &= a^2 +{(a + \rho \sin \phi )}^2 + 2a(a + \rho \sin \phi ) \\ &= 4a^2 + \rho ^2 \sin ^2 \phi + 4 a \rho \sin \phi \end{align*}

\begin{align*} k^2 &= \frac{4a^2 + a\rho \sin \phi }{4a^2 + \rho ^2 \sin ^2 \phi + 4 a \rho \sin \phi } \\ &= \frac{4 + \frac{\rho }{a} \sin \phi }{4 + \frac{\rho ^2}{a^2}\sin ^2\phi + 4 \frac{\rho }{a}\sin \phi } \\ &= 1 - 3 \frac{\rho }{4a} + \frac{\rho ^2}{2a^2} + \mathcal{O}\left (\frac{\rho ^3}{a^3}\right ) \end{align*}

\begin{align*} A_\phi (r, \theta ) &= \frac{\mu _0}{4\pi } \frac{4Ia}{\sqrt{4a^2 + \rho ^2 \sin ^2 \phi + 4 a \rho \sin \phi }} \left [ \frac{(2 - k^2)K(k) - 2E(k)}{k^2} \right ] \\ &= \frac{\mu _0}{4\pi } \frac{4I}{\sqrt{4 + \frac{\rho ^2}{a^2} \sin ^2 \phi + 4 \frac{\rho }{a} \sin \phi }} \left [ \frac{(1 + \frac{3\rho }{4a})K(k) - 2E(k)}{1 - \frac{3\rho }{4a}} \right ] \end{align*}

\begin{align*} K(k) = \int _0^{\frac{\pi }{2}} \frac{d\theta }{\sqrt{1 - k^2 \sin ^2 \theta }} &= \frac{\pi }{2} \left \{ 1 + \frac{1}{4}\left (1 - \frac{3 \rho }{4 a}\right ) \right \} \\ &= \frac{\pi }{8} \left ( 5 - \frac{3\rho }{4a} \right ) \\ E(k) = \int _0^{\frac{\pi }{2}} d\theta \;\sqrt{1 - k^2 \sin ^2 \theta } &= \frac{\pi }{2} \left \{ 1 - \frac{1}{4}\left (1 - \frac{3 \rho }{4 a}\right ) \right \} \\ &= \frac{3\pi }{8} \left ( 1 + \frac{\rho }{4a} \right ) \end{align*}

\begin{align*} A_\phi (r, \theta ) &= \frac{\mu _0}{4\pi } \frac{4I}{\sqrt{4 + \frac{\rho ^2}{a^2} \sin ^2 \phi + 4 \frac{\rho }{a} \sin \phi }} \left [ \frac{\pi }{8} \frac{(1 + \frac{3\rho }{4a}) (5 - 3\rho / 4 a) - 2(3 + 3\rho / 4a)}{1 - \frac{3\rho }{4a}} \right ] \\ &= \frac{\mu _0 I}{8} \left ( \frac{1}{2} - \frac{1}{4}\frac{\rho }{a}\sin \phi \right ) \left [ \frac{(1 + \frac{3\rho }{4a}) (5 - 3\rho / 4 a) - 2(3 + 3\rho / 4a)}{1 - \frac{3\rho }{4a}} \right ] \\ &= \frac{\mu _0 I}{8} \left ( \frac{1}{2} - \frac{1}{4}\frac{\rho }{a}\sin \phi \right ) \left [ \frac{-1 + 21 \rho / 4 a}{1 - \frac{3\rho }{4a}} \right ] \\ &= \frac{\mu _0 I}{8} \left ( \frac{1}{2} - \frac{1}{4}\frac{\rho }{a}\sin \phi \right ) \left [ \frac{-1 + 21 \rho / 4 a}{1 - \frac{3\rho }{4a}} \right ] \\ &= \cdots \end{align*}

(b)

Since the vector potential of part a is, apart from a constant, just that outside a straight circular wire carrying a current \(I\), determine the vector potential inside the wire \((\rho < b)\) in the same approximation by requiring continuity of \(A_\phi \) and its radial derivative at \(\rho = b\), assuming that the current is uniform in density inside the wire: \begin{align*} A_\phi &= (\mu _0 I / 4 \pi )(1 - \rho ^2 / b^2) + (\mu _0 I / 2 \pi ) [\ln (8a / b) - 2], & \rho < b \end{align*}

(c)

Use (5.149) to find the magnetic energy, hence the self-inductance, \begin{align*} L = \mu _0a[\ln (8a/b) - 7/4] \end{align*}

Are the corrections of order \(b/a\) or \({(b/a)}^2\)? What is the change in \(L\) if the current is assumed to flow only on the surface of the wire (as occurs at high frequencies when the skin depth is small compared to \(b\))?