Exam 2 Practice

Notes: (v) satisfies Poisson's equation with \(\rho = -4\varepsilon_0\), so it describes a region with charge. (vi) is the potential of a line charge and (vii) of a point charge, which is why they only work away from \(s=0\) and \(r=0\).

The takeaway for separation of variables: the separation constants must add to zero. Directions with exponential or hyperbolic behaviour contribute \(+k^2\) and sinusoidal directions contribute \(-k^2\), so you can never have all three directions sinusoidal.

Solution: On the circle,

\[ V = (x_0+R\cos\phi)^2-(y_0+R\sin\phi)^2 = x_0^2-y_0^2+2R\,(x_0\cos\phi-y_0\sin\phi)+R^2\cos 2\phi \]

using \(\cos^2\phi-\sin^2\phi=\cos2\phi\). The \(\cos\phi\), \(\sin\phi\) and \(\cos 2\phi\) terms each integrate to zero over a full period, so

\[ \langle V\rangle = \frac{1}{2\pi}\int_0^{2\pi}V\,\dd\phi = x_0^2-y_0^2 = V(x_0,y_0). \]

Same idea as Tutorial 6, where you averaged a point charge's potential over a sphere in 3D.

Solution: Suppose \(V\) had a local maximum at \(P\). On a small sphere around \(P\), every point has \(V \lt V(P)\), so the average over the sphere is less than \(V(P)\). That contradicts \(V(P) = \langle V\rangle_{\text{sphere}}\). The same argument rules out a minimum, so the extreme values of \(V\) sit on the boundary.

A test charge \(Q\) has potential energy \(U = QV\). Stable equilibrium needs a local minimum of \(U\), which means a local minimum of \(V\) (if \(Q\gt 0\)) or maximum (if \(Q\lt 0\)). In a charge-free region \(V\) has neither, so no arrangement of fixed charges can trap a charge stably.

Solution: Put an image \(-q\) at \((0,0,-d)\):

\[ V(x,y,z) = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{1}{\sqrt{x^2+y^2+(z+d)^2}}\right] \]

At \(z=0\) the two distances are equal, so \(V = 0\) on the plane, and \(V\to 0\) far away. The only charge in the region \(z \gt 0\) is \(q\). The uniqueness theorem says the potential in a region is fixed by the charge inside it and \(V\) on its boundary. Both of those match the real problem, so this is the answer for \(z\ge 0\).

For \(z\lt 0\) the formula is wrong: that region holds the image charge, which isn't real. The actual potential there is \(V=0\) (inside/behind the grounded conductor).

Solution:

\[ \frac{\partial V}{\partial z} = \frac{q}{4\pi\varepsilon_0}\left[\frac{-(z-d)}{\left(x^2+y^2+(z-d)^2\right)^{3/2}}+\frac{z+d}{\left(x^2+y^2+(z+d)^2\right)^{3/2}}\right] \]

At \(z=0\) both denominators become \((x^2+y^2+d^2)^{3/2}\):

\[ \left.\frac{\partial V}{\partial z}\right|_{z=0} = \frac{q}{4\pi\varepsilon_0}\,\frac{2d}{(x^2+y^2+d^2)^{3/2}} \quad\Longrightarrow\quad \sigma(x,y) = -\frac{q\,d}{2\pi\,(x^2+y^2+d^2)^{3/2}} \]

Total induced charge:

\[ Q_{\text{ind}} = \int_0^{2\pi}\!\!\int_0^{\infty}\frac{-q\,d}{2\pi\,(s^2+d^2)^{3/2}}\,s\,\dd s\,\dd\phi = -q\,d\left[\frac{-1}{\sqrt{s^2+d^2}}\right]_0^{\infty} = -q\,d\cdot\frac{1}{d} = -q \]

Makes sense: every field line leaving \(q\) ends on the plane.

Solution: Above the plane the field is the same as the two-charge field, so the force is just the pull of the image:

\[ \mathbf F = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2d)^2}\,\zhat \]

It's attractive (toward the plane) whatever the sign of \(q\).

Solution: You apply \(+\frac{q^2}{16\pi\varepsilon_0 z^2}\zhat\) while moving from \(z=\infty\) to \(z=d\):

\[ W = \int_{\infty}^{d}\frac{q^2}{16\pi\varepsilon_0 z^2}\,\dd z = \frac{q^2}{16\pi\varepsilon_0}\left[-\frac{1}{z}\right]_{\infty}^{d} = -\frac{q^2}{16\pi\varepsilon_0\,d} \]

Two real charges would give \(-\frac{q^2}{4\pi\varepsilon_0(2d)} = -\frac{q^2}{8\pi\varepsilon_0 d}\), twice as much. In the real problem the field only exists for \(z\gt 0\) (it's zero in the conductor), so \(\frac{\varepsilon_0}{2}\int E^2\,\dd\tau\) covers half the space the two-charge field fills. Another way to say it: the image moves along with \(q\), so you never do the work of bringing in a second charge.

Solution: Three images: \(-q\) at \((-a,b,0)\), \(-q\) at \((a,-b,0)\), and \(+q\) at \((-a,-b,0)\).

Check \(x=0\): it's the mirror plane between \(q\) and \(-q\) at \((\mp a, b)\), and between \(-q\) and \(+q\) at \((\pm a,-b)\). Each pair cancels there. The same pairing works for \(y=0\). All images are outside the physical region \(x,y \gt 0\), as they must be.

Solution: Add the three Coulomb forces on \(q\):

• From \(-q\) at \((-a,b)\), separation \(2a\): \(\;-\dfrac{q^2}{16\pi\varepsilon_0 a^2}\,\xhat\)
• From \(-q\) at \((a,-b)\), separation \(2b\): \(\;-\dfrac{q^2}{16\pi\varepsilon_0 b^2}\,\yhat\)
• From \(+q\) at \((-a,-b)\), separation vector \((2a,2b)\) of length \(2\sqrt{a^2+b^2}\): \(\;\dfrac{q^2}{4\pi\varepsilon_0}\dfrac{2a\,\xhat+2b\,\yhat}{\left(2\sqrt{a^2+b^2}\right)^3} = \dfrac{q^2\,(a\,\xhat+b\,\yhat)}{16\pi\varepsilon_0\,(a^2+b^2)^{3/2}}\)

\[ \mathbf F = \frac{q^2}{16\pi\varepsilon_0}\left[\left(\frac{a}{(a^2+b^2)^{3/2}}-\frac{1}{a^2}\right)\xhat+\left(\frac{b}{(a^2+b^2)^{3/2}}-\frac{1}{b^2}\right)\yhat\right] \]

Both components are negative (since \(a/(a^2+b^2)^{3/2} \lt 1/a^2\)), so the charge is pulled into the corner.

Solution: The \(\yhat\) part goes to zero and \(\mathbf F \to -\dfrac{q^2}{16\pi\varepsilon_0 a^2}\,\xhat = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2a)^2}\,\xhat\), the force from a single grounded plane a distance \(a\) away (Problem 2 with \(d=a\)). Far from the second plate, only the first one matters.

Solution: Repeated reflections only close on themselves when the angle is \(180^\circ/n\) for an integer \(n\) (that takes \(2n-1\) images: 3 for \(90^\circ\), 5 for \(60^\circ\)). For \(70^\circ\) the reflections never close, and some images would land inside the physical region, where you're not allowed to add charge.

Solution: Take a Gaussian sphere of radius slightly larger than \(R\) (but smaller than \(a\)). The flux through it is the same in both problems. In the image problem it encloses \(q'\); in the real problem it encloses the induced charge. So

\[ Q_{\text{ind}} = q' = -\frac{R}{a}\,q \]

Less than \(q\) in magnitude: unlike the infinite plane, not every field line from \(q\) lands on the sphere.

Solution: The distance between \(q\) and \(q'\) is

\[ a-b = a-\frac{R^2}{a} = \frac{a^2-R^2}{a} \] \[ \mathbf F = \frac{1}{4\pi\varepsilon_0}\frac{q\,q'}{(a-b)^2}\,\zhat = -\frac{1}{4\pi\varepsilon_0}\frac{q^2\,R\,a}{(a^2-R^2)^2}\,\zhat \]

Attractive. Near the surface, \(a^2-R^2 \approx 2R\delta\) and \(a\approx R\):

\[ F \approx -\frac{1}{4\pi\varepsilon_0}\frac{q^2 R^2}{4R^2\delta^2} = -\frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2\delta)^2} \]

That's the flat-plane result: up close, the sphere looks flat.

Solution: Add \(q'' = +\frac{R}{a}q\) at the centre. Total charge on the sphere is \(q'+q''=0\), and the surface is still an equipotential (at \(V = \frac{q''}{4\pi\varepsilon_0 R} = \frac{q}{4\pi\varepsilon_0 a}\)).

\[ \mathbf F = \frac{q}{4\pi\varepsilon_0}\left[\frac{q'}{(a-b)^2}+\frac{q''}{a^2}\right]\zhat = \frac{q^2R}{4\pi\varepsilon_0}\left[\frac{1}{a^3}-\frac{a}{(a^2-R^2)^2}\right]\zhat \]

For \(a\gg R\), \(\frac{a}{(a^2-R^2)^2} \approx \frac{1}{a^3}\left(1+\frac{2R^2}{a^2}\right)\), so

\[ \mathbf F \approx -\frac{1}{4\pi\varepsilon_0}\frac{2q^2R^3}{a^5}\,\zhat \]

Still attractive, but it falls off like \(1/a^5\): \(q\) induces a dipole \(\propto 1/a^2\) on the sphere, whose field back at \(q\) goes like \(1/a^3\).

Solution: Keep \(q'\) at \(b\) and add \(Q_0 = 4\pi\varepsilon_0 R V_0\) at the centre. That central charge raises the whole surface uniformly by \(\frac{Q_0}{4\pi\varepsilon_0 R} = V_0\).

Solution: \(V = X(x)Y(y)\) with \(\frac{X''}{X} = -\frac{Y''}{Y} = k^2\). \(Y\) has to vanish at \(y = 0\) and \(y = b\), so it's sinusoidal: \(Y = \sin(n\pi y/b)\), \(k = n\pi/b\). Then \(X = Ce^{kx}+De^{-kx}\), and \(X(0) = 0\) forces \(D=-C\), i.e. \(X\propto\sinh(kx)\).

\[ V(x,y) = \sum_{n=1}^{\infty} C_n \sinh\!\left(\frac{n\pi x}{b}\right)\sin\!\left(\frac{n\pi y}{b}\right) \]

Solution: At \(x = a\): \(V_0 = \sum_n C_n\sinh(n\pi a/b)\sin(n\pi y/b)\). Fourier's trick gives

\[ C_n\sinh\!\left(\frac{n\pi a}{b}\right) = \frac{2}{b}\int_0^b V_0\sin\!\left(\frac{n\pi y}{b}\right)\dd y = \frac{2V_0}{n\pi}\left(1-\cos n\pi\right) = \begin{cases}\dfrac{4V_0}{n\pi} & n \text{ odd}\\[4pt] 0 & n\text{ even}\end{cases} \] \[ V(x,y) = \frac{4V_0}{\pi}\sum_{n=1,3,5,\dots}\frac{1}{n}\,\frac{\sinh(n\pi x/b)}{\sinh(n\pi a/b)}\,\sin\!\left(\frac{n\pi y}{b}\right) \]

Solution: At \(x=a\) the sinh ratio is 1, and \(\sin(n\pi/2) = +1,-1,+1,\dots\) for \(n = 1,3,5,\dots\):

\[ V\!\left(a,\tfrac{b}{2}\right) = \frac{4V_0}{\pi}\left(1-\frac13+\frac15-\cdots\right) = \frac{4V_0}{\pi}\cdot\frac{\pi}{4} = V_0 \;\checkmark \]

Solution: The boundary data is already a single term of the series (\(n=2\)), so every other \(C_n\) is zero:

\[ V(x,y) = V_0\,\frac{\sinh(2\pi x/b)}{\sinh(2\pi a/b)}\,\sin\!\left(\frac{2\pi y}{b}\right) \]

Solution: \(Y\) is still \(\sin(n\pi y/b)\). Now \(X\) must die off at infinity, so \(C = 0\) and \(X = e^{-n\pi x/b}\) (a sinh would blow up):

\[ V(x,y) = \sum_n C_n\,e^{-n\pi x/b}\sin\!\left(\frac{n\pi y}{b}\right),\qquad C_n = \frac{2}{b}\int_0^b V_0(y)\sin\!\left(\frac{n\pi y}{b}\right)\dd y \]

For constant \(V_0\), \(C_n = 4V_0/(n\pi)\) for odd \(n\). (This is Griffiths Example 3.3.)

Solution: \(X\) vanishes at \(x=0,a\) and \(Y\) at \(y=0,b\), so both are sinusoidal: \(X = \sin(n\pi x/a)\), \(Y = \sin(m\pi y/b)\). The constants must add to zero, so \(Z'' = \gamma^2 Z\) with

\[ \gamma_{nm} = \pi\sqrt{\left(\frac{n}{a}\right)^2+\left(\frac{m}{b}\right)^2} \]

and \(Z(0) = 0\) gives \(Z = \sinh(\gamma_{nm} z)\).

\[ V = \sum_{n,m} C_{nm}\sin\!\left(\frac{n\pi x}{a}\right)\sin\!\left(\frac{m\pi y}{b}\right)\sinh(\gamma_{nm}z) \]

At \(z=c\):

\[ C_{nm}\sinh(\gamma_{nm}c) = \frac{4}{ab}\int_0^a\!\!\int_0^b V_0\sin\!\left(\frac{n\pi x}{a}\right)\sin\!\left(\frac{m\pi y}{b}\right)\dd y\,\dd x = \frac{4V_0}{ab}\cdot\frac{2a}{n\pi}\cdot\frac{2b}{m\pi} = \frac{16V_0}{\pi^2\,nm} \]

for \(n, m\) both odd, zero otherwise:

\[ V = \frac{16V_0}{\pi^2}\sum_{n,m\ \text{odd}}\frac{1}{nm}\sin\!\left(\frac{n\pi x}{a}\right)\sin\!\left(\frac{m\pi y}{b}\right)\frac{\sinh(\gamma_{nm}z)}{\sinh(\gamma_{nm}c)} \]

Solution: \(X\) and \(Y\) each need two zeros, so both must be sinusoidal, contributing \(-k_x^2 - k_y^2\). The only way to cancel that is \(+\gamma^2\) from \(Z\), which makes \(Z\) exponential/hyperbolic. Rule of thumb: pairs of grounded walls get sines; the direction with the live face gets sinh, cosh or exponentials.

Solution: Superposition. Let \(V_{\text{top}}\) be the answer to (b). Then \(V_{\text{top}}(x,y,c-z)\) solves the problem with only the bottom face live. Their sum satisfies Laplace's equation and all six boundary conditions, so by uniqueness it's the answer.

Solution: From \(P_2 = \frac12(3\cos^2\theta-1)\): \(\cos^2\theta = \frac{2P_2+1}{3} = \frac13P_0+\frac23P_2\).

Solution: Inside, \(B_\ell = 0\); outside, \(A_\ell = 0\). Matching each Legendre term at \(r = R\):

\[ V_{\text{in}} = V_0\left[\frac13+\frac23\left(\frac{r}{R}\right)^2P_2(\cos\theta)\right],\qquad V_{\text{out}} = V_0\left[\frac13\frac{R}{r}+\frac23\left(\frac{R}{r}\right)^3P_2(\cos\theta)\right] \]

\[ \left.\frac{\partial V_{\text{out}}}{\partial r}\right|_R = -\frac{V_0}{R}\left[\frac13+2P_2\right],\qquad \left.\frac{\partial V_{\text{in}}}{\partial r}\right|_R = \frac{V_0}{R}\cdot\frac43P_2 \] \[ \sigma(\theta) = \frac{\varepsilon_0V_0}{R}\left[\frac13+2P_2+\frac43P_2\right] = \frac{\varepsilon_0V_0}{3R}\Big[1+10\,P_2(\cos\theta)\Big] \]

Only the \(\ell=0\) piece survives integration over the sphere, since \(\int_0^\pi P_2(\cos\theta)\sin\theta\,\dd\theta = 0\):

\[ Q = 4\pi R^2\cdot\frac{\varepsilon_0V_0}{3R} = \frac{4\pi\varepsilon_0RV_0}{3} \]

Check: far away \(V_{\text{out}}\to\frac{V_0R}{3r}\), which equals \(\frac{Q}{4\pi\varepsilon_0 r}\). ✓

Solution: Boundary conditions: \(V(R,\theta) = 0\) and \(V\to -E_0r\cos\theta\) as \(r\to\infty\). The far condition picks \(A_1 = -E_0\) and all other \(A_\ell = 0\). Then \(V(R,\theta)=0\) for every \(\theta\) needs \(B_1 = E_0R^3\) and every other \(B_\ell = 0\) (\(B_0 = 0\) also because the sphere is uncharged).

\[ V(r,\theta) = -E_0\left(r-\frac{R^3}{r^2}\right)\cos\theta \] \[ \sigma = -\varepsilon_0\left.\frac{\partial V}{\partial r}\right|_R = \varepsilon_0E_0\left(1+\frac{2R^3}{r^3}\right)\Big|_{r=R}\cos\theta = 3\varepsilon_0E_0\cos\theta \]

Solution: \(\frac{E_0R^3\cos\theta}{r^2}\) is a dipole potential \(\frac{p\cos\theta}{4\pi\varepsilon_0r^2}\) with \(\mathbf p = 4\pi\varepsilon_0R^3E_0\,\zhat\). The induced charge (positive on top, negative on the bottom) acts like a dipole at the centre pointing along the applied field.

Solution: \(Q = q-2q+q = 0\) and \(\mathbf p = qa\,\zhat-qa\,\zhat+0 = 0\). The quadrupole term: the charge at the origin has \(r'=0\), and \(P_2(\cos(\pi-\theta)) = P_2(-\cos\theta) = P_2(\cos\theta)\), so both outer charges contribute \(qa^2P_2(\cos\theta)\):

\[ V \approx \frac{1}{4\pi\varepsilon_0}\frac{2qa^2}{r^3}P_2(\cos\theta) = \frac{1}{4\pi\varepsilon_0}\frac{qa^2}{r^3}\left(3\cos^2\theta-1\right) \]

\[ V_{\text{exact}}(z=r) = \frac{q}{4\pi\varepsilon_0}\left[\frac{1}{r-a}-\frac{2}{r}+\frac{1}{r+a}\right] = \frac{q}{4\pi\varepsilon_0}\,\frac{2a^2}{r\,(r^2-a^2)} \approx \frac{1}{4\pi\varepsilon_0}\frac{2qa^2}{r^3} \]

At \(\theta = 0\) the expansion gives \(\frac{qa^2}{4\pi\varepsilon_0r^3}(3-1)\), the same. ✓

Solution: \(Q = 2q-q-q = 0\). \(\mathbf p = 2q(a\,\zhat)+(-q)(-a\,\zhat) = 3qa\,\zhat\).

\[ V \approx \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0}\frac{3qa\cos\theta}{r^2} \]

\(E_r = -\partial V/\partial r\) and \(E_\theta = -\frac1r\partial V/\partial\theta\):

\[ \mathbf E \approx \frac{p}{4\pi\varepsilon_0r^3}\left(2\cos\theta\,\rhat+\sin\theta\,\thetahat\right) = \frac{3qa}{4\pi\varepsilon_0r^3}\left(2\cos\theta\,\rhat+\sin\theta\,\thetahat\right) \]

Solution: Shift the origin by \(\mathbf d\): \(\mathbf p' = \sum q_i(\mathbf r_i'-\mathbf d) = \mathbf p-Q\,\mathbf d\). With \(Q = 0\), \(\mathbf p\) doesn't depend on the origin. A single charge \(q\) at the origin has \(\mathbf p = 0\), but measured from a shifted origin it has \(\mathbf p = -q\mathbf d \ne 0\). The dipole moment is only origin-independent when the total charge is zero.

\[ Q = \int_{-L}^{L}kz\,\dd z = 0,\qquad \mathbf p = \zhat\int_{-L}^{L}z\,(kz)\,\dd z = \frac{2kL^3}{3}\,\zhat,\qquad V\approx\frac{1}{4\pi\varepsilon_0}\frac{2kL^3}{3}\frac{\cos\theta}{r^2} \]

Solution: The magnetic force supplies the centripetal force: \(qvB = mv^2/r\), so

\[ r = \frac{mv}{qB}\quad(\text{so } p = qBr),\qquad \omega = \frac{v}{r} = \frac{qB}{m} \]

The cyclotron frequency doesn't depend on speed. Work: \(\dd W = \mathbf F\cdot\dd\mathbf l = q(\mathbf v\times\mathbf B)\cdot\mathbf v\,\dd t = 0\), since \(\mathbf v\times\mathbf B\perp\mathbf v\). Magnetic forces change the direction of motion, not the speed.

Solution: \(\mathbf v\times\mathbf B = vB\,(\yhat\times\xhat) = -vB\,\zhat\), so \(\mathbf F = q(E-vB)\,\zhat\). It vanishes when \(v = E/B\).

\[ m\ddot y = qB\dot z,\qquad m\ddot z = qE-qB\dot y\quad\Longrightarrow\quad \ddot y = \omega\dot z,\qquad \ddot z = \omega\left(\frac{E}{B}-\dot y\right) \]

Then \(\dddot y = \omega\ddot z = \omega^2\left(\frac EB-\dot y\right)\). With \(u = \dot y-\frac EB\), \(\ddot u = -\omega^2u\). The initial conditions \(\dot y(0) = 0\) and \(\ddot y(0) = \omega\dot z(0) = 0\) give \(u = -\frac EB\cos\omega t\):

\[ \dot y = \frac EB(1-\cos\omega t),\qquad \dot z = \frac{\ddot y}{\omega} = \frac EB\sin\omega t \] \[ y(t) = \frac{E}{\omega B}\left(\omega t-\sin\omega t\right),\qquad z(t) = \frac{E}{\omega B}\left(1-\cos\omega t\right) \]

A cycloid: a point on a wheel of radius \(E/(\omega B)\) rolling along \(y\) at speed \(E/B\), the velocity-selector speed. It reaches a maximum height of \(2mE/(qB^2)\).

Solution: The wire's field is \(\frac{\mu_0I}{2\pi s}\). The near side (parallel current) is attracted with force \(\frac{\mu_0II'a}{2\pi s}\). The far side (antiparallel, at \(s+a\)) is repelled with \(\frac{\mu_0II'a}{2\pi(s+a)}\). The two perpendicular sides feel equal and opposite forces that cancel.

\[ F = \frac{\mu_0II'a}{2\pi}\left(\frac1s-\frac{1}{s+a}\right) = \frac{\mu_0II'a^2}{2\pi s(s+a)}\quad\text{toward the wire} \]

Solution: The angle between \(\dd\mathbf l'\) and the separation vector is \(90^\circ-\theta\), so the cross product with the unit separation vector has magnitude \(\dd z'\cos\theta\). Then

\[ B = \frac{\mu_0I}{4\pi}\int\frac{\dd z'\cos\theta}{s^2\sec^2\theta} = \frac{\mu_0I}{4\pi}\int_{\theta_1}^{\theta_2}\frac{s\sec^2\theta\cos\theta}{s^2\sec^2\theta}\,\dd\theta = \frac{\mu_0I}{4\pi s}\left(\sin\theta_2-\sin\theta_1\right) \]

Square loop: each side has \(s = a/2\) and runs from \(-45^\circ\) to \(45^\circ\), so each gives \(\frac{\mu_0I}{4\pi(a/2)}\sqrt2\). All four point the same way:

\[ B = 4\cdot\frac{\sqrt2\,\mu_0I}{2\pi a} = \frac{2\sqrt2\,\mu_0I}{\pi a} \]

\[ B_z = \frac{\mu_0I}{4\pi}\oint\frac{\dd l'}{R^2+z^2}\cdot\frac{R}{\sqrt{R^2+z^2}} = \frac{\mu_0I}{4\pi}\frac{R\,(2\pi R)}{(R^2+z^2)^{3/2}} = \frac{\mu_0IR^2}{2\,(R^2+z^2)^{3/2}} \]

At \(z=0\): \(B = \frac{\mu_0I}{2R}\). For \(z\gg R\): \(B\approx\frac{\mu_0IR^2}{2z^3} = \frac{\mu_0}{4\pi}\frac{2m}{z^3}\) with \(m = I\pi R^2\), the on-axis field of a magnetic dipole.

Contrast with the tire: its \(E\) vanished at the centre, but \(B\) from the current ring is largest there. Far away, the tire's \(E\) falls like \(1/z^2\) because it has net charge; the loop's \(B\) falls like \(1/z^3\) because there are no magnetic monopoles, so the leading term is a dipole.

Solution: The semicircle gives half a loop: \(\frac{\mu_0I}{4R}\). Each straight wire ends right next to the centre, at perpendicular distance \(R\), and runs from \(\theta = 0\) to \(90^\circ\): \(\frac{\mu_0I}{4\pi R}(1-0)\). The current circulates the same way around the point in all three pieces, so they add:

\[ B = \frac{\mu_0I}{4R}+2\cdot\frac{\mu_0I}{4\pi R} = \frac{\mu_0I}{4R}\left(1+\frac2\pi\right) \]

Solution: The ring has charge \(\sigma\,2\pi s\,\dd s\) and goes around \(\omega/2\pi\) times per second, so \(\dd I = \sigma\omega s\,\dd s\) (equivalently, \(K = \sigma v = \sigma\omega s\)). From (b) with \(z=0\), it contributes \(\frac{\mu_0\,\dd I}{2s} = \frac{\mu_0\sigma\omega}{2}\dd s\).

\[ B = \int_0^R\frac{\mu_0\sigma\omega}{2}\,\dd s = \frac{\mu_0\sigma\omega R}{2}\quad\text{along }\boldsymbol\omega\ (\sigma\gt 0) \]

\[ I = \int_0^a ks\,(2\pi s)\,\dd s = \frac{2\pi ka^3}{3} \]

Inside: \(B\,(2\pi s) = \mu_0\int_0^s ks'\,2\pi s'\,\dd s' = \frac{2\pi\mu_0ks^3}{3}\). Outside: all of \(I\) is enclosed.

\[ \mathbf B_{\text{in}} = \frac{\mu_0ks^2}{3}\,\phihat = \frac{\mu_0Is^2}{2\pi a^3}\,\phihat,\qquad \mathbf B_{\text{out}} = \frac{\mu_0I}{2\pi s}\,\phihat \]

Both give \(\frac{\mu_0I}{2\pi a}\) at \(s = a\). ✓

\[ \mathbf B = \begin{cases}\dfrac{\mu_0Is}{2\pi a^2}\,\phihat & s\lt a\quad(I_{\text{enc}} = Is^2/a^2)\\[8pt] \dfrac{\mu_0I}{2\pi s}\,\phihat & a\lt s\lt b\\[8pt] 0 & s\gt b\quad(I_{\text{enc}} = 0)\end{cases} \]

Like the coax \(\mathbf E\) in Tutorial 5, the field outside a balanced cable is zero.

Solution: For current along \(+\xhat\), the right-hand rule puts \(\mathbf B\) along \(-\yhat\) above the slab and \(+\yhat\) below. The loop gives \(\oint\mathbf B\cdot\dd\mathbf l = 2B(z)\,l\), and for \(z\lt d\)

\[ I_{\text{enc}} = l\int_{-z}^{z}\alpha z'^2\,\dd z' = \frac{2\alpha l z^3}{3}\quad\Longrightarrow\quad B = \frac{\mu_0\alpha z^3}{3} \] \[ \mathbf B = \begin{cases}-\dfrac{\mu_0\alpha d^3}{3}\,\yhat & z\gt d\\[8pt] -\dfrac{\mu_0\alpha z^3}{3}\,\yhat & \lvert z\rvert\le d\\[8pt] +\dfrac{\mu_0\alpha d^3}{3}\,\yhat & z\lt -d\end{cases} \]

Compare Tutorial 3: \(\mathbf E = \frac{\alpha z^3}{3\varepsilon_0}\,\zhat\). Same \(z^3\) shape, but \(\mathbf B\) is parallel to the slab and perpendicular to \(\mathbf J\). Outside, it's a sheet with \(K = 2\alpha d^3/3\): \(B = \mu_0K/2\). ✓

Solution: Solenoid: \(B = \mu_0nI\) along the axis inside, 0 outside. Toroid: \(\mathbf B = \frac{\mu_0NI}{2\pi s}\,\phihat\) inside, 0 outside.

Ampère's law fails to give \(B\) for (i) and (iii): there's no loop along which \(B\) is constant and parallel (and a finite segment isn't a closed steady current on its own). Use Biot–Savart for those. (ii) and (iv) have enough symmetry.

\[ (\nabla\times\mathbf A)_z = \frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y} = \frac{B_0}{2}+\frac{B_0}{2} = B_0 \]

The \(x\) and \(y\) components involve \(\partial/\partial z\) or \(A_z\), which are all zero. \(\nabla\cdot\mathbf A = \partial_x(-B_0y/2)+\partial_y(B_0x/2) = 0\). Since \(-y\,\xhat+x\,\yhat = s\,\phihat\),

\[ \mathbf A = \frac{B_0s}{2}\,\phihat\qquad\left(\text{also } \mathbf A = \tfrac12\mathbf B\times\mathbf r\right) \]

Solution: Inside, the flux through the circle is \(\mu_0nI\,\pi s^2\); outside it's \(\mu_0nI\,\pi R^2\):

\[ \mathbf A_{\text{in}} = \frac{\mu_0nIs}{2}\,\phihat,\qquad \mathbf A_{\text{out}} = \frac{\mu_0nIR^2}{2s}\,\phihat \]

Inside matches (a) with \(B_0 = \mu_0nI\). Outside, \(\mathbf B = 0\) but \(\mathbf A\ne0\): \(\frac1s\frac{\dd}{\dd s}(sA) = 0\) there. ✓

\[ \mathbf B = \nabla\times\mathbf A = -\frac{\partial A_z}{\partial s}\,\phihat = -2ks\,\phihat \] \[ \mu_0\mathbf J = \nabla\times\mathbf B = \frac1s\frac{\partial}{\partial s}\left(sB_\phi\right)\zhat = \frac1s\frac{\partial}{\partial s}\left(-2ks^2\right)\zhat = -4k\,\zhat\quad\Longrightarrow\quad \mathbf J = -\frac{4k}{\mu_0}\,\zhat \]

Check: \(\nabla^2A_z = \frac1s\frac{\dd}{\dd s}\left(s\cdot2ks\right) = 4k = -\mu_0J_z\). ✓ And \(\nabla\cdot\mathbf A = 0\) since \(A_z\) doesn't depend on \(z\).

Solution: \(B_\perp\) is continuous (from \(\nabla\cdot\mathbf B = 0\)). The parallel part jumps:

\[ \mathbf B_{\text{above}}-\mathbf B_{\text{below}} = \mu_0\,\mathbf K\times\nhat \]

with \(\nhat\) pointing from below to above. \(\mathbf A\) is continuous; its normal derivative jumps: \(\frac{\partial\mathbf A_{\text{above}}}{\partial n}-\frac{\partial\mathbf A_{\text{below}}}{\partial n} = -\mu_0\mathbf K\).

Sheet check: \(\mathbf B = \mp\frac{\mu_0K}{2}\,\yhat\) above/below, so the jump is \(-\mu_0K\,\yhat\). And \(\mu_0K\,\xhat\times\zhat = -\mu_0K\,\yhat\). ✓ Compare \(E\): there \(E_\perp\) jumps by \(\sigma/\varepsilon_0\) and \(E_\parallel\) is continuous, the other way around.