SMT359 - 2019

General remarks: these solutions were sourced from this GitHub repository, with permission. Users are encouraged to make amendments, comment on subjective points, make suggestions, and develop solutions for questions not yet solved.
== Question 1 ==
=== Part (i) ===
: 
 \text{div }\mathbf{B}=0
where \mathbf{B} is the magnetic field.
Reference: Equations booklet, page 1
=== Part (ii) ===
The electric field \mathbf{E} around a closed loop, C, enclosing an open surface, S, is given by:
: 
 \oint_C \mathbf{E} \cdot \mathrm{d} \mathbf{l} = - \dfrac{\text{d}}{\text{d}t} \int_S \mathbf{B} \cdot \mathrm{d} \mathbf{S}
where \mathbf{B} is the magnetic field.
Reference: Equations booklet, page 1
=== Part (iii) ===
A current density \mathbf{J} and time-varying electric field \mathbf{E} give rise to a magnetic field, \mathbf{B}, as given by:
: 
 \text{curl }\mathbf{B} = \mu_0 \mathbf{J} + \varepsilon_0 \mu_0 \frac{\partial\mathbf{E}}{\partial t}
where \mu_0 = 4 \pi \times 10^{-7} \text{ N A}^{-2} is the permeability of free space, and \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} is the permittivity of free space.
Reference: Equations booklet, page 1
=== Part (iv) ===
A closed surface, S, containing a volume, V, of charge density, \rho , gives rise to an electric field, \mathbf{E} , as given by:
: 
 \int_S \mathbf{E} \cdot \mathrm{d} \mathbf{S} = \frac{1}{\epsilon_0} \int_V \rho \,\mathrm{d} V
where \epsilon_0 = 8.85\times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} is the permittivity of free space.
Reference: Equations booklet, page 1
== Question 2 ==
=== Part (a) ===
 \mathbf{G} could represent an electrostatic field if its curl is equal to zero.
: 
\begin{align}
 \text{curl }\mathbf{G} & =\left|\begin{array}{rrr}\mathbf{e}_x&\mathbf{e}_y&\mathbf{e}_z\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\G_x&G_y&G_z\end{array} \right|\\
 &=\left(\frac{\partial G_z}{\partial y}-\frac{\partial G_y}{\partial z}\right)\mathbf{e}_x +
 \left(\frac{\partial G_x}{\partial z}-\frac{\partial G_z}{\partial x}\right)\mathbf{e}_y + 
 \left(\frac{\partial G_y}{\partial x}-\frac{\partial G_x}{\partial y}\right)\mathbf{e}_z\\
 & = \left(x^2 - x^2\right)\mathbf{e}_x + (2xy - 2xy)\mathbf{e}_y + (2xz - 2xz)\mathbf{e}_z\\
 &= \mathbf{0}
\end{align}
Hence \mathbf{G} could represent an electrostatic field.
Reference: Book 1, Chapter 5, section 5.1
=== Part (b) ===
\mathbf{G} could represent a magnetic field if its divergence is equal to zero, as per the no-monopole law.
: 
\begin{align}
 \text{div }\mathbf{G} &= \frac{\partial G_x}{\partial x}+\frac{\partial G_y}{\partial y}+\frac{\partial G_z}{\partial z}\\
 & = 2yz + 0 + 0\\
 & \neq 0
\end{align}
Hence \mathbf{G} could not represent a magnetic field.
Reference: Book 1, Chapter 4, section 4.1
== Question 3 ==
We use the vector form of Coulomb's law to find the electrostatic forces acting on C due to A, and due to B
: \mathbf{F}_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{|\mathbf{r}_1-\mathbf{r}_2|^3}(\mathbf{r}_1-\mathbf{r}_2)
:
\begin{align}
 \mathbf{F}_{\textrm{CA}} &= \frac{1}{4 \pi \epsilon_0} \frac{-7q^2}{(2a^2)^{3/2}}(-a \mathbf{e}_x - a \mathbf{e}_y)
 &\mathbf{F}_{\textrm{CB}} &= \frac{1}{4 \pi \epsilon_0} \frac{-q^2}{(2a^2)^{3/2}}(-a \mathbf{e}_x + a \mathbf{e}_y)\\
 &= \frac{1}{4 \pi \epsilon_0} \frac{7q^2}{2 \sqrt{2} a^2}(\mathbf{e}_x + \mathbf{e}_y)
 &&= \frac{1}{4 \pi \epsilon_0} \frac{-q^2}{2 \sqrt{2} a^2}(-\mathbf{e}_x + \mathbf{e}_y)
\end{align}
: 
\begin{align}
 \mathbf{F}_{\textrm{C}} &= \mathbf{F}_{\textrm{CA}} + \mathbf{F}_{\textrm{CB}}\\
 &=\frac{1}{4 \pi \epsilon_0} \frac{1}{2 \sqrt{2} a^2} (7q^2 (\mathbf{e}_x + \mathbf{e}_y) - q^2 (- \mathbf{e}_x + \mathbf{e}_y))\\
 &= \frac{q^2}{4 \sqrt{2} \pi \epsilon_0 a^2} (4 \mathbf{e}_x + 3 \mathbf{e}_y) = \frac{5q^2}{4 \sqrt{2} \pi \epsilon_0 a^2} \left( \tfrac{4}{5} \mathbf{e}_x + \tfrac{3}{5}\mathbf{e}_y\right)
\end{align}
Reference: Book 1, Chapter 1, Worked Example 1.2
== Question 4 ==
The average charge density can be found by finding the total charge Q contained in the sphere, and dividing by its volume V.
: 
\begin{align}
 Q & = \int_V \rho(\mathbf{r})\,\mathrm{d}V = \int_0^R 2Cr^3 \times 4 \pi r^2 \mathrm{d}r = 8 \pi C \int_0^R r^5 \mathrm{d}r\\
 & = 8 \pi C \left[\tfrac{1}{6} r^6\right]_0^R = 8 \pi C \left(\frac{R^6}{6}\right) = \frac{4}{3} \pi C R^6\\
 V & = \frac{4}{3} \pi R^3
\end{align}
So the average charge density is
: 
 \overline{\rho} = \frac{Q}{V}= \frac{\tfrac{4}{3} \pi C R^6}{\tfrac{4}{3} \pi R^3} = CR^3
Reference: Book 1, Chapter 2, Worked Example 2.1
== Question 5 ==
Given the symmetry and uniformity involved, we apply this equation from page 1 of the Equations booklet to the parallel-plate capacitor:
: 
\begin{align}
V (\mathbf{r}_2)-V(\mathbf{r}_1) & =-\int_{\mathbf{r}_1}^{\mathbf{r}_2}\mathbf{E\cdot}\text{d}l\\
& =-\int_0^d(-E)\text{d}z\\
V &=Ed
\end{align}
Since we're dealing with an LIH dielectric, we use the following two equations from pages 1 and 4 of the Equations booklet:
: 
\begin{align}
 \mathbf{D} & = \epsilon \epsilon_0 \mathbf{E}\\
 u & = \tfrac{1}{2} \mathbf{D} \cdot \mathbf{E}\\
 & = \tfrac{1}{2} \epsilon \epsilon_0 E^2, \text{ and substitute } E=\frac{V}{d} \text{ from above, giving:}\\
 u & = \tfrac{1}{2} \epsilon \epsilon_0 \frac{V^2}{d^2}
\end{align}
Hence the energy density, u , is independent of A . Putting in the values from the question and relevant constants:
\begin{align}
 u &= \tfrac{1}{2} \times 2.5 \times 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2} \times \frac{(3\text{ V})^2}{(0.001\text{ m})^2}\\
 & = 1.0 \times 10^{-4} \text{ J m}^{-3}
\end{align}
References: Book 1, Chapter 5, Subsection 5.4.1, and Book 2, Chapter 8, Subsection 8.1.3
== Question 6 ==
=== Part (a) ===
As there are no surface currents (i_s=0), we have these results from page 4 of the Equations booklet
: 
\begin{align}
 H_{1\parallel} & = H_{2\parallel} & B_{1\bot} & = B_{2\bot}
\end{align}
=== Part (b) ===
We're dealing with an LIH dielectric so we start with this equation from page 1 of the Equations booklet
: 
\begin{align}
 \mathbf{B} & = \mu \mu_0 \mathbf{H}\\
 H_1 \sin{\theta_1} & = H_2 \sin{\theta_2} & B_1 \cos{\theta_1} & = B_2 \cos{\theta_2}\\
 \frac{B_1}{\mu_1 \mu_0}\sin{\theta_1} & = \frac{B_2}{\mu_2 \mu_0}\sin{\theta_2} & B_2 & = \frac{B_1 \cos{\theta_1}}{\cos{\theta_2}}\\
 \mu_2 & = \frac{\mu_1 B_2 \sin{\theta_2}}{B_1 \sin{\theta_1}}\\
 & = \frac{\mu_1 \frac{B_1 \cos{\theta_1}}{\cos{\theta_2}} \sin{\theta_2}}{B_1 \sin{\theta_1}}\\
 & = \frac{\mu_1 \tan{\theta_2}}{\tan{\theta_1}}\\
 & = \frac{\mu_1 \tan 25^{\circ}}{\tan 15^{\circ}}\\
 & = 1.74 \text{ } \mu_1 \text{ (to 3 s.f.)}
\end{align}
References: Book 2, Chapter 2, Subsection 2.5, and Book 3, Chapter 5, Subsection 5.2
== Question 7 ==
As there is no deflection the net force on the particle is zero. So:
: 
 \mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) = \mathbf{0}
As the electric field, magnetic field and velocity are all perpendicular:
: 
\begin{align}
 |q \mathbf{E}| & = |q| |\mathbf{v}| |\mathbf{B}|\\
 E & = vB\\
 B & = \frac{E}{v}\\
 & = \frac{1 \times 10^4 \text{ V m}^{-1}}{1 \times 10^5 \text{m s}^{-1}}\\
 & = 0.1 \text{ T}
\end{align}
== Question 8 ==
=== Part (a) ===
: 
\begin{align}
 \delta & = \sqrt{\frac{2}{\mu_0 \sigma \omega}}\\
 & = \sqrt{\frac{2}{\mu_0 \sigma 2 \pi f}}\\
 & = \sqrt{\frac{1}{\mu_0 \sigma \pi f}}\\
 & = 7.5 \times 10^{-6} \text{ m}
\end{align}
=== Part (b) ===
Plane wave penetration through a conductor falls off exponentially characterised by the skin depth. Given that 0.1 mm is several skin depths the amplitude of the wave will be negligible when it emerges. So, 0.1 mm of gold is an effective shield for radio signals with this frequency.
== Question 9 ==
=== Part (a) ===
A transverse wave will have a scalar product of its vector field and direction of travel equal to zero.
: 
\begin{align}
 \frac{1}{\sqrt{2}} (-\mathbf{e}_x + \mathbf{e}_y) \cdot \frac{1}{\sqrt{3}} (\mathbf{e}_x + \mathbf{e}_y - \mathbf{e}_z) & = \frac{1}{\sqrt{6}} ((-\mathbf{e}_x) \cdot \mathbf{e_x} + \mathbf{e}_y \cdot \mathbf{e}_y + \mathbf{0} \cdot (\mathbf{e}_z)) \\
 & = \frac{1}{\sqrt{6}} \left(-1 + 1 + 0\right)\\
 & = 0
\end{align}
Hence the wave is transverse.
=== Part (b) ===
: 
\begin{align}
 \mathbf{E}_{\textrm{phys}} & = \frac{E}{\sqrt{3}} \cos{(\mathbf{k} \cdot \mathbf{r} - \omega t}) (\mathbf{e}_x + \mathbf{e}_y - \mathbf{e}_z)\\
 \mathbf{k} \cdot \mathbf{r} & = \frac{k}{\sqrt{2}} (- \mathbf{e}_x + \mathbf{e}_y) \cdot (x \mathbf{e}_x + y \mathbf{e}_y + z \mathbf{e}_z)\\
 & = \frac{k}{\sqrt{2}} (-x + y)\\
 \mathbf{E}_{\textrm{phys}} & = \frac{E}{\sqrt{3}} \cos \left( \frac{k}{\sqrt{2}} (-x + y) -\omega t \right) (\mathbf{e}_x + \mathbf{e}_y - \mathbf{e}_z)
\end{align}
shrug
== Question 10 ==
=== Part (a) ===
Anticlockwise.
=== Part (b) ===
Assume ohmic materials so:
: 
I = \frac{V}{R}
Faraday's law:
: 
\begin{align}
 V_{\textrm{emf}} & = - \frac{\text{d}}{\text{d}t} \int_S \mathbf{B} \cdot \mathrm{d} \mathbf{S}\\
 |V_{\textrm{emf}}| & = \frac{\text{d}}{\text{d}t} B(t) \times \pi r^2\\
 \frac{\text{d}B}{\text{d}t} & = \frac{6.0\times10^{-3}\text{ T} - 3.0\times10^{-3}\text{ T}}{2\text{ s}} = 1.5\times10^{-3}\text{ T s}^{-1}\\
 I & = \frac{1.5\times10^{-3}\text{ T s}^{-1} \times \pi \times (15\times10^{-2}\text{ m})^2}{5.0\times10^{-3}\,\Omega}\\
 & = 21\text{ mA (to 2 s.f.)}
\end{align}
== Question 11 ==
=== Part (a) ===
==== Part (i) ====
No time-varying electric field so the Ampère-Maxwell law is:
: 
\begin{align}
 \oint_C \mathbf{H} \cdot \mathrm{d} \mathbf{l} & = \int_S \mathbf{J}_{\textrm{f}} \cdot \mathrm{d} \mathbf{S} = I_{\textrm{enclosed}} = I \frac{\pi r^2}{\pi a^2}\\
 \frac{B_{\phi} (r)}{\mu \mu_0} \times 2 \pi r & = I \frac{r^2}{a^2}\\
 \mathbf{B} & = \frac{\mu_0 I r}{2 \pi a^2} \mathbf{e}_{\phi} & r \leq a
\end{align}
==== Part (ii) ====
: 
\begin{align}
 \oint_C \mathbf{H} \cdot \mathrm{d} \mathbf{l} & = \int_S \mathbf{J}_{\textrm{f}} \cdot \mathrm{d} \mathbf{S} = I_{\textrm{enclosed}} = I\\
 \frac{B_{\phi} (r)}{\mu \mu_0} \times 2 \pi r & = I\\
 \mathbf{B} & = \frac{\mu_0 I}{2 \pi r} \mathbf{e}_{\phi} & r > a
\end{align}
=== Part (b) ===
Given |\mathbf{E}| = 0 the Lorentz force law gives:
: 
\begin{align}
 \mathbf{F} & = q (\mathbf{E} + \mathbf{v} \times \mathbf{B})\\
 & = -e (\mathbf{v} \times \mathbf{b})\\
 & = -evB (\mathbf{e}_z \times \mathbf{e}_{\phi})\\
 & = -evB (- \mathbf{e}_r)\\
 & = evB \mathbf{e}_r\\
 & = \frac{\mu_0 e I v}{2 \pi b}\mathbf{e}_r
\end{align}
=== Part (c) ===
: 
\begin{align}
 \delta \mathbf{F} & = I \delta \mathbf{l} \times \mathbf{B}\\
 \frac{\delta \mathbf{F}}{\delta l} & = 3I (- \mathbf{e}_z) \times B \mathbf{e}_{\phi}\\
 & = 3IB ((- \mathbf{e}_z) \times \mathbf{e}_{\phi})\\
 & = 3IB \mathbf{e}_r\\
 & = \frac{3 \mu_0 I^2}{2 \pi d} \mathbf{e}_r
\end{align}
== Question 12 ==
=== Part (a) ===
They represent static fields if their curl is zero.
: 
\begin{align}
 \mathbf{\nabla\times E_1} & = (0 - 0)\mathbf{e}_r + (0 - 0)\mathbf{e}_{\phi} + (0 - 0)\mathbf{e}_z\\
 & = \mathbf{0}\\
 \mathbf{\nabla\times E_2} & = (0 - 0)\mathbf{e}_r + (0 - 0)\mathbf{e}_{\phi} + (0 - 0)\mathbf{e}_z\\
 & = \mathbf{0}
\end{align}
=== Part (b) ===
==== Part (i) ====
From the differential version of Gauss' law:
: 
\begin{align}
 \rho & = \epsilon_0 \mathbf{\nabla \cdot E_1}\\
 & = \epsilon_0 \left(\frac{1}{r} \left(\frac{\partial}{\partial r} \frac{Cr^2}{a}\right)\right)\\
 & = \epsilon_0 \left(\frac{1}{r} \left(\frac{2Cr}{a}\right)\right)\\
 & = \frac{2C \epsilon_0}{a}
\end{align}
==== Part (ii) ====
: 
\begin{align}
 \rho & = \epsilon_0 \mathbf{\nabla \cdot E_2} \\
 & = \epsilon_0 \left(\frac{1}{r} \left(\frac{\partial}{\partial r} K\right)\right)\\
 & = 0
\end{align}
=== Part (c) ===
Assuming at r = a , \mathbf{E}_1 = \mathbf{E}_2 :
: 
\begin{align}
 \frac{Ca}{a} & = \frac{K}{a}\\
 K & = Ca
\end{align}
=== Part (d) ===
Assumptions:
 q is positive
 zero of potential is at r = \infty 
 
: 
\begin{align}
 W & = q (V (r = R) - V (r = 0))\\
 & = q \left(- \int_0^a \mathbf{E}_1 \cdot \mathrm{d} \mathbf{l} - \int_a^R \mathbf{E}_2 \cdot \mathrm{d} \mathbf{l}\right)\\
 & = q \left(- \frac{C}{a} \int_0^a r \,\mathrm{d} r - K \int_a^R \frac{1}{r} \mathrm{d}r\right)\\
 & = q \left(- \frac{C}{a} \left[ \tfrac{1}{2} r^2 \right]_0^a - K [ \ln(r) ]_a^R\right)\\
 & = q \left(- \frac{Ca}{2} - K \ln \left(\frac{R}{a}\right)\right)
\end{align}
shrug
== Question 13 ==
=== Part (a) ===
Biot–Savart is given by:
: 
 \delta \mathbf{B} = \frac{\mu_0}{4 \pi} \frac{I \delta \mathbf{l} \times (\mathbf{r} - \mathbf{r}_0)}{|\mathbf{r} - \mathbf{r}_0|^3}
 \delta \mathbf{B} can be split into a component parallel to the wire and a component perpendicular to the wire.
 Around the whole loop the perpendicular components cancel.
 Around the whole loop the parallel components sum together.
 |\mathbf{r} - \mathbf{r}_0| = \sqrt{R^2 + z^2} 
 Parallel component is: I \delta \mathbf{l} \sqrt{R^2 + z^2} \cos{\theta} .
 \cos{\theta} = \frac{R}{\sqrt{R^2 + z^2}} .
 Around the whole loop: I \delta \mathbf{l} = 2 \pi I R 
Hence:
: 
\begin{align}
 \mathbf{B} & = \frac{\mu_0}{4 \pi} \frac{2 \pi I R \sqrt{R^2 + z^2} \cos{\theta}}{(R^2 + z^2)^{3/2}} \mathbf{e}_z\\
 & = \frac{\mu_0}{2} \frac{IR^2}{(R^2 + z^2)^{3/2}} \mathbf{e}_z\text{ ,}
\end{align}
as required.
=== Part (b) ===
==== Part (i) ====
==== Part (ii) ====
: 
\begin{align}
 \mathbf{B}_{\textrm{sol}} & = n_1 l \int_{-\infty}^{\infty} \frac{\mu_0}{2} \frac{IR^2}{(R^2 + z^2)^{3/2}} \mathbf{e}_z \mathrm{d} z\\
 & = \frac{\mu_0 I n_1 l}{2} \int_{-\infty}^{\infty} \frac{R^2}{(R^2 + z^2)^{3/2}}\\
 & = \mu_0 I n_1 l
\end{align}
shrug
=== Part (c) ===
: 
\begin{align}
 \Phi_{21} & = \int_S \mathbf{B} \cdot \mathrm{d} \mathbf{S} = B_1 \pi 4 R^2 n_2 l\\
 B_1 & = \mu_0 I n_1 l \pi R^2\\
 \Phi_{21} & = 4 \pi^2 \mu_0 I n_1 n_2 R^4 l^2\\
 M_{21} & = \frac{\text{d}\Phi_{21}}{\text{d}I_1} = 4 \pi^2 \mu_0 n_1 n_2 R^4 l^2
\end{align}
shrug
== Question 14 ==
=== Part (a) ===
==== Part (i) ====
Take a Gaussian cylinder of cross sectional area A with one end inside the plate and other end outside such that the long axis of the cylinder is normal to the plane of the plate.
: 
\begin{align}
 \int_S \mathbf{E} \cdot \mathrm{d} \mathbf{S} & = \frac{Q}{\epsilon_0}\\
 EA & = \frac{\sigma A}{\epsilon_0}\\
 E & = \frac{\sigma}{\epsilon_0}
\end{align}
==== Part (ii) ====
The reversal of charge sign cancels with the reversal of the direction of the normal on this plate so the electric fields simply sum:
: 
 E = \frac{2 \sigma}{\epsilon_0}
=== Part (b) ===
==== Part (i) ====
: 
\begin{align}
 |\mathbf{E}| & = \frac{V}{d} = \frac{100\text{ V}}{0.1\times10^{-3}\text{ m}} = 1\times10^6\text{ V m}^{-1}\\
 \mathbf{D} & = \epsilon \epsilon_0 \mathbf{E}\\
 |\mathbf{D}| & = 4.0 \times 8.85\times10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2} \times 1\times10^6\text{ V m}^{-1}\\
 & = 3.5\times10^{-5}\text{ C m}^{-2} \text{ (to 2 s.f.)}
\end{align}
==== Part (ii) ====
: 
\begin{align}
 C & = \epsilon \epsilon_0 \frac{A}{d}\\
 & = \frac{4.0 \times 8.85\times10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2} \times 1\times10^{-3}\text{ m}^2}{0.1\times10^{-3}\text{ m}}\\
 & = 3.5\times10^{-10}\text{ F (to 2 s.f.)}
\end{align}
==== Part (iii) ====
: 
\begin{align}
 Q & = CV = 3.5\times10^{-10}\text{ F} \times 100\text{ V}\\
 & = 3.5\times10^{-8}\text{ C}
\end{align}
==== Part (iv) ====
: 
\begin{align}
 \sigma = \frac{Q}{2A} = \frac{3.5\times10^{-8}\text{ C}}{2 \times 1\times10^{-3}\text{ m}^2}\\
 & = 1.77\times10^{-5}\text{ C m}^{-2}
\end{align}
==== Part (v) ====
== Question 15 ==
=== Part (a) ===
: 
\begin{align}
 \mathbf{\nabla\times E} & = - \frac{\partial\mathbf{B}}{\partial t}\\
 \mathbf{\nabla\times}(\mathbf{\nabla\times E}) & = - \frac{\partial}{\partial t} (\mathbf{\nabla\times B}) & \text{curl both sides}\\
 \mathbf{\nabla\times}(\mathbf{\nabla\times E}) & = - \epsilon_0 \mu_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} & \text{From Ampère-Maxwell: } \mathbf{\nabla\times B} = \epsilon_0 \mu_0 \frac{\partial \mathbf{E}}{\partial t}\\
 \mathbf{\nabla}(\nabla\cdot E) - \nabla^2\mathbf{E} & = - \epsilon_0 \mu_0 \frac{\partial^2\mathbf{E}}{\partial t^2} & \text{From: } \mathbf{\nabla\times}(\mathbf{\nabla\times F}) = \mathbf{\nabla(\nabla \cdot F)} - \nabla^2\mathbf{F} \\
 - \nabla^2\mathbf{E} & = - \epsilon_0 \mu_0 \frac{\partial^2\mathbf{E}}{\partial t^2} & \text{From Gauss' law: } \mathbf{\nabla \cdot E} = 0 \text{ , in empty space}\\
 \nabla^2\mathbf{E} & = \epsilon_0 \mu_0 \frac{\partial^2\mathbf{E}}{\partial t^2}\,,
\end{align}
which is the wave equation.
=== Part (b) ===
From Faraday's law:
: 
\begin{align}
 \mathbf{\nabla\times E} & = - \frac{\partial \mathbf{B}}{\partial t}\\
 \mathbf{\nabla\times E} & = (\text{i}k E_0 \exp{[\text{i}(ky - \omega t)]} - 0)\mathbf{e}_x + (0 - 0) \mathbf{e}_y + (0 - 0) \mathbf{e}_z\\
 & = \text{i}k E_0 \exp{[\text{i}(ky - \omega t)]} \mathbf{e}_x\\
 - \mathbf{B} & = \int \text{i}k E_0 \exp{[\text{i}(ky - \omega t)]} \mathbf{e}_x \mathrm{d} t\\
 & = - \frac{k E_0}{\omega} \exp{[\text{i}(ky - \omega t)]} \mathbf{e}_x\\
 \mathbf{B} & = \frac{E_0}{c} \exp{[\text{i}(ky - \omega t)]} \mathbf{e}_x\\
 & = B_0 \exp{[\text{i}(ky - \omega t)]} \mathbf{e}_x \text{ , where } B_0 = \tfrac{1}{c} E_0 
\end{align}
=== Part (c) ===
: 
\begin{align}
 \mathbf{N} & = \mathbf{E}_{\textrm{phys}} \times \mathbf{H}_{\textrm{phys}}\\
 \mathbf{B} & = \mu \mu_0 \mathbf{H}\\
 \mu & = 1\\
 E_0 & = c B_0\\
 \mathbf{N} & = E_0 \cos{(ky - \omega t)} \frac{B_0}{\mu \mu_0} \cos{(ky - \omega t)} (\mathbf{e}_z \times \mathbf{e}_x)\\
 & = \frac{c B_0^2}{\mu_0} \cos^2(ky - \omega t) \,\mathbf{e}_y\\
 \overline{\mathbf{N}} & = \frac{c B_0^2}{\mu_0} \left\langle\cos^2(ky - \omega t)\right\rangle\,\mathbf{e}_y\\
 & = \frac{c B_0^2}{2 \mu_0} \,\mathbf{e}_y\\
 & = \frac{(1\times10^{-13}\text{ T})^2 \times 3\times10^8\text{ m s}^{-1}}{2 \times 4 \times \pi \times 10^{-7}\text{ N A}^{-2}} \,\mathbf{e}_y\\
 & = 1.2\times10^{-12}\text{ W m }^{-2} \,\mathbf{e}_y
\end{align}
== Question 16 ==
=== Part (a) ===
Given
: c = \frac{\omega}{k}\text{ ; } k = \frac{\omega}{c} \text{ ; } \omega = 2 \pi f 
The phase of the wave at r = 2\text{ m} and t = 0 is
: 
\begin{align}
 kr - (\omega \times 0) & = \frac{2 \pi fr}{c}\\
 & = \frac{2 \times \pi \times 2\times10^9\text{ Hz} \times 2\text{ m}}{3\times10^8\text{ m s}^{-1}}\\
 & = \frac{80}{3} \pi = 84 \text{ (to 2 s.f.)}
\end{align}
=== Part (b) ===
Spatially the wave is only dependant on the r coordinate so for a given instant of fixed r (of fixed phase in an instant of time) the wave front sweeps out a sphere centred on the origin (the dipole).
=== Part (c) ===
Using Faraday's law:
: 
\begin{align}
 \mathbf{\nabla\times E} & = - \frac{\partial \mathbf{B}}{\partial t}\\
 \mathbf{\nabla\times E} & = (0 - 0)\mathbf{e}_r + (0 - 0)\mathbf{e}_{\theta} + \frac{1}{r}\left(\frac{\partial}{\partial r} (E_0 \sin{\theta}\sin{(kr - \omega t)}) - 0\right)\mathbf{e}_{\phi}\\
 & = k E_0 \frac{\sin{\theta}}{r} \cos{(kr - \omega t)}\mathbf{e}_{\phi}\\
 - \mathbf{B} & = \int k E_0 \frac{\sin{\theta}}{r} \cos{(kr - \omega t)}\mathbf{e}_{\phi} \mathrm{d} t\\
 & = - \frac{k E_0}{\omega} \frac{\sin{\theta}}{r} \sin{(kr - \omega t}) \mathbf{e}_{\phi}\\
 \mathbf{B} & = \frac{k E_0}{\omega} \frac{\sin{\theta}}{r} \sin{(kr - \omega t}) \mathbf{e}_{\phi}\text{ ,}
\end{align}
which is consistent with \mathbf{E}, and gives
: 
\begin{align}
 B_0 & = \frac{k E_0}{\omega}\\
 E_0 & = \frac{B_0 \omega}{k}
\end{align}
=== Part (d) ===
The total power, P, over an area A is given by:
: P = \overline{\mathbf{N}} \times A\text{ ,}
where \overline{\mathbf{N}} is the time-averaged Poynting vector.
: 
\begin{align}
 \mathbf{N} & = \mathbf{E}_{\textrm{phys}} \times \mathbf{H}_{\textrm{phys}}\\
 \mathbf{B} & = \mu \mu_0 \mathbf{H}\\
 \mu & = 1\\
 A & = 4 \pi r^2 \\
 \mathbf{N} & = E_0 \frac{\sin{\theta}}{r} \sin{(kr - \omega t)}\frac{B_0}{\mu_0} \frac{\sin{\theta}}{r} \sin{(kr - \omega t)} (\mathbf{e}_{\theta} \times \mathbf{e}_{\phi})\\
 & = \frac{E_0 B_0 \sin^2{\theta}}{\mu_0 r^2} \sin^2{(kr - \omega t)}\mathbf{e}_r\\
 \overline{\mathbf{N}} & = \frac{E_0 B_0 \sin^2{\theta}}{\mu_0 r^2} \left\langle\sin^2{(kr - \omega t)}\right\rangle\mathbf{e}_r\\
 & = \frac{E_0 B_0 \sin^2{\theta}}{2 \mu_0 r^2} \mathbf{e}_r\\
 P & = \frac{E_0 B_0 \sin^2{\theta}}{2 \mu_0 r^2} \mathbf{e}_r \times 4 \pi r^2\\
 & = \frac{2 \pi E_0 B_0 \sin^2{\theta}}{\mu_0}\mathbf{e}_r\text{ ,}
\end{align}
which is independent of radial distance, r.