SMT359 - 2015

General remarks: the solutions contained on this page have been provided by former SMT359 students according to their best endeavours. Users are encouraged to make amendments, comment on subjective points, make suggestions, and develop solutions for questions not yet solved.

Question 1
We apply Coulomb's law to find the position of $$C$$ in order for the net force on $$A$$ to be zero. Since all three charges have the same sign, $$C$$ will need to be on the opposite side of the $$x$$-axis as $$B$$.



\begin{align} &\textbf{F}_{AB}+\textbf{F}_{AC}=\textbf{0}\\ &\dfrac{q(2q)}{4\pi\epsilon_0a^2}(-\textbf{e}_x)+\dfrac{q(3q)}{4\pi\epsilon_0c^2}\textbf{e}_x=\textbf{0}\\ &\dfrac{2q^2}{4\pi\epsilon_0a^2}=\dfrac{3q^2}{4\pi\epsilon_0c^2}\\ &c^2=\dfrac{3q^2a^2}{2q^2} \end{align} $$

So $$C$$ should be placed at: $$x=-\sqrt{\tfrac{3}{2}}(a)$$.

Part (a)
Electrostatic fields must satisfy $$\text{curl }\textbf{E}=\textbf{0}$$. For the vector field $$\textbf{E}$$,



\begin{align} \text{curl }\textbf{E}&=\begin{vmatrix}\textbf{e}_x&\textbf{e}_y&\textbf{e}_z\\\tfrac{\partial}{\partial x}&\tfrac{\partial}{\partial y}&\tfrac{\partial}{\partial z}\\(y^2-z^2)&2xy&-2xz\end{vmatrix}\\ &=(0-0)\textbf{e}_x-(-2z-(-2z))\textbf{e}_y+(2y-2y)\textbf{e}_z\\ &=\textbf{0}, \end{align} $$

so $$\textbf{E}$$ could represent an electrostatic field.

Part (b)
Since $$\textbf{E}$$ is an electrostatic field, we can find the work done by using the following integral

W=-\!\!\!\int_{(0,0,0)}^{(1,2,0)}\!\!\!q\mathbf{E}\cdot\mathrm{d}\mathbf{l} $$

We will use a path from (0,0,0) to (1,0,0), then from (1,0,0) to (1,2,0):



\begin{align} W&=-q\left(\int_{x=0}^{x=1}\!\!\!E_x\cdot\mathrm dx+\int_{y=0}^{y=2}\!\!\!E_y\cdot\mathrm dy+\int_{z=0}^{z=0}z\!\!\!E_z\cdot\mathrm dz\right)\\ &=-q\left(\int_{x=0}^{x=1}\!\!\!(y^2-z^2)\mathrm dx+\int_{y=0}^{y=2}\!\! 2xy\ \mathrm dy+0\right)\\ &=-q\left(\left[(y^2-z^2)x\right]_{x=0}^{x=1}+\left[xy^2\right]_{y=0}^{y=2}\right)\\ &=-q((y^2-z^2)+4x) \end{align} $$

As we move from (0,0,0) to (1,0,0) then $$y$$ and $$z$$ are both zero, so   $$y^2-z^2=0$$.

As we move from (1,0,0) to (1,2,0) then   $$x=1\Rightarrow4x=4$$.

So the work done is   $$W=-q(0+4)=-4q$$.

Question 3
[ This is a similar system to Book 2, page 94, figure 5.2, but with different orientation, e.g. $$x$$ upwards, and $$z$$ towards page. ]

We use Ampère's law   $$\int_C\textbf{B}\cdot\mathrm{d}\textbf{l}=\mu_0\int_S\textbf{J}\cdot\mathrm{d}\textbf{S}$$,

with a closed rectangular loop $$C$$, with length $$l$$ parallel to $$y$$ axis, enclosing a surface $$S$$, giving


 * $$\textbf{B}=B_y\textbf{e}_y \Rightarrow \int_C\textbf{B}\cdot\mathrm{d}\textbf{l}=2B_yl$$
 * $$\int_S\textbf{J}\cdot\mathrm{d}\textbf{S}=\eta l$$

So by Ampère's law:   $$2B_yl=\mu_0\eta l \Rightarrow B_y=\dfrac{\mu_0\eta}{2}$$

So the magnetic field is   $$\textbf{B}=\pm\dfrac{\mu_0\eta}{2}\textbf{e}_y$$

Question 4
Applying Ampère's law, the magnetic field around each wire is:

\begin{align} \oint_C\textbf{B}\cdot\mathrm{d}\textbf{l}=\mu_0\int_S\textbf{J}\cdot\mathrm{d}\textbf{S}\\ \Rightarrow 2\pi rB=\mu_0I\\ \Rightarrow \textbf{B}=\dfrac{\mu_0I}{2\pi r}\textbf{e}_\phi \end{align} $$

So the magnetic field on a particle with distance $$a$$ from one wire, and distance $$2a$$ from the other is:



\begin{align} \textbf{B}&=B_\phi\textbf{e}_\phi=B_z\textbf{e}_z\\ &=\dfrac{\mu_0I}{2\pi a}(-\textbf{e}_z)+\dfrac{\mu_0I}{2\pi2a}\textbf{e}_z\\ \Rightarrow \textbf{B}&=-\dfrac{\mu_0I}{4\pi a}\textbf{e}_z \end{align} $$

Hence
 * $$\textbf{F}=q(\textbf{v}\times\textbf{B})=(-q)\left(v\textbf{e}_x\times\left(-\dfrac{\mu_0I}{4\pi a}\textbf{e}_z\right)\right)$$
 * $$=\dfrac{qv\mu_0I}{4\pi a}(-\textbf{e}_y)$$

N.B. Depending on which wire the particle is assumed to be closest to, the magnetic field could instead have been positive (but with same magnitude), giving a force vector in the $$+\textbf{e}_y$$ direction, but with same magnitude.

Part (a)
The required figure is similar to the one in Book 2, page 257, exercise answer 4.7, with A replaced with (1,1), B with (-1,1), C with (-1,-1) and D with (1,-1).

Part (b)
The electrostatic potential is:



V=\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_A}-\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_B}+\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_C}-\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{r_D} $$

where
 * $$r_A$$ is the distance of (2,2) from (1,1),
 * $$r_B$$ is the distance of (2,2) from (-1,1),
 * $$r_C$$ is the distance of (2,2) from (-1,-1),
 * $$r_D$$ is the distance of (2,2) from (1,-1).

$$\therefore V=\dfrac{q}{4\pi\epsilon_0\sqrt{2}}-\dfrac{q}{4\pi\epsilon_0\sqrt{10}}+\dfrac{q}{4\pi\epsilon_0\sqrt{18}}-\dfrac{q}{4\pi\epsilon_0\sqrt{10}} $$

Question 6
We use the boundary conditions for the electric field and electric displacement at the interface between different dielectric materials:


 * $$E_{R\parallel}=E_{S\parallel}$$
 * $$\Rightarrow E_R\cos30^{\circ}=E_S\cos\theta$$   (Equation 1)

And $$D_{R\perp}=D_{S\perp}$$, and since the dielectrics are LIH materials, $$D=\epsilon\epsilon_0E$$, giving


 * $$D_R=2\epsilon_0E_R$$ and $$D_S=3\epsilon_0E_S$$
 * $$\Rightarrow 2\epsilon_0E_{R\perp}=3\epsilon_0E_{S\perp}$$
 * $$\Rightarrow 2\epsilon_0E_R\sin30^{\circ}=3\epsilon_0E_S\sin\theta$$   (Equation 2)

Dividing Equation 2 by Equation 1 gives:



\begin{align} 2\epsilon_0\tan30^{\circ}&=3\epsilon_0\tan\theta\\ \Rightarrow \tan\theta&=\tfrac{2}{3}\tan30^{\circ}\\ \Rightarrow \theta&=\tan^{-1}(\tfrac{2}{3}\tan30^{\circ})\\ \end{align} $$

So   $$\theta=21.1^{\circ}$$ (1 d.p.).

Question 7
Applying the following equation from the Equations Booklet:

\begin{align} &V(\textbf{r}_2)-V(\textbf{r}_1)=-\int_{\textbf{r}_1}^{\textbf{r}_2}\textbf{E}\cdot\mathrm{d}\textbf{l}\\ \Rightarrow &V-0=Ed \Rightarrow V=Ed\\ \end{align} $$

And applying the following equation from the Equations Booket, with $$\textbf{B}=\textbf{0}$$, since there's no magnetic field:

\begin{align} &\textbf{J}=\sigma(\textbf{E}+\textbf{v}\times\textbf{B})\\ \Rightarrow\text{ } &J=\sigma E\\ \end{align} $$

And since $$J=\dfrac{I}{A}$$, then
 * $$\dfrac{I}{A}=\sigma E \Rightarrow I=\sigma EA$$.

Now we can use Ohm's law to find the resistance of the arrangement:

R=\dfrac{V}{I}=\dfrac{Ed}{\sigma EA}=\dfrac{d}{\sigma A} $$

Part (a)
A superconductor exhibits the Meissner effect when in its superconducting state, i.e. when a magnetic field is applied, the field is excluded ($$\textbf{B}=\textbf{0}$$) throughout its interior. A perfect conductor does not exhibit the Meissner effect.

Part (b)
For a magnetic field being applied parallel to a thin specimen of the superconductor, when it reaches the critical magnetic field strength $$B_c$$, the specimen stops being superconductive. The critical current $$I_C$$ is the maximum current a superconducting wire can carry with zero resistance.

Part (a)
$$ \begin{align} \text{curl }\textbf{E}&=\begin{vmatrix}\textbf{e}_x&\textbf{e}_y&\textbf{e}_z\\\tfrac{\partial}{\partial x}&\tfrac{\partial}{\partial y}&\tfrac{\partial}{\partial z}\\E_0\exp[i(ky-\omega t)]&0&0\end{vmatrix}\\ &=(0-0)\textbf{e}_x-(0-0)\textbf{e}_y+(0-ikE_0\exp[i(ky-\omega t)])\textbf{e}_z\end{align} $$

We use the differential form of Faraday's law:

\begin{align} &\text{curl }\textbf{E}=-\dfrac{\partial\textbf{B}}{\partial t}\\ &\Rightarrow -ikE_0\exp[i(ky-\omega t)]\textbf{e}_z=-\dfrac{\partial\textbf{B}}{\partial t}\\ &\therefore \textbf{B}=\int ikE_0\exp[i(ky-\omega t)]\mathrm{d}t\,\,\textbf{e}_z=-\dfrac{ik}{i\omega}E_0\exp[i(ky-\omega t)]\textbf{e}_z \end{align} $$

And since $$c=\dfrac{w}{k}$$, then $$\textbf{B}=-\dfrac{E_0}{c}\exp[i(ky-\omega t)]\textbf{e}_z$$.

Part (b)
Using the formula for the mean energy flux density vector from the Equations Booklet:

\begin{align} &\overline{\textbf{N}}=\tfrac{1}{2}\epsilon\epsilon_0E_0^2\dfrac{c}{n}\hat{\textbf{k}}\\ &N=\tfrac{1}{2}\epsilon_0E_0^2c\\ & \therefore E_0=\sqrt{\dfrac{2N}{\epsilon_0c}}=\sqrt{\dfrac{2\times0.2}{8.85\times10^{-12}\times3\times10^{8}}}=12.3 \text{ NC}^{-1} (1\text{ d.p.}) \end{align} $$

Part (a)
We know from the inverse square law for electromagnetic radiation that the power per unit area (i.e. "dose") falls off as the inverse square of the radial distance, so


 * $$P\propto\dfrac{1}{r^2} \Rightarrow P=\dfrac{k}{r^2}$$, where $$k$$ is a constant

Let the power at the detector be $$P_d$$ and at the team be $$P_t$$, let $$d$$ be the distance of the team from the source, and assume $$\tfrac{P_t}{P_d}=\tfrac{1}{10000}$$.

Then $$P_d=\dfrac{k}{0.04^2}$$ and $$P_t=\dfrac{k}{d^2}$$, so


 * $$\dfrac{P_t}{P_d}=\dfrac{k}{d^2}\div\dfrac{k}{0.04^2}=0.0001$$
 * $$\therefore \dfrac{0.04^2}{d^2}=0.0001\Rightarrow d=4$$.

So the team should be instructed to stand 4 metres away from the source.

Part (b)
We have assumed that none of the radiation is absorbed before it reaches the team, so the energy is conserved as it radiates outwards.

Part (a)


\begin{align}&\epsilon=(n_{re}+in_{im})^2=40+10i\\ &n_{re}^2-n_{im}^2+2n_{re}n_{im}i=40+10i\\ \end{align} $$

Equating imaginary parts: $$2n_{re}n_{im}=10\Rightarrow n_{im}=\dfrac{5}{n_{re}}$$.

Equating real parts, and substituting the above expression for $$n_{im}$$:

\begin{align} &n_{re}^2-\left(\dfrac{5}{n_{re}}\right)^2=40 \Rightarrow n_{re}^4-40n_{re}^2-25=0\\ &n_{re}^2=\dfrac{-(-40)\pm\sqrt{(-40)^2-4(1)(-25)}}{2(1)} \Rightarrow n_{re}^2=20\pm\sqrt{425}\\ \end{align} $$ We use the non-negative value for $$n_{re}^2$$ to find $$n_{re}$$:

\begin{align} &n_{re}=\sqrt{20+\sqrt{425}}\approx6.37\\ &v_{phase}=c/n_{re}=\dfrac{3\times10^8}{6.37}=4.7\times10^7\text{ ms}^{-1}\end{align} $$.

Part (b)
The absorption length is: $$\dfrac{1}{k_{im}}$$

From part (a) we know that $$n_{im}=\dfrac{5}{n_{re}}\approx\dfrac{5}{6.37}\approx0.78$$


 * $$k_{im}=\dfrac{\omega}{c}n_{im}$$ and  $$\omega=2\pi f$$, so: $$k_{im}=\dfrac{2\pi f}{c}n_{im}$$

Hence the absorption length is:
 * $$\dfrac{1}{k_{im}}=\dfrac{c}{2\pi fn_{im}}=\dfrac{3\times10^8}{2\pi\times2.45\times10^9\times0.78}\approx0.025\text{ m}$$.

Question 12
The pulse is broader due to dispersion, caused by the frequency dependence of the real part of the fibre's relative permittivity, $$\epsilon_{real}(\omega)$$. This also causes its intensity (amplitude) to decrease.

However, the pulse's intensity has been reduced by a greater factor than its increase in width, so absorption has also taken place. As the light wave travels through the fibre its electromagnetic energy is being transformed into thermal energy. This is due to the complex part of the fibre's refractive index.

Part (a)
Since we have spherical symmetry, $$\textbf{E}=E_r(\textbf{r})\textbf{e}_r$$.

Applying the integral form of Gauss's law, for $$r>R$$:

\begin{align} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow &E_r(\textbf{r})\times4\pi r^2=\dfrac{1}{\epsilon_0}\int_0^{2\pi}\int_0^{\pi}\int_0^R\rho r^2\sin\theta\ \mathrm dr\ \mathrm d\theta\ \mathrm d\phi\\ \Rightarrow &E_r(\textbf{r})=\dfrac{\rho 4\pi R^3}{3\epsilon_0}\div 4\pi r^2\Rightarrow \textbf{E}=\dfrac{\rho R^3}{3\epsilon_0 r^2}\textbf{e}_r\\ \end{align} $$ So the electrostatic potential energy required to take a unit test charge from infinity to a point outside the sphere at $$r$$ is:


 * $$V_1(r)=-\int_{\textbf{r}_1}^{\textbf{r}_2}\textbf{E}\cdot\mathrm d\textbf{l}=-\int_\infty^r\dfrac{\rho R^3}{3\epsilon_0r^2}\ \mathrm dr=-\dfrac{\rho R^3}{3\epsilon_0}\left[-\dfrac{1}{r}\right]_\infty^r=\dfrac{\rho R^3}{3\epsilon_0r}$$

Part (b)
Applying the integral form of Gauss's law, for $$r\le R$$:

\begin{align} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow &E_r(\textbf{r})\times4\pi r^2=\dfrac{1}{\epsilon_0}\int_0^{2\pi}\int_0^{\pi}\int_0^r\rho r^2\sin\theta\ \mathrm dr\ \mathrm d\theta\ \mathrm d\phi\\ \Rightarrow &E_r(\textbf{r})=\dfrac{\rho 4\pi r^3}{3\epsilon_0}\div 4\pi r^2\Rightarrow \textbf{E}=\dfrac{\rho r}{3\epsilon_0}\textbf{e}_r\\ \end{align} $$

So the electrostatic potential difference between a point on the surface of the sphere to a point on the interior of the sphere at $$r$$ is:


 * $$V(r) - V(R)=-\int_{\textbf{r}_1}^{\textbf{r}_2}\textbf{E}\cdot\mathrm d\textbf{l}=-\int_R^r\dfrac{\rho r}{3\epsilon_0}\ \mathrm dr=-\dfrac{\rho}{3\epsilon_0}\left[\dfrac{r^2}{2}\right]_R^r=\dfrac{\rho}{6\epsilon_0}(R^2-r^2)$$

The electrostatic potential energy is then this potential difference multipied by the test charge, $$q$$:


 * $$\dfrac{q \rho}{6\epsilon_0}(R^2-r^2)$$

Part (c)
The electrostatic potential at infinity is $$V=0$$. So the electrostatic potential inside the sphere is the potential energy needed to take a test charge from infinity to the surface of the sphere, added to the potential energy to take it from the surface to a point on the interior at $$r$$:
 * $$V_{in}=V_1(R)+V_2(r)=\dfrac{\rho R^3}{3\epsilon_0R}+\dfrac{\rho}{6\epsilon_0}(R^2-r^2)=\dfrac{\rho}{6\epsilon_0}(3R^2-r^2)$$

The electrostatic potential outside the sphere is simply the potential energy needed to take a test charge from infinity to a point outside the sphere at $$r$$, i.e.
 * $$V_{out}=V_1(r)=\dfrac{\rho R^3}{3\epsilon_0r}$$

Part (d)
From the Equations Booklet, $$\textbf{E}=-\mathrm{grad}\ V$$, so by using the answers from part (c):

The electric field inside the sphere is:


 * $$-\mathrm{grad}\ V_{in}=-\dfrac{\partial V_{in}}{\partial r}\textbf{e}_r=-\dfrac{\partial}{\partial r}\left(\dfrac{\rho}{6\epsilon_0}(3R^2-r^2)\right)\textbf{e}_r=\dfrac{\rho r}{3\epsilon_0}\textbf{e}_r$$

and the electric field outside the sphere is:


 * $$-\mathrm{grad}\ V_{out}=-\dfrac{\partial V_{out}}{\partial r}\textbf{e}_r=-\dfrac{\partial}{\partial r}\left(\dfrac{\rho R^3}{3\epsilon_0r}\right)\textbf{e}_r=\dfrac{\rho R^3}{3\epsilon_0r^2}\textbf{e}_r$$.

Part (a)
An electromagnetic wave transports energy as it travels through space, with its electric and magnetic waves having equal energy densities. Since there is a magnetic field inside the solenoid, then it has an energy density.

From the Equations Booklet, the magnetic field inside the solenoid is:
 * $$\textbf{B}=\mu_0NI\textbf{e}_z=\mu_0Nkt\textbf{e}_z\ \Rightarrow\ B=\mu_0Nkt

$$

From the Equations Booklet, the energy density associated with a magnetic field is:
 * $$\ u=\tfrac{1}{2}\textbf{B}\cdot\textbf{H}$$.

We assume that the solenoid is made from an LIH material with magnetic permeability $$\mu$$, so
 * $$B=\mu\mu_0H\ \Rightarrow\ H=\dfrac{B}{\mu\mu_0}$$
 * $$u=\tfrac{1}{2}BH=\dfrac{B^2}{2\mu\mu_0}$$,

and substituting $$B=\mu_0Nkt$$ gives:
 * $$u=\dfrac{(\mu_0Nkt)^2}{2\mu\mu_0}=\dfrac{\mu_0N^2k^2t^2}{2\mu}$$.

Part (b)
Inside the solenoid

We use the differential form of Faraday's Law, with $$\textbf{B}=\mu_0Nkt\textbf{e}_z$$


 * $$\mathrm{curl}\ \textbf{E}=-\dfrac{\partial\textbf{B}}{\partial t}=-\mu_0Nk\textbf{e}_z$$

And with cylindrical coordinates, since $$\textbf{E}=E_\phi(r)\textbf{e}_\phi$$
 * $$\mathrm{curl}\ \textbf{E}=\dfrac{1}{r}\begin{vmatrix}\textbf{e}_r&r\textbf{e}_\phi&\textbf{e}_z\\\tfrac{\partial}{\partial r}&\tfrac{\partial}{\partial\phi}&\tfrac{\partial}{\partial z}\\0&rE_\phi(r)&0\end{vmatrix}=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(rE_\phi(r)\right)\textbf{e}_z$$

And so
 * $$\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(rE_\phi(r)\right)\textbf{e}_z=-\mu_0Nk\textbf{e}_z$$
 * $$\Rightarrow\ E_\phi(r)=-\dfrac{\mu_0Nkr}{2}$$

So the electric field inside the solenoid is: $$\ \textbf{E}=-\dfrac{\mu_0Nkr}{2}\textbf{e}_\phi$$.

Outside the solenoid

There is no magnetic field outside the solenoid ($$\textbf{B}=0$$), so the differential form of Faraday's law gives


 * $$\mathrm{curl}\ \textbf{E} = -\dfrac{\partial\textbf{B}}{\partial t}=\textbf{0}$$
 * $$\Rightarrow\ \dfrac{1}{r}\dfrac{\partial}{\partial r}\left(rE_\phi(r)\right)\textbf{e}_z=\textbf{0}$$
 * $$\Rightarrow\ E_\phi(r)=\dfrac{C}{r}$$, where $$C$$ is a constant

Since the electric field is continuous at the solenoid's surface, we can find $$C$$ by equating the expressions for $$E_\phi(r)$$ inside and outside the solenoid, with $$r=R$$:


 * $$-\dfrac{\mu_0NkR}{2}=\dfrac{C}{R}\ \Rightarrow\ C=-\dfrac{\mu_0NKR^2}{2}$$.

So the electric field outside the solenoid is: $$\ \textbf{E}=-\dfrac{\mu_0NKR^2}{2r}\textbf{e}_\phi$$

Part (c)
From the Lorentz force law in the Equations Booklet, we know that the total electromagnetic force on a point charge $$q$$ is:
 * $$\textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B})\ $$,

We know from parts (a) and (b) that there is both a constant electric field and an increasing magnetic field inside the solenoid.

Since the particle is charged, the electric field of $$\ \textbf{E}=-\dfrac{\mu_0Nkr}{2}\textbf{e}_\phi$$ will give a constant force $$q\textbf{E}$$. Hence the particle will be accelerating in an azimuthal direction, either $$-\textbf{e}_\phi$$ if positively-charged, or $$\textbf{e}_\phi$$ if negatively-charged.

Since the electric field gives the particle velocity in an $$\textbf{e}_\phi$$ direction, the Lorentz force law means that the increasing magnetic field (with $$\textbf{e}_z$$ component) will also contribute an increasing force. By the right-hand rule, this force $$q(\textbf{v}\times\textbf{B})$$ will have an $$\textbf{e}_\phi\times \textbf{e}_z=\textbf{e}_r$$ component, so the particle will be moving towards or away from the solenoid's axis, depending on whether the particle is positively- or negatively-charged.

So the particle will be moving around the solenoid's axis, with either increasing or decreasing distance from the axis, depending on the sign of its charge.

Part (a)
The symmetry means that we should use cylindrical coordinates, with the $$z$$-axis aligned with the axis of the rod. The symmetry also means that the magnetic intensity field $$\textbf{H}$$ is in the azimuthal direction and depends only on $$r$$, i.e. $$\textbf{H}=H_\phi(r)\textbf{e}_\phi$$, inside and outside the rod.

If the rod is viewed with the current moving to the right, then by the right-hand grip rule the field $$\textbf{H}$$ is coming out of the page above the rod's axis, and into the page below it.

Part (b)
Inside the rod

We use the integral version of Ampère's law in media, with the Ampèrian surface being a disk of radius $$r$$, and no electric field intensity, i.e. $$\textbf{D}=0$$


 * $$\oint_C\textbf{H}\cdot\mathrm{d}\mathbf{l}=\int_S\textbf{J}_f\cdot\mathrm{d}\textbf{S}$$
 * $$\Rightarrow\ H_\phi(r)\times2\pi r=\dfrac{\pi r^2}{\pi R^2}I$$
 * $$\Rightarrow\ H_\phi(r)=\dfrac{rI}{2\pi R^2}$$,     so   $$\textbf{H}_{in}=\dfrac{rI}{2\pi R^2}\textbf{e}_\phi$$

Outside the rod

We use the integral version of Ampère's law again:


 * $$\oint_C\textbf{H}\cdot\mathrm{d}\mathbf{l}=\int_S\textbf{J}_f\cdot\mathrm{d}\textbf{S}$$
 * $$\Rightarrow\ H_\phi(r)\times2\pi r=I$$
 * $$\Rightarrow\ H_\phi(r)=\dfrac{I}{2\pi r}$$,     so    $$\textbf{H}_{out}=\dfrac{I}{2\pi r}\textbf{e}_\phi$$

Part (c)
We know from the Equations Booklet that for LIH materials,
 * $$\textbf{B}=\mu\mu_0\textbf{H}$$

Since the rod is an LIH material with $$\mu=\mu_1$$, then inside the rod we have
 * $$\textbf{B}_{in}=\mu_1\mu_0\textbf{H}_{in}=\dfrac{\mu_1\mu_0rI}{2\pi R^2}\textbf{e}_\phi$$

Since the insulator is an LIH material with $$\mu=\mu_2$$, then outside the rod we have
 * $$\textbf{B}_{out}=\mu_2\mu_0\textbf{H}_{out}=\dfrac{\mu_2\mu_0I}{2\pi r}\textbf{e}_\phi$$

Part (d)
We know from the Equations Booklet that the energy density for a magnetic field is:

\begin{align} u&=\tfrac{1}{2}\textbf{B}\cdot\textbf{H}\\ &=\tfrac{1}{2}\textbf{B}_{in}\cdot\textbf{H}_{in}\\ &=\tfrac{1}{2}\left(\dfrac{\mu_1\mu_0rI}{2\pi R^2}\textbf{e}_\phi\right)\cdot\left(\dfrac{rI}{2\pi R^2}\textbf{e}_\phi\right)\\ &=\dfrac{\mu_1\mu_0r^2I^2}{8\pi^2R^4} \end{align} $$

We can find the total energy of the magnetic field inside the rod per unit length by finding
 * $$U=\int u\ \mathrm{d}\tau$$,

where $$\mathrm{d}\tau$$ is a cylindrical shell with radius $$r$$, thickness $$\partial r$$ and length 1, so $$\mathrm{d}\tau=2\pi r\ \mathrm{d}r$$, giving

\begin{align} U&=\int_0^R\dfrac{\mu_1\mu_0r^2I^2}{8\pi^2R^4}\ 2\pi r\mathrm{d}r\\ &=\dfrac{\mu_1\mu_0I^2}{4\pi R^4}\int_0^Rr^3.\mathrm{d}r\\ &=\dfrac{\mu_1\mu_0I^2}{4\pi R^4}\left[\dfrac{r^4}{4}\right]_0^R=\dfrac{\mu_1\mu_0I^2}{16\pi} \end{align} $$

Part (a)
We will use Maxwell's equations to derive the equation of continuity for electric charge,
 * $$\dfrac{\partial\rho}{\partial t}+\mathrm{div}\ \textbf{J}=0\ $$.

We start with the differential form of the Ampère-Maxwell law
 * $$\mathrm{curl}\ \textbf{B}=\mu_0\textbf{J}+\epsilon_0\mu_0\dfrac{\partial\textbf{E}}{\partial t}$$,

and take the divergence of both sides
 * $$\mathrm{div}\ (\mathrm{curl}\ \textbf{B})=\mathrm{div}\ \left(\mu_0\textbf{J}+\epsilon_0\mu_0\dfrac{\partial\textbf{E}}{\partial t}\right).$$

Since the divergence of any curl is equal to zero
 * $$\mathrm{div}\left(\mu_0\textbf{J}+\epsilon_0\mu_0\dfrac{\partial\textbf{E}}{\partial t}\right)=0$$
 * $$\Rightarrow\ \mu_0\ \mathrm{div}\ \textbf{J}+\epsilon_0\mu_0\ \mathrm{div}\ \dfrac{\partial\textbf{E}}{\partial t}=0$$.

We can divide both sides by $$\mu_0$$, since $$\mu_0\ne0$$
 * $$\mathrm{div}\ \textbf{J}+\epsilon_0\ \mathrm{div}\ \dfrac{\partial\textbf{E}}{\partial t}=0$$,

then use a divergence identity:
 * $$\mathrm{div}\ \textbf{J}+\epsilon_0\dfrac{\partial}{\partial t}(\mathrm{div}\ \textbf{E})=0$$.

Now we apply the differential form of Gauss' law, $$\mathrm{div}\ \textbf{E}=\dfrac{\rho}{\epsilon_0}$$
 * $$\mathrm{div}\ \textbf{J}+\epsilon_0\dfrac{\partial}{\partial t}\left(\dfrac{\rho}{\epsilon_0}\right)=0$$
 * $$\Rightarrow\ \mathrm{div}\ \textbf{J}+\epsilon_0\left(\dfrac{1}{\epsilon_0}\dfrac{\partial\rho}{\partial t}\right)=0$$
 * $$\Rightarrow\ \dfrac{\partial\rho}{\partial t}+\mathrm{div}\ \textbf{J}=0$$, which is the equation of continuity.

Part (b)
If magnetic monopoles were discovered, the no-monopole law could be analogous to Gauss's law, giving
 * $$\mathrm{div}\ \textbf{B}=\dfrac{\rho_m}{\mu_0}$$ ,

where $$\rho_m$$ is the magnetic charge density (similar to the electric charge density $$\rho$$).

Part (c)
In the derivation of the equation of continuity for electric charge in part (a), we used only the Ampère-Maxwell law and Gauss's law from Maxwell's equations. Since the no-monopole law was not used at all, then the derivation would still be valid with a modified 'no-monopole law'.

Part (d)
Faraday's law states that $$\mathrm{curl}\ \textbf{E}$$ is equal to the negative change of the magnetic field with time. With magnetic monopoles, this expression must also now include the effect on $$\mathrm{curl}\ \textbf{E}$$ of the flowing magnetic charge per unit area (new term $$\textbf{J}_m$$), combined with the permittivity of free space ($$\epsilon_0$$)
 * $$\mathrm{curl}\ \textbf{E}=-\dfrac{\partial\textbf{B}}{\partial t}-\epsilon_0\textbf{J}_m$$.

Please could someone check the above answer, and amend or expand it. Many thanks.

Part (a)
The sketch is similar to Figure 3.10(a) on page 81 of Book 3, but change:
 * the $$x$$-axis to be the $$z$$-axis, positive in the same direction,
 * the $$z$$-axis to be the $$x$$-axis, with positive/negative reversed
 * the vector subscripts $$_i$$, $$_r$$ and $$_t$$ to $$_{inc}$$, $$_{ref}$$ and $$_{trs}$$ respectively
 * "index $$n_1$$" to "index $$n$$",
 * "index $$n_2$$ to "vacuum",

and add:
 * the angles $$\theta_{inc}$$, $$\theta_{ref}$$ and $$\theta_{trs}$$ that the vectors $$\textbf{K}_{inc}$$, $$\textbf{K}_{ref}$$ and $$\textbf{K}_{trs}$$ respectively each make with the $$z$$-axis.

Part (b)
Since
 * $$\textbf{k}_{inc}=\dfrac{n\omega}{c}(-\cos\theta_{inc}\textbf{e}_x+\sin\theta_{inc}\textbf{e}_y)$$ ,

then the wave vector of the reflected and transmitted waves are

\begin{align} &\textbf{k}_{ref}=\dfrac{n\omega}{c}(\cos\theta_{ref}\textbf{e}_x+\sin\theta_{ref}\textbf{e}_y)\\ &\textbf{k}_{trs}=\dfrac{\omega}{c}(-\cos\theta_{trs}\textbf{e}_x+\sin\theta_{trs}\textbf{e}_y) \end{align} $$

Applying the boundary conditions for the electric field, we consider the net electric fields for
 * $$x>0$$,   which is     $$\textbf{E}_{inc}+\textbf{E}_{ref}$$, and
 * $$x<0$$,   which is     $$\textbf{E}_{trs}$$,

and these must be equal at the boundary:
 * $$E_{inc0}\exp[i(\textbf{k}_{inc}\cdot \textbf{r}-\omega t)]\textbf{e}_z+E_{ref0}\exp[i(\textbf{k}_{inc}\cdot \textbf{r}-\omega t)]\textbf{e}_z$$
 * $$=E_{trs0}\exp[i(\textbf{k}_{inc}\cdot \textbf{r}-\omega t)]\textbf{e}_z$$.

And hence for all points in the plane $$x=0$$,
 * $$\textbf{k}_{inc}\cdot\textbf{r}=\textbf{k}_{ref}\cdot\textbf{r}=\textbf{k}_{trs}\cdot\textbf{r}$$,

and substituting the expressions we have for these vectors gives:
 * $$\dfrac{n\omega}{c}(-\cos\theta_{inc}x+\sin\theta_{inc}y)=\dfrac{n\omega}{c}(\cos\theta_{ref}x+\sin\theta_{ref}y)$$
 * $$=\dfrac{\omega}{c}(-\cos\theta_{trs}x+\sin\theta_{trs}y)$$

Since all the points lie in the plane $$x=0$$ , this becomes
 * $$\dfrac{n\omega}{c}(\sin\theta_{inc}y)=\dfrac{n\omega}{c}(\sin\theta_{ref}y)=\dfrac{\omega}{c}(\sin\theta_{trs}y)$$
 * $$\Rightarrow\ n\sin\theta_{inc}=n\sin\theta_{ref}=\sin\theta_{trs}$$,   as required.

Part (c)
Assuming no free charges or currents in the medium (or vacuum), the boundary conditions on $$\textbf{E}$$, $$\textbf{D}$$, $$\textbf{B}$$ and $$\textbf{H}$$ at the interface are the same as when dealing with static fields. But since the waves are travelling normal to the scattering plane, the components of $$\textbf{D}$$ and $$\textbf{B}$$ perpendicular to the interface are both zero, so are not useful.

The components of $$\textbf{H}$$ and $$\textbf{E}$$ are also continuous as they cross the boundary, and crucially are both perpendicular to the interface, so we can use the property $$E_{inc}+E_{ref}=E_{trs}$$ and $$H_{int}+H_{ref}=H_{trans}$$.

We can find expressions for the three $$H$$ fields from $$E_{inc}$$, $$n$$ and $$c$$, and using the property $$H=B/\mu_o$$.

This will finally give us two simultaneous equations we can use to find: the amplitude transmission ratio from the ratio of amplitudes of the transmitted and incident fields, and the amplitude reflection ratio from the ratio of the amplitudes of the reflected and incident fields.

Part (d)
When light (an electromagnetic wave) is incident on the surface between two dielectric media at the Brewster angle $$\theta_B$$, the part of the wave polarised in the scattering plane is completely transmitted. This means that the reflected wave is completely polarised normal to the scattering plane.

An expression for $$\theta_B$$ between two media with refractive indices $$n_1$$ and $$n_2$$ is, from the Equations booklet:
 * $$\theta_B=\mathrm{tan}^{-1}(n_2/n_1)$$

Part (a)
We know from the Equations booklet that

\begin{align} &\text{curl}\ \textbf{H}=\textbf{J}_f+\dfrac{\partial\mathbf{D}}{\partial t}\\ &\textbf{J}=\sigma(\textbf{E}+\textbf{v}\times\textbf{B}),\qquad \text{ and } \textbf{B}=0 \Rightarrow \textbf{J}=\sigma\textbf{E}\\ &\textbf{D}=\epsilon\epsilon_0\textbf{E} \end{align} $$

Combining these equations gives:
 * $$\text{curl}\ \textbf{H}=\sigma\textbf{E}+\epsilon\epsilon_0\dfrac{\partial\textbf{E}}{\partial t}\qquad$$ (Equation 1)

Also from the Equations booklet
 * $$\textbf{B}=\mu\mu_0\textbf{H}\Rightarrow\textbf{H}=\dfrac{\textbf{B}}{\mu\mu_0}$$ ,

and substituting this expression into Equation 1 gives
 * $$\text{curl}\ \textbf{B}=\mu\mu_0\left(\sigma\textbf{E}+\epsilon\epsilon_0\dfrac{\partial\textbf{E}}{\partial t}\right)$$

We assume that $$\mu=1$$ for the conducting media, so
 * $$\text{curl}\ \textbf{B}=\mu_0\left(\sigma\textbf{E}+\epsilon\epsilon_0\dfrac{\partial\textbf{E}}{\partial t}\right)$$   (Equation 2)

We assume that
 * $$\textbf{E}=E_0\exp[i(kz-\omega t)]\textbf{e}_x$$
 * $$\Rightarrow\ \dfrac{\partial\textbf{E}}{\partial t}=-i\omega E_0\exp[i(kz-\omega t)]\textbf{e}_x=-i\omega\textbf{E}$$,

and substituting this into Equation 2 gives
 * $$\text{curl}\ \textbf{B}=\mu_0\left(\sigma\textbf{E}-\epsilon\epsilon_0i\omega\textbf{E}\right)$$

Since $$\omega\ll1/\tau_{collision}$$ then $$\epsilon=1$$ so
 * $$\text{curl}\ \textbf{B}=\mu_0\left(\sigma\textbf{E}-\epsilon_0i\omega\textbf{E}\right)=\mu_0\epsilon_0\textbf{E}\left(\dfrac{\sigma}{\epsilon_0}-i\omega\right)$$

And since $$\omega\ll\dfrac{\sigma}{\epsilon_0}$$ then the $$-i\omega$$ term can be disregarded, so
 * $$\text{curl}\ \textbf{B}=\mu_0\sigma\textbf{E}$$

Part (b)
$$\text{curl }\textbf{B}=\mu_0\sigma\textbf{E}$$, so


 * $$\text{curl }(\text{curl }\textbf{B})=\text{curl }(\mu_0\sigma\textbf{E})$$

From the Equations booklet, we use the vector calculus identity,
 * $$\text{curl }(\text{curl }\textbf{B})=\text{grad }(\text{div }\textbf{B})-\nabla^2\textbf{B}$$, so
 * $$\text{grad}(\text{div }\textbf{B})-\nabla^2\textbf{B}=\text{curl}(\mu_0\sigma\textbf{E})$$

and from the no-monopole law, $$\text{div }\textbf{B}=0$$, so
 * $$-\nabla^2\textbf{B}=\text{curl }(\mu_0\sigma\textbf{E})=\mu_0\sigma \text{curl }(\textbf{E})$$.

We use Faraday's Law, $$\text{curl }\textbf{E}=-\dfrac{\partial\textbf{B}}{\partial t}$$, so


 * $$-\nabla^2\textbf{B}=-\mu_0\sigma\dfrac{\partial\textbf{B}}{\partial t}\ \Rightarrow\ \nabla^2\textbf{B}=\mu_0\sigma\dfrac{\partial\textbf{B}}{\partial t}$$.

Part (c)
The characteristic penetration distance $$\delta$$ (the skin depth) is the distance at which the magnetic field (or electric field) has been reduced by a factor of $$1/e=0.37$$, and is equal to
 * $$\delta=\dfrac{1}{k_{im}}$$,

so we need to find $$k_{im}$$.

In part (a) we assumed that $$\textbf{E}=E_0\exp[i(kz-\omega)]\textbf{e}_x$$, so
 * $$\textbf{B}=\dfrac{E_0n}{c}\exp[i(kz-\omega t)]\textbf{e}_y$$,

and we will substitute this into each side of the equation,
 * $$\nabla^2\textbf{B}=\mu_0\sigma\dfrac{\partial\textbf{B}}{\partial t}$$.

By using one of the cartesian differential operations from the Equations booklet,
 * $$\nabla^2\textbf{B}=\dfrac{\partial^2\textbf{B}}{\partial x^2}+\dfrac{\partial^2\textbf{B}}{\partial y^2}+\dfrac{\partial^2\textbf{B}}{\partial z^2}$$.
 * $$=\dfrac{\partial^2}{\partial z^2}\dfrac{E_0n}{c}\exp[i(kz-\omega t)]$$
 * $$=-\dfrac{k^2E_0n}{c}\exp[i(kz-\omega t)]$$.

And
 * $$\mu_0\sigma\dfrac{\partial\textbf{B}}{\partial t}=\mu_0\sigma\dfrac{\partial}{\partial t}\left(\dfrac{E_0n}{c}\exp[i(kz-\omega t)]\right)$$
 * $$=-i\mu_0\sigma\left(\dfrac{E_0n}{c}\exp[i(kz-\omega t)]\right)$$.

So we have
 * $$-\dfrac{k^2E_0n}{c}\exp[i(kz-\omega t)]=-it\mu_0\sigma\left(\dfrac{E_0n}{c}\exp[i(kz-\omega t)]\right)$$
 * $$\Rightarrow k^2=i\mu_0\sigma$$.

Now
 * $$k^2=(k_{re}+ik_{im})^2=i\mu_0\sigma$$
 * $$\Rightarrow k_{re}^2-k_{im}^2=0$$ and $$2ik_{re}k_{im}=i\mu_0\sigma$$
 * $$\Rightarrow k_{re}=k_{im}=\sqrt{\dfrac{\mu_0\sigma\omega}{2}}$$.

So
 * $$\delta=\dfrac{1}{k_{im}}=\sqrt{\dfrac{2}{\mu_0\sigma\omega}}$$.

Part (d)
(i) Since $$\omega=2\pi f$$, and $$f=\dfrac{c}{\lambda}$$, then $$\omega=\dfrac{2\pi c}{\lambda}$$.

So the skin depth for the specified radio waves in seawater is:

\begin{align} \delta&=\sqrt{\dfrac{2}{\mu_0\sigma\omega}}=\sqrt{\dfrac{2\lambda}{\mu_0\sigma2\pi c}}\\ &=\sqrt{\dfrac{2\times1500\text{ m}}{4\pi\times10^{-7}\text{ NA}^{-2}\times2.0\times10^{-2}\ \Omega^{-1}\text{ m}^{-1}\times2\pi\times3\times10^8\text{ ms}^{-1}}}\\ &=7.96\text{ m}\\ \end{align} $$

(ii) Since $$\omega=2\pi f$$, then the skin depth for the specified radio waves in the plasma is

\begin{align} \delta&=\sqrt{\dfrac{2}{\mu_0\sigma\omega}}=\sqrt{\dfrac{2}{\mu_0\sigma2\pi f}}\\ &=\sqrt{\dfrac{2}{4\pi\times10^{-7}\text{ NA}^{-2}\times100\ \Omega^{-1}\text{ m}^{-1}\times2\pi\times13.6\times10^6\text{ MHz}}}\\ &=0.01\text{ m}\\ \end{align} $$