SM381 - June 2023

General remarks: the solutions contained on this page have been provided by former SMT359/SM381 students according to their best endeavours. Users are encouraged to make amendments, comment on subjective points, make suggestions, and develop solutions for questions not solved yet.
= Part 1 =
== Question 1 ==
We use the vector form of Coulomb's law to find the total electric force \textbf{F} on charge Q due to charges q_1 and q_2.
: \mathbf{F}_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2_{12}}\mathbf{\hat{r}}_{12}
:
\begin{align}
 \mathbf{F}_{Qq_1} &= \frac{1}{4 \pi \epsilon_0} \frac{Qq_1}{a^2}(-\mathbf{e}_y)
 &\mathbf{F}_{Qq_2} &= \frac{1}{4 \pi \epsilon_0} \frac{Qq_2}{a^2+b^2}\frac{(-b \mathbf{e}_x -a \mathbf{e}_y)}{\sqrt{a^2+b^2}}\\
 &&&= \frac{1}{4 \pi \epsilon_0} \frac{Qq_2}{(a^2+b^2)^{3/2}}(-b\mathbf{e}_x - a\mathbf{e}_y)
\end{align}
:\begin{align}
 \mathbf{F} &= \mathbf{F}_{Qq_1} + \mathbf{F}_{Qq_2}\\
 &= \frac{1}{4 \pi \epsilon_0} \frac{Qq_1}{a^2}(-\mathbf{e}_y) + \frac{1}{4 \pi \epsilon_0} \frac{Qq_2}{(a^2+b^2)^{3/2}}(-b\mathbf{e}_x - a\mathbf{e}_y)\\
 &= \frac{Q}{4 \pi \epsilon_0}\left[\frac{q_2b}{(a^2+b^2)^{3/2}}(-\mathbf{e}_x) + \left(\frac{q_1}{a^2}+\frac{q_2a}{(a^2+b^2)^{3/2}}\right)(-\mathbf{e}_y)\right]
\end{align}
N.B. The following alternative version of Coulomb's law can be used instead. It produces the same result, but removes the need to find unit vectors.
:\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)
== Question 2 ==
=== Part (a) ===
\mathbf{K} could represent a magnetic field if its divergence is equal to zero, as per the no-monopole law.
:
\begin{align}
 \text{div }\mathbf{K} &= \frac{\partial K_x}{\partial x}+\frac{\partial K_y}{\partial y}+\frac{\partial K_z}{\partial z}\\
 & = 0 + 0 + 0 = 0
\end{align}
Hence \mathbf{K} could be a magnetic field, irrespective of the values of a_x, a_y and a_z.
=== Part (b) ===
\mathbf{K} could represent an electrostatic field if \text{curl }\mathbf{K} = \mathbf{0}.
:
\begin{align}
 \text{curl }\mathbf{K} & =\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}\\K_x&K_y&K_z\end{array} \right|\\
 &=\left(\frac{\partial K_z}{\partial y}-\frac{\partial K_y}{\partial z}\right)\mathbf{e}_x +
 \left(\frac{\partial K_x}{\partial z}-\frac{\partial K_z}{\partial x}\right)\mathbf{e}_y + 
 \left(\frac{\partial K_y}{\partial x}-\frac{\partial K_x}{\partial y}\right)\mathbf{e}_z\\
 & = \left(0 - 0\right)\mathbf{e}_x + (0 - a_z)\mathbf{e}_y + (-a_y - a_x)\mathbf{e}_z
\end{align}
Hence for \mathbf{K} to be an electrostatic field, a_x=-a_y and a_z=0.
== Question 3 ==
The electrostatic field \mathbf{E} has no E_z component, so we only need to consider the xy plane. 
The potential difference between (L,L) and (0,0) can be found by adding the potential difference between (L,L) and (0,L), to the potential difference between (0,L) and (0,0).
First we find the potential difference between (L,L) and (0,L)
:
\begin{align}
\Delta V &= -\int\mathbf{E}\cdot d\mathbf{l}\\
\Delta V_1 &= -\int_{(L,L)}^{(0,L)}(a(x+y)\textbf{e}_x+a(x+y)\textbf{e}_y)\cdot \textbf{e}_xdx\\
&=-\int_{x=L}^{x=0}a(x+L)dx\\
&= -a\left[\frac{x^2}{2}+Lx\right]_{x=L}^{x=0}\\
&=-a\left(\left(\frac{0^2}{2}+0L\right)-\left(\frac{L^2}{2}+L^2\right)\right)= a\left(\frac{L^2}{2}+L^2\right)
\end{align}
Then we find the potential difference between (0,L) and (0,0)
:
\begin{align}
\Delta V_2 &= -\int_{(0,L)}^{(0,0)}(a(x+y)\textbf{e}_x+a(x+y)\textbf{e}_y)\cdot \textbf{e}_ydy\\
&=-\int_{y=L}^{y=0}a(0+y)dy\\
&= -a\left[\frac{y^2}{2}\right]_{y=L}^{y=0}\\
&=-a\left(\frac{0^2}{2}-\frac{L^2}{2}\right)=\frac{aL^2}{2}
\end{align}
So the electrostatic potential difference between (L,L,0) and the origin is 
:\Delta V_1+\Delta V_2=a\left(\frac{L^2}{2}+L^2\right)+\frac{aL^2}{2}=2aL^2
== Question 4 ==
We use the Lorentz force law to find the total electromagnetic force acting on the point charge q
:
\begin{align}
\mathbf{F}&=q(\mathbf{E}+\mathbf{v\times B})\\
&=q(E_0\textbf{e}_x+(v_x\mathbf{e}_x+v_y\mathbf{e}_y)\times B_0\mathbf{e}_z)\\
&=q(E_0\textbf{e}_x+v_xB_0\mathbf{e}_y+v_yB_0(-\mathbf{e}_x))
\end{align}
If the net electromagnetic force on the charge vanishes, then
\mathbf{F}=q(E_0\textbf{e}_x+v_xB_0\mathbf{e}_y+v_yB_0(-\mathbf{e}_x))=\mathbf{0},
and hence
v_x=0 and v_y=\frac{E_0}{B_0} .
== Question 5 ==
First we find the time-dependent magnetic flux through the coil
:
\begin{align}
\Phi&=\int_S\mathbf{B}\cdot\mathbf{\hat{n}}\ \mathrm{d}S\\
&=\int_S B_0\mathbf{e}_x\cdot(\cos(\omega t)\mathbf{e}_x+\sin(\omega t)\mathbf{e}_y)\ \mathrm{d}S\\
&=\int_S B_0\cos(\omega t)\ \mathrm{d}S\\
&=a^2B_0\cos(\omega t)
\end{align}
The voltage induced in a circuit moving with respect to a magnetic field is given by 
:
\begin{align}
V_{\text{emf}}&=-\frac{\mathrm{d}\Phi}{\mathrm{d}t}\\
&=-\tfrac{\mathrm{d}}{\mathrm{d}t}(a^2B_0\cos(\omega t))\\
&=\omega a^2B_0\sin(\omega t)
\end{align}
So the magnitude of the time-dependent induced voltage is \omega a^2B_0|\sin(\omega t)|.
== Question 6 ==
This answer is being reviewed - it may need correcting.
The complex impedance \tilde{Z} of the circuit is
:\tilde{Z}=\dfrac{\tilde{V}}{\tilde{I}}=\dfrac{V_0\exp(\mathrm{i}\omega t)}{KV_0\frac{1}{\sqrt{2}}\exp[\mathrm{i}(\omega t-\pi/4)]}=\dfrac{\sqrt{2}}{K}\exp(i\pi/4)
We convert this to polar form
:\tilde{Z}=\frac{\sqrt{2}}{K}\exp(\mathrm{i}\pi/4)=\frac{\sqrt{2}}{K}\left(\cos\left(\frac{\pi}{4}\right)+\mathrm{i}\sin\left(\frac{\pi}{4}\right)\right)
The real part of this complex impedance represents the circuit's resistance, and the imaginary part is the reactance from the inductor and capacitor.
So the resistance is R=\frac{\sqrt{2}}{K}(\cos\frac{\pi}{4})=\dfrac{1}{K}
== Question 7 ==
The polarisation direction of an electromagnetic wave is the same as the direction of its electric field vector, so it is in the x-direction.
In the "cos" expression from \mathbf{E}_{\text{phys}}, the k_0\dfrac{z-y}{\sqrt{2}} term suggests the direction of propagation, but since the +\omega t term is positive, it is in the opposite direction. Hence the wave's direction of propagation is
:\mathbf{\hat{k}}=\dfrac{\mathbf{e}_y-\mathbf{e}_z}{\sqrt{2}}.
The wave's amplitude is given by the other factors in \textbf{E}_{\text{phys}},
:E_0\times\exp\left[k_1\dfrac{z-y}{\sqrt{2}}\right].
Since the wave starts near the z=0 plane, \dfrac{z-y}{\sqrt{2}} is at its greatest, so the amplitude is at its greatest.
As the wave propagates away from the z=0 plane in the \textbf{e}_y-\textbf{e}_z direction, y keeps increasing and z becomes increasingly negative. Hence \dfrac{z-y}{\sqrt{2}} becomes increasingly negative, so the wave's amplitude will keep decreasing as it travels.
== Question 8 ==
First we find the group speed,
v_{\text{group}}=\dfrac{\mathrm{d}\omega}{\mathrm{d}k}=\dfrac{\mathrm{d}}{\mathrm{d}k}\left(\omega_0\left(\dfrac{k}{k_0}\right)^{\alpha}\right)=\dfrac{\alpha\omega_0k^{\alpha-1}}{(k_0)^\alpha}
Then we find the phase speed,
v_{\text{phase}}=\omega/k=\omega_0\left(\dfrac{k}{k_0}\right)^{\alpha}\div k=\dfrac{\omega_0k^{\alpha-1}}{(k_0)^\alpha}
For the group speed to be greater than the phase speed,
v_{\text{group}}>v_{\text{phase}} \implies \dfrac{\alpha\omega_0k^{\alpha-1}}{(k_0)^\alpha}>\dfrac{\omega_0k^{\alpha-1}}{(k_0)^\alpha}\implies \alpha>1
So for the group speed to be greater than the phase speed, we require \alpha>1.
= Part 2 =
== Question 9 ==
=== Part (a) ===
The total charge per unit length in the cylinder is 
:
\begin{align}
\lambda=\int_V\rho\ \mathrm dV&=\int_{r=0}^{r=a}\int_{\phi=0}^{\phi=2\pi}\int_{z=0}^{z=1}A\dfrac{1}{a^2+r^2}\cdot r\ \mathrm dz\ \mathrm d\phi\ \mathrm dr\\
&=2\pi A \int_{r=0}^{r=a} \dfrac{r}{a^2+r^2} \mathrm dr\\
&=2\pi A\left[\dfrac{1}{2}\ln(a^2+r^2)\right]_{r=0}^{r=a}\\
&=\pi A\left[\ln(2a^2)-\ln(a^2)\right]=\pi A\ln2
\end{align}
=== Part (b) ===
The net charge density \rho(r) in the cylinder depends on r, and not on the other cylindrical coordinates \phi or z.
Hence the resulting electric field \mathbf{E}(r) is cylindrically symmetric around the central axis z=0.
The electric field points in a perpendicular direction away from the central axis.
The cylinder is infinitely long, so we can ignore the electric field at the ends.
Hence the electric field has the form
:\mathbf{E}_r=E_r(r)\mathbf{e}_r
=== Part (c) ===
Outside the cylinder we have r>a, and apply the integral form of Gauss' Law to a length l of the cylinder. We make use of the expression \lambda=\pi A\ln 2 for the total charge contained in the cylinder per unit length, from part (b).
:
\begin{align}
\int_S\mathbf{E}(r)\cdot\mathrm d\mathbf{S}&=\frac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\
E_r(r)\times2\pi r\times l&=\frac{1}{\epsilon_0}l\lambda=\frac{1}{\epsilon_0}l\pi A\ln 2\\
E_r(r)&=\frac{A}{2r\epsilon_0}\ln 2
\end{align}
Inside the cylinder, we have r<a, and use Gauss' Law again with a length l of the cylinder:
:
\begin{align}
\int_S\mathbf{E}(r)\cdot\mathrm d\mathbf{S}&=\frac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\
E_r(r)\times2\pi r\times l&=\dfrac{1}{\epsilon_0}\int_{r=0}^{r=r}\int_{\phi=0}^{\phi=2\pi}\int_{z=0}^{z=l}A\dfrac{1}{a^2+r^2}\cdot r\ \mathrm dz\ \mathrm d\phi\ \mathrm dr\\
&=\frac{2\pi lA}{\epsilon_0}\left[\dfrac{1}{2}\ln(a^2+r^2)\right]_{r=0}^{r=r}\\
&=\frac{\pi lA}{\epsilon_0}\left(\ln(a^2+r^2)-\ln(a^2)\right)\\
E_r(r)&=\frac{A}{2r\epsilon_0}\ln\left(1+\dfrac{r^2}{a^2}\right)
\end{align}
=== Part (d) ===
The bound volume charge density in the cylinder is 
:
\begin{align}
\rho_b=-\text{div }\mathbf{P}&=-\frac{1}{r}\frac{\partial}{\partial r}(rP_r)+\frac{1}{r}\frac{\partial P_\phi}{\partial\phi}+\frac{\partial P_z}{\partial z}\\
&=-\frac{1}{r}\frac{\partial}{\partial r}\left(rP_0\frac{1}{r}\ln\left(1+\frac{r^2}{a^2}\right)\right)\\
&=-\frac{2P_0}{1+\frac{r^2}{a^2}}
\end{align}
We have the differential form of Gauss Law,
\text{div }\mathbf{E}=\dfrac{\rho}{\epsilon_0},
where the charge density \rho includes all the charge in the cylinder: both the bound charge density \rho_b from polarisation, and the free charge density \rho_f, and hence
\text{div }\mathbf{E}=(\rho_b+\rho_f)/\epsilon_0 (Equation 1).
We evaluate \text{div }\mathbf{E}, using the expression for E_r(r) inside the cylinder found in part (c), 
:
\begin{align}
\text{div }\mathbf{E}&=\text{div }(E_r(r)\mathbf{e}_r)=\frac{1}{r}\frac{\partial}{\partial r}(rE_r)+\frac{1}{r}\frac{\partial E_\phi}{\partial\phi}+\frac{\partial E_z}{\partial z}\\
&=\dfrac{1}{r}\dfrac{\partial}{\partial r}\left(\dfrac{rA}{2r\epsilon_0}\ln\left(1+\dfrac{r^2}{a^2}\right)\right)
=\dfrac{1}{r}\left(\dfrac{rA}{2r\epsilon_0}\times\dfrac{2r}{1+\frac{r^2}{a^2}}\right)=\dfrac{A}{\epsilon_0(1+\frac{r^2}{a^2})}
\end{align}
Substituting this expression for \text{div }\mathbf{E}, and the expression for \rho_b found above, into Equation 1, we obtain
\dfrac{A}{\epsilon_0(1+\frac{r^2}{a^2})}=\left(-\dfrac{2P_0}{1+\frac{r^2}{a^2}}+\rho_f\right)/\epsilon_0
\rho_f=\dfrac{A}{1+\frac{r^2}{a^2}}+\dfrac{2P_0}{1+\frac{r^2}{a^2}}
Hence a relation between A and P_0 such that the free charge density \rho_f vanishes would be A=-2P_0.
== Question 10 ==
=== Part (a) ===
The current density is constant inside the infinite slab, being \mathbf{J}=J_0\mathbf{e}_y, and outside it is \mathbf{J}=\mathbf{0}, where |z|>d/2.
Hence the resulting magnetic field has translational symmetry: it is constant for all points with the same z co-ordinate.
So the magnitude of the magnetic field at any point depends only on its z co-ordinate.
From the Biot-Savart law, we know that the magnetic field produced by the current is perpendicular to it.
By the right-hand grip rule, since the current flows in the \mathbf{e}_y direction, the magnetic field is in the
+\mathbf{e}_x direction for z>0 (above the the x-y plane), and in the 
-\mathbf{e}_x direction for z<0 (below the x-y plane).
Hence the magnetic field produced is \textbf{B}=B_x(z)\ \mathbf{e}_x.
=== Part (b) ===
==== Outside the slab ====
We will apply Ampère's law to a closed rectangular loop.
One side of the loop has length L, and is outside the slab, parallel to its face, and in the direction of the magnetic field, i.e. parallel to \textbf{e}_x.
The opposite side of the loop (also with length L) is along the slab's central axis (z=0) where there is no magnetic field.
The other two sides of the rectangle are perpendicular to the slab's face, and hence perpendicular to the magnetic field, so their dot product with it is zero.
Hence the only side of the rectangle with any magnetic field is the one outside the slab, parallel with the magnetic field, giving 
:
\begin{align}
&\int_C\mathbf{B}\ \cdot \mathrm d\mathbf{l}=\int_S\mu_0 \mathbf{J}\cdot \mathrm d\mathbf{S}\\
&\int_0^L\mathbf{B}\cdot\ \mathrm d\mathbf{l}=\mu_0J_0\times Ld/2\\
&BL=\dfrac{\mu_0J_0Ld}{2}
\end{align}
So the magnetic field outside the slab is 
:
\begin{align}
\mathbf{B}_{z>d/2}&=\dfrac{\mu_0J_0d}{2}\ \textbf{e}_x& &\text{above the top of the slab,}&\\
\mathbf{B}_{z<-d/2}&=\dfrac{\mu_0J_0d}{2}\ (-\mathbf{e}_x)& &\text{below the bottom of it.}&
\end{align}
==== Inside the slab ====
We apply Ampère's law to a different rectangular loop, located inside the slab.
One side of the loop has length L, is a distance 'z' from the central x-y plane, and is parallel to the magnetic field, i.e. \textbf{e}_x.
The opposite side of the loop (also with length L) is along the slab's central axis (z=0), where there is no magnetic field.
The other two sides of the rectangle are perpendicular to the slab's face, and hence perpendicular to the magnetic field, so their dot product with it is zero.
Hence the only side of the rectangle with any magnetic field is the one a distance z from the central plane, giving 
:
\begin{align}
&\int_C\mathbf{B}\ \cdot \mathrm d\mathbf{L}=\int_S\mu_0 \mathbf{J}\cdot \mathrm d\mathbf{S}\\
&\int_0^L\mathbf{B}\cdot\ \mathrm d\mathbf{L}=\mu_0J_0\times Lz\\
&BL=\mu_0J_0Lz
\end{align}
So the magnetic field inside the slab is 
:
\begin{align}
\mathbf{B}_{z>0}&=\mu_0J_0z\ \textbf{e}_x& &\text{above the x-y plane,}&\\
\mathbf{B}_{z<0}&=\mu_0J_0z\ (-\mathbf{e}_x)& &\text{below the x-y plane.}&
\end{align}
Reference: Determining the Magnitude of the Magnetic Field from a Symmetric Conducting Slab with Given Current Density [Boddie, study.com]
=== Part (c) ===
The general expression for the energy density produced by a magnetic field is
:u=\frac{1}{2}\mathbf{B}\cdot\mathbf{H} ,
and with \mathbf{B}=\mu\mu_0\mathbf{H}, this becomes
:u=\dfrac{B^2}{2\mu\mu_0} .
Then since we have found in part (b) that the magnetic field outside the slab is
:\mathbf{B}=\dfrac{\mu_0J_0d}{2}\ \mathbf{e}_x ,
then the magnetic energy per unit volume outside the slab, assuming the slab consists of an LIH material with permeability \mu=1 is
u=\dfrac{B^2}{2\mu\mu_0}=\dfrac{\left(\dfrac{\mu_0J_0d}{2}\right)^2}{2\mu_0}=\dfrac{\mu_0J_0^2d^2}{8}
And since we have found that in part (b) the magnetic field inside the slab is
:\mathbf{B}=\mu_0J_0z\ \mathbf{e}_x
then the magnetic energy per unit volume inside the slab is
:u=\dfrac{B^2}{2\mu\mu_0}=\dfrac{(\mu_0J_0z)^2}{2\mu_0}=\dfrac{\mu_0J_0^2z^2}{2}
=== Part (d) ===
==== Between the slabs ====
The new slab's current density \mathbf{J}=-J_0\ \mathbf{e}_y has the same magnitude (J_0) as the original slab, and its material and thickness are the same.
However, its current density is in the opposite direction (-\mathbf{e}_y), so the magnetic field will be in the opposite direction to the original slab.
Hence the new slab can be considered the same as the original slab, with a 180^{\circ} rotation about the x-axis, and translated 3d/2 up in the z-direction.
Hence the magnetic field between the slabs will be 
:\mathbf{B}_{\text{bet}} = \dfrac{\mu_0J_0d}{2}\mathbf{e}_x + \dfrac{\mu_0J_0d}{2}\mathbf{e}_x=\mu_0J_0d\ \mathbf{e}_x
==== Inside the original slab ====
Inside the original slab, the magnetic field wil be
:\mathbf{B}_{\text{ins}}=\mu_0J_0z\ \mathbf{e}_x+\dfrac{\mu_0J_0d}{2}\ \mathbf{e}_x=\mu_0J_0\left(z+\dfrac{d}{2}\right)\mathbf{e}_x
== Question 11 ==
=== Part (a) ===
A propagation vector \mathbf{k} has magnitude k=\omega/v.
Hence the propagation vector of the incident wave is
:\mathbf{k}_\text{i}=\dfrac{\omega}{c}\textbf{e}_y.
The propagation vector of the wave reflected from the surface at y=0 is 
:\mathbf{k}_\text{r}=\dfrac{\omega}{c}(-\textbf{e}_y).
Inside the dielectric we have n=c/v and n=\sqrt{\epsilon}, so v=c/\sqrt{\epsilon}, and hence the propagation vector of the wave transmitted inside the dielectric is
:\mathbf{k}_\text{t}=\dfrac{\omega\sqrt{\epsilon}}{c}\textbf{e}_y.
The propagation vector of the wave reflected by the surface at y=d is
:\mathbf{k}_\text{r}'=\dfrac{\omega\sqrt{\epsilon}}{c}(-\textbf{e}_y).
The propagation vector of the transmitted wave for y>d is
:\mathbf{k}_\text{t}'=\dfrac{\omega}{c}\textbf{e}_y.
=== Part (b) ===
The electric field of the incident wave is
:\mathbf{E}_\text{i}=E_{i0}\exp[\text{i}(k_iy-\omega t)]\mathbf{e}_z, where k_i=\frac{\omega}{c}.
We modify this, using each respective expression from part (a), to give these complex exponential expressions for the other electric fields:
:\mathbf{E}_\text{r}=E_{r0}\exp[\text{i}(-\frac{\omega}{c}y-\omega t)]\mathbf{e}_z
:\mathbf{E}_\text{t}=E_{t0}\exp[\text{i}(\frac{\omega\sqrt{\epsilon}}{c}y-\omega t)]\mathbf{e}_z
:\mathbf{E}_\text{r}'=E_{r0}'\exp[\text{i}(-\frac{\omega\sqrt{\epsilon}}{c}y-\omega t)]\mathbf{e}_z
:\mathbf{E}_\text{t}'=E_{t0}'\exp[\text{i}(\frac{\omega}{c}y-\omega t)]\mathbf{e}_z
=== Part (c) ===
Making use of \mathbf{B}=\dfrac{1}{c}\mathbf{\hat{k}}\times\mathbf{E}, and substituting c=c/\sqrt\epsilon inside the dielectric, the magnetic fields of the five waves are:
:
\begin{align}
\mathbf{B}_\text{i}=\dfrac{1}{c}\mathbf{\hat{k}_\text{i}}\times\mathbf{E}_\text{i}&=\dfrac{1}{c}\mathbf{e}_y\times E_{i0}\exp[\text{i}(\tfrac{\omega}{c}y-\omega t)]\mathbf{e}_z\\
&=\dfrac{E_{i0}}{c}\exp[\text{i}(\tfrac{\omega}{c}y-\omega t)]\mathbf{e}_x\\
\mathbf{B}_\text{r}=\dfrac{1}{c}\mathbf{\hat{k}_\text{r}}\times\mathbf{E}_\text{r}&=\dfrac{1}{c}(-\mathbf{e}_y)\times E_{r0}\exp[\text{i}(-\tfrac{\omega}{c}y-\omega t)]\mathbf{e}_z\\
&=\dfrac{E_{r0}}{c}\exp[\text{i}(-\tfrac{\omega}{c}y-\omega t)](-\mathbf{e}_x)\\
\mathbf{B}_\text{t}=\dfrac{1}{c}\mathbf{\hat{k}_\text{t}}\times\mathbf{E}_\text{t}&=\dfrac{\sqrt\epsilon}{c}\mathbf{e}_y\times E_{t0}\exp[\text{i}(\tfrac{\omega\sqrt{\epsilon}}{c}y-\omega t)]\mathbf{e}_z\\
&=\dfrac{E_{t0}\sqrt\epsilon}{c}\exp[\text{i}(\tfrac{\omega\sqrt{\epsilon}}{c}y-\omega t)]\mathbf{e}_x\\
\mathbf{B}_\text{r}'=\dfrac{1}{c}\mathbf{\hat{k}_\text{r}'}\times\mathbf{E}_\text{r}'&=\dfrac{\sqrt\epsilon}{c}(-\mathbf{e}_y)\times E_{r0}'\exp[\text{i}(-\tfrac{\omega\sqrt{\epsilon}}{c}y-\omega t)]\mathbf{e}_z\\
&=\dfrac{E_{r0}'\sqrt{\epsilon}}{c}\exp[\text{i}(-\tfrac{\omega\sqrt{\epsilon}}{c}y-\omega t)](-\mathbf{e}_x)\\
\mathbf{B}_\text{t}'=\dfrac{1}{c}\mathbf{\hat{k}_\text{t}'}\times\mathbf{E}_\text{t}'&=\dfrac{1}{c}\mathbf{e}_y\times E_{t0}'\exp[\text{i}(\tfrac{\omega}{c}y-\omega t)]\mathbf{e}_z\\
&=\dfrac{E_{t0}'}{c}\exp[\text{i}(\tfrac{\omega}{c}y-\omega t)]\mathbf{e}_x\\
\end{align}
=== Part (d) ===
==== Boundary conditions at y=0 ====
For y<0, the net electric field is equal to \mathbf{E}_i+\mathbf{E}_r.
For 0\le y\le d, the electric field is \mathbf{E}_t+\mathbf{E}_t'.
Continuity requires that, for all times,
:\mathbf{E}_i(0,t)+\mathbf{E}_r(0,t)=\mathbf{E}_t(0,t)+\mathbf{E}_r'(0,t)
:\implies E_{i0}\exp[\text{i}(-\omega t)]+E_{r0}\exp[\text{i}(-\omega t)]=E_{t0}\exp[\text{i}(-\omega t)]+E_{r0}'\exp[\text{i}(-\omega t)]
So the boundary conditions require that
:E_{i0}+E_{r0}=E_{t0}+E_{r0}'
Similarly, matching the net magnetic fields at the boundary y=0 gives
:\mathbf{B}_i(0,t)+\mathbf{B}_r(0,t)=\mathbf{B}_t(0,t)+\mathbf{B}_r'(0,t)
:\implies \dfrac{E_{i0}}{c}\exp[\text{i}(-\omega t)]-\dfrac{E_{r0}}{c}\exp[\text{i}(-\omega t)]=\dfrac{E_{t0}\sqrt\epsilon}{c}\exp[\text{i}(-\omega t)]-\dfrac{E_{r0}'\sqrt\epsilon}{c}\exp[\text{i}(-\omega t)]
So the boundary conditions require that
:E_{i0}-E_{r0}=\sqrt{\epsilon}(E_{t0}-E_{r0}')
==== Boundary conditions at y=d ====
In a similar way, at the boundary y=d, we have
:\mathbf{E}_t(d,t)+\mathbf{E}_r'(d,t)=\mathbf{E}_t'(d,t)
:\implies E_{t0}\exp[\text{i}(\tfrac{\omega\sqrt{\epsilon}}{c}d-\omega t)]+E_{r0}'\exp[\text{i}(-\tfrac{\omega\sqrt{\epsilon}}{c}d-\omega t)]=E_{t0}'\exp[\text{i}(\tfrac{\omega}{c}d-\omega t)]
The time-dependence cancels, since the same factor of '-\omega t' appears in each exponential term.
Hence the boundary conditions require that
:E_{t0}\exp(\text{i}\sqrt{\epsilon}\omega d/c)+E_{r0}'\exp(-\text{i}\sqrt{\epsilon}\omega d/c)=E_{t0}'\exp(\text{i}\omega d/c)
Similarly, matching the net magnetic fields at the boundary y=d gives
:\mathbf{B}_t(d,t)+\mathbf{B}_r'(d,t)=\mathbf{B}_t'(d,t)
:\implies \dfrac{E_{t0}\sqrt\epsilon}{c}\exp[\text{i}(\tfrac{\omega\sqrt\epsilon}{c}y-\omega t)]-\dfrac{E_{r0}'\sqrt{\epsilon}}{c}\exp[\text{i}(-\tfrac{\omega\sqrt\epsilon}{c}y-\omega t)]=\dfrac{E_{t0}'}{c}\exp[\text{i}(\tfrac{\omega\sqrt\epsilon}{c}y-\omega t)]
Again, the time-dependence factor cancels, so the boundary conditions require that 
:\sqrt{\epsilon}[E_{t0}\exp(\text{i}\sqrt{\epsilon}\omega d/c)-E_{r0}'\exp(-\text{i}\sqrt{\epsilon}\omega d/c)]=E_{t0}'\exp(\text{i }\omega d/c)