SMT359 - 2016


 * 1) 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.

Part (i)
Electrostatic fields must satisfy

$$\text{curl }\textbf{G}=\textbf{0}$$ . For the vector field

$$\textbf{G}$$ ,



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

$$\textbf{G}$$ could not represent an electrostatic field.

Part (ii)
From the no-monopole law, a magnetic field must be divergence-free, so we check whether

$$\text{div }\textbf{G}=0$$ .



\begin{align} \text{div }\mathbf{G}&=\dfrac{\partial{G_x}}{\partial{x}}+\dfrac{\partial{G_y}}{\partial{y}}+\dfrac{\partial{G_z}}{\partial{z}}\\ &=-y+0+y\\ &=0 \end{align} $$ so

$$\textbf{G}$$ could represent a magnetic field.

Question 2
The magnitudes of the forces on A due to B and due to C are equal, and given by



F=\dfrac{2q^2}{4\pi\epsilon_0(\sqrt{2}a)^2}=\dfrac{q^2}{4\pi\epsilon_0a^2} $$ The components of these forces in the

$$y$$ -direction cancel. The forces have equal components in the

$$-x$$ -direction,

$$F\cos45^{\circ}$$ , so the resultant force has magnitude



2\times\dfrac{q^2}{4\pi\epsilon_0a^2}\cos45^{\circ}=\dfrac{q^2}{2\sqrt{2}\pi\epsilon_0a^2} $$, and is in the

$$-x$$ -direction.

Part (a)
By Lenz's law and the right-hand grip rule the current induced in the loop is anticlockwise when viewed from above.

Part (b)
The magnetic flux through the loop's closed area S is given by

$$\Phi=\int_S\textbf{B}.\text{d}S=\pi r^2B$$

The induced emf around the loop is the rate of decrease of the magnetic flux:

$$V=-\dfrac{\text{d}\Phi}{\text{d}t}=-\dfrac{\pi(0.1)^2\times(2.0\times10^{-3})-\pi(0.1)^2\times(1.0\times10^{-3})}{1.0}$$ $$=-\pi(0.01\times10^{-3})=-\pi(0.00001)$$

Then by Ohm's law the magnitude of the induced current in the loop is

$$I=\dfrac{V}{R}=\dfrac{\pi(0.00001)}{3.0\times10^{-3}}=0.01\text{ A}$$

Question 4
First we use the integral form of Gauss's law to find the total charge contained in the cylinder:



\begin{align} Q&=\int\rho\ \mathrm dV=\int_0^a\int_0^{2\pi}\int_0^L2Cr^2.r\ \mathrm dz\ \mathrm d\phi\ \mathrm dr\\ &=4C\pi l\int_0^a r^3\ \mathrm dr=C\pi la^4 \end{align} $$

Then we divide this total charge by the volume of the cylinder, to find the average charge density:


 * $$\bar{\rho}=\dfrac{Q}{V}=\dfrac{C\pi la^4}{\pi a^2l}=Ca^2$$

Question 5
We use the boundary conditions for the electric field and electric displacement at the interface between different dielectric materials:


 * $$E_{A\parallel}=E_{B\parallel}$$
 * $$\Rightarrow E_A\cos\theta=E_B\cos40^{\circ}$$(Equation 1)

And

$$D_{A\perp}=D_{B\perp}$$ , and since the dielectrics are LIH materials,

$$D=\epsilon\epsilon_0E$$ , giving


 * $$D_A=2.5\epsilon_0E_A$$and$$D_B=1.5\epsilon_0E_B$$
 * $$\Rightarrow 2.5\epsilon_0E_{A\perp}=1.5\epsilon_0E_{B\perp}$$
 * $$\Rightarrow 2.5\epsilon_0E_A\sin\theta=1.5\epsilon_0E_B\sin40^{\circ}$$(Equation 2)

Dividing Equation 2 by Equation 1 gives:



\begin{align} 2.5\epsilon_0\tan\theta&=1.5\epsilon_0\tan40^{\circ}\\ \Rightarrow \tan\theta&=\tfrac{1.5}{2.5}\tan40^{\circ}\\ \Rightarrow \theta&=\tan^{-1}(\tfrac{1.5}{2.5}\tan40^{\circ})=26.7^{\circ} \text{(1 d.p.)}\\ \end{align} $$

Part (a)
The integral form of Ampere's law gives us
 * $$\int{\mathbf{J\left(r\right)}\cdot\mathrm{d}\mathbf{S}}=I$$,

Since the flow of current is radially outward then current density is also radially outward, and we have spherical symmetry, so
 * $$J\left(r\right)\times 2\pi rl=I$$
 * $$\Rightarrow\mathbf{J\left(r\right)}=\dfrac{I}{2\pi rl}\mathbf{e}_r$$

Part (b)
We know that

$$\mathbf{J}=\sigma\textbf{E}$$ , from the 'Miscellaneous formula' section of the exam paper, so we have
 * $$\mathbf{E}=\dfrac{\mathbf{J}}{\sigma}=\dfrac{I}{2\pi rl\sigma}\mathbf{e}_r$$

Part (c)
We will use the equation

$$R=\dfrac{V}{I}$$ . First we find

$$V$$ by integrating the electric field radially outwards between the wire and the tube.



\begin{align} V&=\int_a^b{\mathbf{E}\cdot\mathrm{d}\mathbf{r}}\\ &=\dfrac{I}{2\pi l\sigma}\int_a^b{\dfrac{1}{r}\ \mathrm{dr}}\\ &=\dfrac{I}{2\pi l\sigma}(\ln b-\ln a)=\dfrac{I.\ln(\tfrac{b}{a})}{2\pi l\sigma}\end{align} $$ Then we divide this expression for

$$V$$ by

$$I$$ to find the resistance:

R=\dfrac{V}{I}=\dfrac{\ln(\tfrac{b}{a})}{2\pi l\sigma} $$

Part (i)
Since we have spherical symmetry,

$$\textbf{E}=E_r(r)\textbf{e}_r$$ . Applying the integral form of Gauss's law, inside the region bounded by

$$0<r<R$$ ,



\begin{aligned} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow \qquad &E_r(r)\times4\pi r^2=\dfrac{1}{\epsilon_0}\int_0^r\dfrac{C}{r^2}\ 4\pi r^2\ \mathrm dr\\ \Rightarrow \qquad &E_r(r)=\dfrac{4\pi rC}{4\pi r^2\epsilon_0}=\dfrac{C}{\epsilon_0r}\\ \end{aligned} $$

Reference: Book 1, Chapter 2, exercise 2.4

Part (ii)
Applying the integral form of Gauss's law, outside the region bounded by

$$0<r<R$$ ,



\begin{aligned} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow \qquad &E_r(r)\times4\pi r^2=\dfrac{1}{\epsilon_0}\int_0^R\dfrac{C}{r^2}\ 4\pi r^2\ \mathrm dr\\ \Rightarrow \qquad &E_r(r)=\dfrac{4\pi RC}{4\pi r^2\epsilon_0}=\dfrac{CR}{\epsilon_0r^2}\\ \end{aligned} $$

Reference: Book 1, Chapter 2, exercise 2.4

Part (a)
At the interface between the slab and the air we have the boundary conditions

$$D_{air\perp}=D_{slab\perp}$$ and since

$$\textbf{D}=\epsilon\epsilon_0\textbf{E}$$ , and

$$e_{air}=1$$ , this becomes



\begin{aligned} D_{slab} = D_{air} = \epsilon_0E &= 8.85\times10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2}\times110\text{ Vm}^{-1}\\ & = 9.7\times10^{-10}\text{ CM}^{-2}\\ \textbf{D}_{slab} &= 9.7\times10^{-10}\text{ CM}^{-2} \mathbf{e}_z \end{aligned} $$

Part (b)
Assuming that the slab is an LIH material then

$$\textbf{D}=\epsilon\epsilon_0\textbf{E}$$ , so



\begin{aligned} &D_{slab} = D_{air}\\ &\epsilon\epsilon_0E_{slab} = \epsilon_0E_{air}\\ &E_{slab} = \dfrac{E_{air}}{\epsilon} = \dfrac{110\text{ Vm}^{-1}}{2.2}=50\text{ VM}^{-1}\\ &\textbf{E}_{slab} = 50\text{ VM}^{-1}\textbf{e}_z \end{aligned} $$

Part (c)
Since

$$\textbf{D}=\epsilon_0\textbf{E} + \textbf{P}$$ , then



\begin{aligned} P_{slab} &= D_{slab}-\epsilon_0E_{slab}\\ &= 9.7\times10^{-10}\text{ CM}^{-2} - (8.85\times10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2} \times50\text{ VM}^{-1})\\ &= 5.3\times10^{-10}\text{ CM}^{-2}\\ \textbf{P}_{slab} &= 5.3\times10^{-10}\text{ CM}^{-2}\textbf{e}_z \end{aligned} $$

Reference: Book 2, Chapter 2, exercise 2.7

Question 9
We substitute the values, including

$$\omega = \dfrac{f}{2\pi}$$ , into the general complex expression for a polarized electromagnetic wave,



\begin{aligned} \textbf{E}(\textbf{r},t) &= \textbf{E}_0\exp [i(\textbf{k}\cdot \textbf{r}-\omega t + \phi)]\\ &=E_0\dfrac{(\textbf{e}_x+\textbf{e}_y)}{\sqrt{2}}\exp \left[i\left(k\left(\dfrac{-\textbf{e}_x+\textbf{e}_y+\textbf{e}_z}{\sqrt{3}}\right)\cdot(x\textbf{e}_x+y\textbf{e}_y+z\textbf{e}_z)-\dfrac{ft}{2\pi}\right)\right]\\ &=E_0\dfrac{(\textbf{e}_x+\textbf{e}_y)}{\sqrt{2}}\exp \left[i\left(\dfrac{k(-x+y+z)}{\sqrt{3}}-\dfrac{ft}{2\pi}\right)\right] \end{aligned} $$ To get the physical electric field vector we need the real parts only of the above expression, from the identity

$$\exp i\theta=\cos\theta+i\sin\theta$$ ,


 * $$\textbf{E}_{phys}=\text{Re}(\textbf{E})=E_0\dfrac{(\textbf{e}_x+\textbf{e}_y)}{\sqrt{2}}\cos \left[\dfrac{k(-x+y+z)}{\sqrt{3}}-\dfrac{ft}{2\pi}\right]$$

Reference: Book 3, Chapter 1, section 1.3

The wave is transverse if the scalar product of its propagation direction and polarisation direction is equal to zero,


 * $$\dfrac{(-\textbf{e}_x+\textbf{e}_y+\textbf{e}_z)}{\sqrt{3}}\cdot\dfrac{(\textbf{e}_x+\textbf{e}_y)}{\sqrt{2}}=-\dfrac{1}{\sqrt{6}}+\dfrac{1}{\sqrt{6}}=0,$$

so the wave is a transverse wave.

Reference: Book 3, Chapter 1, worked example 1.5 and below

Question 10
We use the skin depth formula from the Equations Booklet, and substitute

$$\omega=2\pi f$$ ,



\begin{aligned} \delta&=\sqrt{\dfrac{2}{\mu_0\sigma\omega}}=\sqrt{\dfrac{2}{\mu_0\sigma2\pi f}}\\ f&=\dfrac{2}{\mu_0\sigma2\pi\delta^2}\\ &=\dfrac{2}{4\pi\times10^{-7}\text{ NA}^{-2}\times3\times10^7\ \Omega^{-1}\text{ m}^{-1}\times2\pi\times(0.001\text{ m})^2}\\ &\approx8.4\text{ kHz}\\ \end{aligned} $$

Reference: Book 3, Chapter 5, Exercise 5.2

Question 11
Using the mean energy flux density vector formula from the Equations Booklet, with

$$\epsilon=1$$ for air,



\begin{align} \overline{\textbf{N}}&=\tfrac{1}{2}\epsilon\epsilon_0E_0^2\dfrac{c}{n}\hat{\textbf{k}}\\ \Rightarrow \qquad N&=\tfrac{1}{2}\epsilon_0E_0^2c\\ \Rightarrow \qquad E_0&=\sqrt{\dfrac{2N}{\epsilon_0c}}=\sqrt{\dfrac{2\times1330\text{ W m}^{-2}}{8.85\times10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}\times3\times10^{8}\text{ m s}^{-1}}}\\ &\simeq 1000 \text{ NC}^{-1} \end{align} $$

Reference: Book 3, Chapter 1, Exercise 1.10

We know from the Equations Booklet that the magnetic and electric fields are related by the formula



\begin{align} \textbf{B}&=\dfrac{1}{c}\hat{\textbf{k}}\times\textbf{E}\\ \Rightarrow \qquad B_0&=\dfrac{E_0}{c}=\dfrac{1000 \text{ NC}^{-1}}{3\times10^{8}\text{ m s}^{-1}}\simeq 3.3\times10^{-6}\text{ NC}^{-1} \end{align} $$ The wavenumber is

$$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{5.55\times10^{-7}\text{ m}}\simeq 1.1\times10^7\text{ rad m}^{-1}$$ , and since

$$c=\dfrac{\omega}{k}$$ , then the wave's angular frequency is


 * $$\omega=ck=3\times10^{8}\text{ m s}^{-1}\times1.1\times10^7\text{ rad m}^{-1}\simeq3.3\times10^{15}\text{ rad s}^{-1}$$

Reference: Book 1, Chapter 7, Section 7.4.1

Question 12
Since the permittivity

$$\epsilon$$ is a complex number, then the refractive index

$$n=\sqrt{\epsilon}$$ , and the propagation vector's magnitude

$$k=n\omega/c$$ , will also both be complex numbers


 * $$n=n_{real}+\text{i }n_{imag} \qquad \text{ and } \qquad k=k_{real}({\omega})+\text{i }k_{imag}(\omega)$$.

So the magnitude of the electric field of propagating waves will be



\begin{align} E&=E_0\exp[\text{i}(kz-\omega t)]\\ &=E_0\exp[\text{i}(\{k_{real}+\text{i }k_{imag}\}z-\omega t)]\\ &=E_0\exp[-k_{imag}z]\exp[\text{i}(k_{real}z-\omega t)] \end{align} $$ The magnitude of the physical electric field is the real part of the above expression, using the identity

$$\exp i\theta=\cos\theta+i\sin\theta$$ ,


 * $$E_{phys}=E_0\exp[-k_{imag}z]\cos(k_{real}z-\omega t)$$.

The

$$\exp(-k_{imag}z)$$ factor represents absorption: the amplitude of propagating waves decreases exponentially, as the electromagnetic energy being transferred to other forms, especially thermal energy.

Reference: Book 3, Chapter 4, Section 4.4.1.

Part (a)
From the Lorentz force law

$$\textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B})$$ , with no electric field present, the proton experiences a force equal to

$$\textbf{F}=q(\textbf{v}\times\textbf{B})$$ . Since the magnetic field is uniform, the proton moves in an anti-clockwise helical path around a helix axis in the

$$z$$ -direction. This type of circular motion is called cyclotron motion.

See Figure 6.4 in Section 6.2 of Book 3 which shows a particle experiencing this type of motion, except that our proton begins its path at the origin.

Reference: Book 3, Chapter 6, Section 6.2.3.

Part (b)
(i) We find the cyclotron frequency using the formula from the Equations Booklet,


 * $$\omega_c=\dfrac{|q|B}{m}=\dfrac{|1.6\times10^{-19}\text{ C }|\times2.0\text{ T}}{1.67\times10^{-27}\text{ kg}}\simeq1.9\times10^{8}\text{ rad s}^{-1}$$.

To find the velocity vector of the proton we use the following general solution, which is the same as equation 6.6 from Book 3, Subsection 6.2.3, but with

$$\dfrac{qb}{m}$$ replaced by

$$\omega_c$$ ,


 * $$\textbf{v}=v_\perp\sin(\omega_ct+\phi_0)\textbf{e}_x+v_{\perp}\cos(\omega_ct+\phi_0)\textbf{e}_y+v_\parallel\textbf{e}_z$$,

and use these values:
 * $$v_\perp$$is the constant speed in the plane perpendicular to the magnetic field, which is$$\sqrt{v_x^2+v_y^2}=\sqrt{(2.0\times10^6\text{ m s}^{-1})^2+2.0\times10^6\text{ m s}^{-1})^2}=2.8\times10^6\text{ m s}^{-1}$$,
 * $$\omega_c=1.9\times10^{8}\text{ rad s}^{-1}$$,
 * $$t=3.2\times10^{-8}\text{ s}$$,
 * $$\phi_0$$is the phase offset for the proton's angular velocity at time$$t=0$$, which is$$\tan^{-1}{\dfrac{v_x}{v_y}}=\tan^{-1}(1)=\dfrac{\pi}{4}$$,
 * $$v_\parallel=v_z=2.0\times10^{6}\text{ m s}^{-1}$$,

giving:



\begin{align} \textbf{v}&=2.8\times10^6\times\sin\left(1.9\times10^{8}\times3.2\times10^{-8}\text{ s}+\dfrac{\pi}{4}\right)\textbf{e}_x\\ &+2.8\times10^6\times\cos\left(1.9\times10^{8}\times3.2\times10^{-8}\text{ s} +\dfrac{\pi}{4}\right)\textbf{e}_y\\ &+2.0\times10^{6}\text{ m s}^{-1}\textbf{e}_z\\ &=1.7\times10^6\textbf{e}_x+2.3\times10^6\textbf{e}_y+2\times10^6\textbf{e}_z \end{align} $$ (ii) Since the component of the proton's velocity parallel to the field is constant, being

$$v_z=2.0\times10^6\text{ m s}^{-1}$$ , then the

$$z$$ -coordinate of the proton at time

$$t=3.2\times10^{-8}\text{ s}$$ is:



\begin{align} \text{Distance }&=0+\text{Speed}\times\text{Time}\\ &=2.0\times10^6\text{ m s}^{-1}\times3.2\times10^{-8}\text{ s }\\ &=0.064\text{ m} \end{align} $$

(iii) We find the general solution for the proton's position by integrating the general solution for its velocity from part (ii), giving



\textbf{r}=\left[x_0-\dfrac{v_\perp}{\omega_c}\cos(\omega_ct+\phi_0)\right]\textbf{e}_x+\left[y_0+\dfrac{v_\perp}{\omega_c}\sin(\omega_ct+\phi_0)\right]\textbf{e}_y+(z_0+v_{\parallel}t)\textbf{e}_z $$. The proton's circular motion is centred on

$$(x_0,y_0)$$ , with radius

$$r_c$$ given by the amplitude of the cosine and sine terms, being

$$v_\perp/\omega_c$$ . Since the proton starts at the origin, then from Pythagoras' Theorem,
 * $$|x_0|=|y_0|=\sqrt{r_c^2/2}=\sqrt{(v_\perp/\omega_c)^2/2}=\sqrt{\left(\dfrac{2.8\times10^6}{1.9\times10^8}\right)^2\div2}=0.010$$.
 * The centre of the circular motion is always a right-angle anti-clockwise from its circular velocity, and we know its velocity at time$$t=0$$, so this centre is$$(x_0=0.010,y_0=-0.010)$$

Since the proton's velocity

$$v_z$$ in the

$$\textbf{e}_z$$ direction is constant, the

$$z$$ -coordinate of the proton at

$$t=1.6\times10^{-8}$$ will be half that from part (ii) above, being 0.032. Substituting the values into the general solution above for the proton's position at

$$t=1.6\times10^{-8}$$ gives:



\begin{align} \textbf{r}&=\left[0.010-\dfrac{2.8\times10^6}{1.9\times10^{8}}\cos\left((1.9\times10^{8})\times(1.6\times10^{-8})+\dfrac{\pi}{4}\right)\right]\textbf{e}_x\\ &+\left[-0.010+\dfrac{2.8\times10^6}{1.9\times10^{8}}\sin\left((1.9\times10^{8})\times(1.6\times10^{-8})+\dfrac{\pi}{4}\right)\right]\textbf{e}_y\\ &+0.032\textbf{e}_z\\ &=0.021\textbf{e}_x-0.019\textbf{e}_y+0.032\textbf{e}_z \end{align} $$

Part (c)
The two particles start in nearly the same place, with the same constant velocity

$$v_z$$ in the

$$\textbf{e}_z$$ direction, so their

$$z$$ -coordinates are always equal. We need to check whether there are any times after

$$t=0$$ when their respective

$$x$$ - and

$$y$$ -coordinates will be the same. The

$$\alpha$$ -particle has twice the charge of the proton, and four times the mass, so its cyclotron frequency from the formula

$$w_c=\dfrac{|q|B}{m}$$ will be half that of the proton's.

The general solution for a particle's position is,


 * $$\textbf{r}=\left[x_0-\dfrac{v_\perp}{\omega_c}\cos(\omega_ct+\phi_0)\right]\textbf{e}_x+\left[y_0+\dfrac{v_\perp}{\omega_c}\sin(\omega_ct+\phi_0)\right]\textbf{e}_y+(z_0+v_{\parallel}t)\textbf{e}_z$$

The radius of the circular motion is equal to the amplitude of the cosine and sine terms. Since the

$$\alpha$$ -particle's circular speed

$$v_\perp$$ is the same as the proton's, and its frequency

$$\omega_c$$ is half, then its radius will be twice as much. So the particles will meet each time the

$$\alpha$$ -particle performs a complete revolution after

$$t=0$$ .

Part (a)
We use the integral form of the Ampère-Maxwell law in media to find the magnetic intensity, with no changing electric flux, and choose a cylindrical coordinate system to exploit the arrangement's symmetry.



\begin{align} \oint_C\textbf{H}\cdot\mathrm{d}\textbf{l}&=\int_S\textbf{J}_f\cdot\mathrm{d}\textbf{S}\\ 2\pi rH&=I\\ \textbf{H}&=\dfrac{I}{2\pi r}\textbf{e}_\phi \end{align} $$

From the Equations booklet we know that $$\textbf{B}=\mu\mu_0\textbf{H}$$ for LIH materials, and since $$\mu=1$$ in air, the magnetic field is

$$\textbf{B}=\dfrac{\mu_0I}{2\pi r}\textbf{e}_\phi$$

Reference: Book 1, Chapter 4, Section 4.2 and Book 2, Chapter 3, Subsection 3.4.1

Part (i)
See sketch. From the right-hand rule, the magnetic flux is going through the loop in the $$\textbf{e}_\theta$$ direction.

Part (ii)
We know from the `Miscellaneous formulae' of the Equations booklet that the magnetic flux is



\begin{align} \Phi &= \int_S\textbf{B}\cdot\mathrm{d}\textbf{S}\\ &= \int_{z=0}^{z=l}\int_{r=d}^{r=d+w}\dfrac{\mu_0I}{2\pi r}\ \mathrm{d}r\ \mathrm{d}z\\ &= \dfrac{\mu_0Il}{2\pi}\left[\ln\text{ r}\right]_{r=d}^{r=d+w}\\ &= \dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{d+w}{d}\right) \end{align} $$

Part (i)
See sketch.

Part (ii)
First we calculate the total magnetic flux that loop 1 produces through loop 2. We assume loop 1 has current $$I$$ in the anti-clockwise direction.

The flux produced by the right side of loop 1 is simply the same as part (b)

$$\Phi_{21r} = \dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{d+w}{d}\right)$$

The current in the left side of loop 1 flows in the opposite direction to part (a), so its magnetic field is its negative

$$\textbf{B}_{1l}=-\dfrac{\mu_0I}{2\pi r}\textbf{e}_\phi$$ ,

which produces this flux in loop 2



\begin{align} \Phi_{21l} &= \int_S\textbf{B}_l\cdot\mathrm{d}\textbf{S}\\ &= \int_{z=0}^{z=l}\int_{r=w+d}^{r=2w+d}-\dfrac{\mu_0I}{2\pi r}\ \mathrm{d}r\ \mathrm{d}z\\ &= -\dfrac{\mu_0Il}{2\pi}\left[\ln\text{ r}\right]_{r=w+d}^{r=2w+d}\\ &= -\dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{2w+d}{w+d}\right) \end{align} $$

We ignore the flux produced by the current in the smaller sides of loop 1, because $$l$$ is long compared to $$w$$.

So the combined magnetic flux produced in loop 2 is



\begin{align} \Phi_{21}&=\Phi_{21l}+\Phi_{21r}=-\dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{2w+d}{w+d}\right)+\dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{d+w}{d}\right)\\ &=\dfrac{\mu_0Il}{2\pi}\left(\ln\left(\dfrac{d+w}{d}\right)-\ln\left(\dfrac{2w+d}{w+d}\right)\right)\\ &=\dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{(d+w)^2}{d(2w+d)}\right)=\dfrac{\mu_0Il}{2\pi}\ln\left(\dfrac{d+w}{2d}\right)\\ \end{align} $$

Since the current $$I$$ is constant, then from the `Miscellaneous formulae' of the Equations booklet the mutual inductance is

$$M=M_{21}=\dfrac{d\Phi}{dI_1}=\dfrac{\Phi}{I}=\dfrac{\mu_0l}{2\pi}\ln\left(\dfrac{d+w}{2d}\right)$$

Reference: Book 2, Chapter 7, Subsection 7.2.2

Part (a)
We use the integral form of Gauss's law in media to find the electric displacement between the plates

\begin{align} \int_S\textbf{D}\cdot\mathrm{d}\textbf{S}&=\int_V\rho_fdV=Q\\ D\times\Delta S&=Q\times\dfrac{\Delta S}{A}\\ D&=Q/A \end{align} $$

From the Equations booklet we have $$D=\epsilon\epsilon_0E$$, so the magnitude of the electric field in the dielectric part is
 * $$E=D/\epsilon\epsilon_0=Q/\epsilon\epsilon_0A$$.

Given a constant electric field we know that $$\Delta V=Ed$$, so the potential difference for this part is
 * $$\Delta V_{di}=Q/\epsilon\epsilon_0A$$

In the air part of the capacitor, since $$\epsilon=1$$, the electric field is
 * $$E=Q/\epsilon_0A$$

so the potential for this part is
 * $$\Delta V_{air}=Qd/\epsilon_0A$$

Hence the combined capacitance for these parallel parts is

\begin{align} C=\dfrac{Q}{V}&=\dfrac{Q_{di}}{\Delta V_{di}}+\dfrac{Q_{air}}{\Delta V_{air}}=\dfrac{Q/2}{Qd/\epsilon\epsilon_0A}+\dfrac{Q/2}{Qd/\epsilon_0A}\\ &=\dfrac{\epsilon\epsilon_0A}{2d}+\dfrac{\epsilon_0A}{2d}=\dfrac{\epsilon_0A}{d}\left(\dfrac{1+\epsilon}{2}\right) \end{align} $$

Reference: Book 2, Chapter 2, Subsection 2.4.1

Part (b)
The dielectric part of the capacitor has the same electric field as in part (a),
 * $$E=Q/\epsilon\epsilon_0A$$

so using the equation $$\Delta V=Ed$$, the potential across this part is
 * $$\Delta V_{di}=Q/\epsilon\epsilon_0A\times d/2=Qd/2\epsilon\epsilon_0A$$

In the air part, the electric field is the same as in part (a),
 * $$E=Q/\epsilon_0A$$

so the potential across this part is
 * $$\Delta V_{air}=E=Q/\epsilon_0A\times d/2=Qd/2\epsilon_0A$$

Hence the combined capacitance for these parts in series is

\begin{align} C &= \dfrac{Q}{\Delta V_{di}+\Delta V_{air}}=\dfrac{Q}{Qd/2\epsilon\epsilon_0A+Qd/2\epsilon_0A}\\ &= \dfrac{2\epsilon}{d/\epsilon_0A(1+\epsilon)}=\dfrac{\epsilon_0A}{d}\left(\dfrac{2\epsilon}{1+\epsilon}\right) \end{align} $$

Part (c)
From part (a), when the dielectric and air are parallel,
 * $$C_p = \dfrac{\epsilon_0A}{d}\left(\dfrac{1+\epsilon}{2}\right)$$

From part (b), when the dielectric and air are in series,
 * $$C_s =\dfrac{\epsilon_0A}{d}\left(\dfrac{2\epsilon}{1+\epsilon}\right)$$

When $$\epsilon=1$$ (i.e. for air) then $$c_p=c_s=\dfrac{\epsilon_0A}{d}$$.

When $$\epsilon=2$$ then $$C_p=\dfrac{\epsilon_0A}{d}(1\tfrac{1}{2})$$ and $$C_s=\dfrac{\epsilon_0A}{d}(1\tfrac{1}{3})$$

As $$\epsilon$$ increases,
 * $$C_p\to\dfrac{\epsilon_0A}{d}\left(\dfrac{\epsilon}{2}\right)$$, and
 * $$C_s\to\dfrac{\epsilon_0A}{d}\left(2\right)$$.

So the arrangement in (a) is more effective at increasing the capacitance than (b), and becomes increasingly so as $$\epsilon$$ increases.

Part (d)
The magnitude of the electric displacement between the plates is still
 * $$D=Q/A$$.

Since the permittivity varies as $$\epsilon(z)=\epsilon_0(2+z/d)$$, then the electric field's magnitude in the dielectric is
 * $$E=Q/\epsilon\epsilon_0A=Q/\epsilon_0(2+z/d)\epsilon_0A$$

We find the potential difference across the plates by using this formula 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}l\\ &=-\int_{z=0}^{z=d}Q/\epsilon_0(2+z/d)\epsilon_0A\ \mathrm{d}z\\ &=-Q/\epsilon_0^2A\left[2z+z^2/2d\right]_{z=0}^{z=d}\\ &=-Q/\epsilon_0^2A(2d+d^2/2d)\\ &=-\tfrac{5}{2}dQ/\epsilon_0^2A \end{align} $$

So the capacitance will be
 * $$C=Q/|V|=Q/(\tfrac{5}{2}dQ/\epsilon_0^2A)=\dfrac{2\epsilon_0^2A}{5d}$$

Reference: Book 1, Chapter 5, Subsection 5.2.1

Part (a)
See sketch.

Using the right-hand grip rule, the magnetic field is in the $$\textbf{e}_z$$ direction.

We use the integral form of the Ampère law in media to find the magnetic intensity, with a cylindrical coordinate system, so $$\textbf{H}(r)=H_z(r)\textbf{e}_z$$.

We consider a rectangular loop of length $$l_1$$ alongside the solenoid, with lengths parallel to the $$z$$-axis.

Since the wire has $$n/l$$ turns per unit length, the total current flowing through the loop is $$(n/l)l_1I$$.



\begin{align} \oint_C\textbf{H}\cdot d\textbf{l}&=\int_S\textbf{J}_f\cdot\mathrm{d}\textbf{S}\\ \textbf{H}l_1&=(n/l)l_1I\ \textbf{e}_z\\ \textbf{H}&=In/l\ \textbf{e}_z \end{align} $$

By treating air as an LIH material, then from the Equations Booklet we have


 * $$\textbf{B}=\mu\mu_0\textbf{H}$$

and so


 * $$\textbf{B}=\mu\mu_0In/l\ \textbf{e}_z$$

Reference: Book 1, Chapter 4, Subsection 4.3.3

Part (b)
From the Miscellaneous Formulae, the energy density is


 * $$u=\tfrac{1}{2}\textbf{B}\cdot\textbf{H}=\tfrac{1}{2}\mu\mu_0In/l\times In/l=\tfrac{1}{2}\mu\mu_0I^2n^2/l^2$$

Reference: Book 2, Chapter 8, Subsection 8.2.3

Part (c)
The energy stored in the solenoid is


 * $$u\times\text{volume}= \mu\mu_0I^2n^2/l^2\times\pi r^2l=\dfrac{\pi r^2\mu\mu_0I^2n^2}{l}$$

Part (d)
From the Miscellaneous Formulae, the flux produced through the coil's area by the solenoid's magnetic field $$\textbf{B}$$, from part (a), is



\begin{align} \Phi_{cs}&=\int_S\textbf{B}_l\cdot\mathrm{d}\textbf{S}=B\times\pi r^2=\mu\mu_0\pi Inr^2/l\\ \end{align} $$

Since the current $$I$$ is constant, then from the Miscellaneous Formulae the mutual inductance is


 * $$M=M_{cs}=\dfrac{d\Phi_{21}}{dI_1}=\dfrac{\Phi_{cs}}{I}=\mu\mu_0\pi nr^2/l$$

Part (e)
Since $$I=I_0\cos\omega t$$, it produces time-varying flux of magnitude


 * $$\Phi_{cs}=B\times\pi r^2=\mu\mu_0\pi I_0\cos(\omega t)nr^2/l$$

From the Miscellaneous Formulae, since the loop is a stationary circuit then its induced emf is



\begin{align} V_{emf}&=-\dfrac{\mathrm{d}\Phi_{cs}}{\mathrm{d}t}\\ &=-\dfrac{d}{dt}\left(\mu\mu_0\pi I_0\cos(\omega t)nr^2/l\right)\\ &=\mu\mu_0\omega\pi I_0\sin(\omega t)nr^2/l \end{align} $$

Since the resistance is $$R$$, then the induced current is


 * $$I=V/R=\mu\mu_0\omega\pi I_0\sin(\omega t)nr^2/Rl$$

Part (a)
From the Miscellaneous Formulae, the plasma frequency is



\begin{align} \omega_{\text{p}}=\sqrt{\dfrac{n_ee^2}{m\epsilon_0}}&=\sqrt{\dfrac{3\times10^4\text{ m}^{-3}\times(1.6\times10^{-19}\text{ C})^2}{9.11\times10^{-31}\text{ kg}\times8.85\times10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}}}\\ &=9800\text{ rad s}^{-1} \end{align} $$

Reference: Book 3, Chapter 6, Subsection 6.1.6

Part (b)
From the Miscellaneous Formulae, in a collisionless regime the relative permittivity function is



\epsilon_{\text{eff}}(\omega)=1-\dfrac{\omega_p^2}{\omega^2} $$

For the plasma frequency of $$\omega=\omega_p$$, $$\epsilon_{\text{eff}}=0$$. As the frequency increases, the relative permittivity increases and approaches a value of 1.

See sketch (to be added).

Reference: Book 3, Chapter 6, Subsection 6.1.7

Part (c)
The speed of the signal's packets is found using the group speed $$v_{group}$$ and $$\omega$$ from the Miscellaneous Formulae,



\begin{align} v_{\text{group}}&=\dfrac{\mathrm{d}\omega}{\mathrm{d}k}=\dfrac{\mathrm{d}}{\mathrm{d}k}\left(\sqrt{\omega_p^2+k^2c^2}\right)\\ &=\dfrac{2kc^2}{2\sqrt{\omega_P^2+k^2c^2}}=\dfrac{kc^2}{\sqrt{\omega_P^2+k^2c^2}}=\dfrac{kc^2}{\omega} \end{align} $$

From the dispersion relation $$\omega=\sqrt{\omega_p^2+k^2c^2}$$, we have


 * $$k=\dfrac{\sqrt{\omega^2-\omega_p^2}}{c}=\dfrac{\sqrt{(2.3\times10^9\times2\pi)^2-9800^2}}{3\times10^8}=48$$, so


 * $$v_{\text{group}}=\dfrac{48\times(3\times10^8)^2}{2.3\times10^9\times2\pi}=3\times10^8\text{ ms}^{-1}$$.

Part (d)
The relative permittivity is


 * $$e=1-\dfrac{\omega_p^2}{\omega^2}=\dfrac{9800^2}{(2.3\times10^9\times2\pi)^2}=1-4.6\times10^{-13}$$

so the refractive index is
 * $$n=\sqrt{e}=\sqrt{1-4.6\times10^{-13}}$$

Since the phase speed of the waves is $$v_{phase}=c/n$$, and $$v_{phase}v_{group}=c^2$$, then $$v_{group}=cn=c\sqrt{1-4.6\times10^{-13}}$$

So the extra time taken for Voyager's signal to arrive compared to visible light is



\begin{align} \dfrac{d}{v_{group}}-\dfrac{d}{c}&=\dfrac{d}{c}\left(\dfrac{1}{\sqrt{1-4.6\times10^{-13}}}-1\right)\\ &=\dfrac{2\times10^{13}}{3\times10^8}\left(\dfrac{1}{\sqrt{1-4.6\times10^{-13}}}-1\right)\\ &=1.5\times10^{-8}\text{ s} \end{align} $$

A high-precision calculator was used to evaluate this value.