SMT359 - 2012

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 would like the force on

$$q_1$$ to be zero, so require that


 * $$\mathbf F=\mathbf 0=\frac{kq_1q_2}{a^2}\hat{\mathbf i}-\frac{kq_1q_3}{(2a)^2}\hat{\mathbf i}.$$

Since

$$q_1=q_2$$ , we can write



\begin{align} \frac{kq_1^2}{a^2}&=\frac{kq_1q_3}{4a^2}\\ q_1&=\frac{q_3}{4}\\ q_3&=4q_1. \end{align} $$

Question 2
To find the average charge density we will first find the total charge in the cylinder and divide this quantity by the volume of the cylinder. The total charge is the volume integral of charge density:



\begin{align} Q&=\int\!\rho\ \mathrm dV\\ &=\int_0^l\!\!\int_0^{2\pi}\!\!\!\!\int_0^a\!\frac{C}{2}r^3\ r\ \mathrm dr\ \mathrm d\theta\ \mathrm dz \\&=\frac{C}{2}\times\int_0^l\!\! \mathrm dz\times\int_0^{2\pi}\!\!\!\mathrm d\theta\times \int_0^a\!\! r^4\ \mathrm dr\\&=\frac{\pi l C a^5}{5}. \end{align} $$

Dividing this by the volume of the cylinder:



\bar\rho=\frac{Q}{V}=\frac{\pi l C a^5}{5}\times\frac{1}{\pi l a^2} =\frac{C a^3}{5}, $$

as required.

Part (a)
An electrostatic field must be conservative, which means that the curl of

$$\mathbf E$$ must be zero.



\begin{align} \mathrm{curl}\ \mathbf E&= \left| \begin{matrix} \mathbf i&\mathbf j&\mathbf k \\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ 2xy&(2z+x^2)&2y \end{matrix} \right| \\&=\mathbf i (2-2)-\mathbf j (0-0)+\mathbf k (2x-2x) \\&=\mathbf 0. \end{align} $$ The curl of the vector field

$$\mathbf E$$ is zero, so it could represent an electrostatic field.

Part (b)
This question can be solved by performing a path integral from the start to end position (noticing that

$$z=0$$ everywhere an the path, so any term containing

$$z$$ must come to zero).



\begin{align} W&=-q\!\!\int_{(0,0,0)}^{(a,a,0)}\!\!\!\mathbf E\cdot\mathrm d\mathbf l\\ &=-q\left(\int_0^a\!\! E_x\ \mathrm dx+\int_0^a\!\! E_y\ \mathrm dy+\int_0^0\!\! E_z\ \mathrm dz\right)\\ &=-qC\left(\int_0^a\!\! 2xy\ \mathrm dx+\int_0^a\!\! x^2\ \mathrm dy\right)\\ &=-qC\left(x^2y\Big|_{x=0}^{x=a}+x^2y\Big|_{y=0}^{y=a}\right)\\ &=-qC(a^2y+x^2a). \end{align} $$ If we choose our path to move along the

$$x$$ -axis first, then

$$y$$ will be equal to zero so the first term (which corresponds to movement along the

$$x$$ -axis) disappears. If we then move parallel to the

$$y$$ -axis then we have

$$x$$ =a along this path, so the work done is the second term with

$$x=a$$ , or

$$-qCa^3$$ . Checking the other path: if we move along the

$$y$$ -axis first, then the second term disappears because

$$x=0$$ along this path. We then move parallel to the

$$x$$ -axis, where we have

$$y=a$$ , so the first term gives

$$-qCa^3$$ as the work done.


 * $$W=-qCa^3.$$

Question 4
From the Equations booklet, the magnetic field around a current-carrying wire is
 * $$\textbf{B}=\dfrac{\mu_0I}{2\pi r}\textbf{e}_\phi$$.

So applying the Lorentz force law from the Equations booklet gives

\begin{align} \textbf{F}&=q(\textbf{E}+\textbf{v}\times\textbf{B})\\ &=q(v\textbf{e}_z\times\dfrac{\mu_0I}{2\pi r}\textbf{e}_\phi)\\ &=\dfrac{qv\mu_0I}{2\pi r}(\textbf{e}_z\times\textbf{e}_\phi)\\ &=\dfrac{-1.6\times10^{-19}\text{ C }\times100000\text{ m s }^{-1}\times4\pi\times10^{-7}\text{ NA }^{-2}}{\pi\times0.01\text{ m }}(-\textbf{e}_r)\\ &=6.4\times10^{-19}\text{ N }\textbf{e}_r\\ \end{align} $$ So the force acting on the electron has magnitude

$$6.4\times10^{-19}$$ N, in an outward radial direction.

Question 5
We use the integral form of Gauss's law in media, from the Equations booklet, to find the electric field inside the sphere (

$$r\le R$$ ). Since we have spherical symmetry,

$$\textbf{D}=D_r(\textbf{r})\textbf{e}_r$$



\begin{align} &\int_S\textbf{D}\cdot\mathrm{d}\textbf{S}=\int_V\rho \mathrm{d}V\\ \Rightarrow\qquad &D_r\times4\pi r^2=\dfrac{\rho4\pi r^3}{3}\\ \Rightarrow\qquad &D_r=\dfrac{\rho r}{3} \end{align} $$ Since the sphere is an LIH material, we use the identity

$$D=\epsilon\epsilon_0E$$ from the Equations booklet to give
 * $$E_r=\dfrac{\rho r}{3\epsilon\epsilon_0}$$

We assume that the potential at the surface is 0, and use a line integral to find the potential

$$V$$ at the centre, using this 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 \qquad V&=-\int_R^0\dfrac{\rho r}{3\epsilon\epsilon_0}\cdot\mathrm{d}r=-\dfrac{\rho}{3\epsilon\epsilon_0}\left[\dfrac{r^2}{2}\right]_R^0=\dfrac{\rho R^2}{6\epsilon\epsilon_0} \end{align} $$

Part (a)
The force q(v X B) acting on the particle only depends on the component of the particle's velocity which is perpendicular to the direction of magnetic field. The cross product of two parallel vectors is always 0. The combination of circular motion in the plane perpendicular to the magnetic field, and uniform motion along the direction of the field, makes a spiral trajectory of a charged particle in a magnetic field, where the field forms the axis of the spiral.

Part (b)
i) Using the cyclotron frequency formula from the Equations booklet:


 * $$\omega_c=\dfrac{|q|B}{m}=\dfrac{1.6\times10^{-19}\text{ C }\times1.0\text{ T }}{4.0\times10^{-26}\text{ kg }}=4\times10^{6}\text{ rad s}^{-1}$$

(ii) Using the cyclotron radius formula from the Equations booklet:


 * $$r_c=\dfrac{mv_{\bot}}{|q|B}=\dfrac{4.0\times10^{-26}\text{ kg}\times4.0\times10^{6}\text{ ms}^{-1}}{1.6\times10^{-19}\text{ C }\times1.0\text{ T}}=1\text{ m}$$

Part (a)
In its superconducting state, a superconductor has zero resistance to the flow of current. It will also exhibit the Meissner effect, i.e. if a magnetic field is applied, the field is excluded (

$$B=0$$ ) throughout the superconductor's interior.

Part (b)
For a type-I superconductor, as you increase the magnetic field being applied, the critical field strength

$$B_c$$ is the point where the material stops being superconductive. And a superconductor's critical current

$$I_C$$ is the maximum current it can carry with zero resistance, which corresponds to the critical field strength.

Question 8
We use the method of images, and consider a different system with the same boundary conditions. We replace the plate with an image charge

$$-q$$ at a distance

$$d$$ in the opposite direction from the plate's original position. This gives the field lines and equipotentials in Figure 4.10 on page 81 of Book 2, but not including the lines on the right-hand side of the line CD.

Reference: Book 2, Chapter 4, Section 4.3.

Question 9
The characteristics of the B-field for an EM wave in vacuum are: (i) the E-field and B-field are in phase; (ii) the magnitude of the E-field is

$$c$$ times the magnitude of the B-field; (iii) the B-field is perpendicular to the E-field and perpendicular to the wave motion (the k-vector), the 180 degree ambiguity in this statement can be resolved by noting that the Poynting vector should also be in the direction of motion.

From the information given:


 * $$B = E_0/c$$
 * $$\phi = \mathbf k\cdot {\mathbf r} - \omega t +\pi/4$$

and the unit vector

$$\mathbf b$$ is perpendicular to

$$\textbf{e}_z$$ and

$$\textbf{e}_x + \textbf{e}_y$$ . Thus it is in the x,y plane and must either be

$$(\textbf{e}_x - \textbf{e}_y)/\sqrt{2}$$ or

$$(-\textbf{e}_x + \textbf{e}_y)\sqrt{2}$$ . To figure out which, we check that the vector product of E and B (which gives the direction of the Poynting vector) is in fact

$$(\textbf{e}_x + \textbf{e}_y)/\sqrt{2}$$ . From this we see that

$$\textbf{e}_z \times (\textbf{e}_x - \textbf{e}_y)/\sqrt{2} = (\textbf{e}_y + \textbf{e}_x)/\sqrt{2}$$ and so


 * $$\mathbf b = (\textbf{e}_x - \textbf{e}_y)/\sqrt{2}$$.

Alternatively, you could blindly use

$$\mathbf B = \dfrac{\mathbf k \times \mathbf E}{kc}$$ to get the same result!

Hence we have


 * $$\textbf{B} = \dfrac{E_0}{c\sqrt{2}}\exp[i(\textbf{k}\cdot\textbf{r}-\omega t+\pi/4)](\textbf{e}_x-\textbf{e}_y)$$.

We obtain the physical magnetic field $$\textbf{B}_{phys}$$ from the real part of the above expression, using the identity $$\exp i\theta=\cos\theta+i\sin\theta$$,


 * $$\textbf{B}_{phys}=\dfrac{E_0}{c\sqrt{2}}\cos[\textbf{k}\cdot\textbf{r}-\omega t+\pi/4](\textbf{e}_x-\textbf{e}_y)$$.

Reference: Book 3, Chapter 1, section 1.3.4

Question 10
From the Equations booklet, the mean radiation flux transported by an electromagnetic wave is:
 * $$\overline{\textbf{N}}=\tfrac{1}{2}\epsilon\epsilon_0E_0^2\dfrac{c}{n}\hat{\textbf{k}}$$

We know that radiation flux falls off as

$$1/r^2$$ and from the above equation that

$$\overline{\textbf{N}}\propto E_0^2$$ , so

$$E_0\propto\dfrac{1}{r}$$ , i.e. the strength of the electric field must fall off as

$$1/r$$ . The electric field created by this dipole does depend on other

$$1/r^n$$ terms, but the

$$1/r$$ term dominates at a distance.

Reference: Book 3, Chapter 2, sections 2.1, 2.3.5 and 2.4.1

Part (a)
From the Equations booklet, we substitute
 * the dispersion relation formula$$\omega=\sqrt{\omega_P^2+k^2c^2}$$
 * into the phase speed formula$$v_{phase}=\omega k$$, giving
 * $$v_{phase}=\dfrac{\sqrt{\omega_P^2+k^2c^2}}{k}$$

And substituting

$$k=2\pi/\lambda$$ and

$$c=f\lambda$$ , the phase speed becomes
 * $$v_{phase}=\dfrac{\sqrt{\omega_P^2+(2\pi/\lambda)^2(f\lambda)^2}}{2\pi/\lambda}=\dfrac{\lambda\sqrt{\omega_P^2+(2\pi f)^2}}{2\pi}$$

The group speed is given by
 * $$v_{group}=\dfrac{\text{d}\omega}{\text{d}k}=\dfrac{\text{d}}{\text{d}k}\left(\omega_P^2+k^2c^2\right)^{1/2}=\dfrac{kc^2}{\sqrt{\omega_P^2+k^2c^2}}$$,

And using the previous substitutions for

$$k$$ and

$$c$$ , the group speed becomes
 * $$v_{group}\dfrac{2\pi\lambda f^2}{\sqrt{\omega_P^2+(2\pi f)^2}}$$.

Reference: Book 3, Chapter 6, section 6.1.7

Part (b)
Although the phase speed exceeds

$$c$$ , the energy travels at the group speed

$$v_{group}$$ , which is less than

$$c$$ , so there is no conflict with special relativity.

Query: I've not been able to confirm the above answer from the course books yet.

Part (i)
Since we have spherical symmetry,

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

$$0<r<R$$

\begin{align} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow \qquad &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 \qquad &E_r(\textbf{r})=\dfrac{2\pi Cr^2}{\epsilon_0}\div 4\pi r^2\\ \end{align} $$ So the electric field intensity is

$$E=\dfrac{C}{2\epsilon_0}$$

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

$$0<r<R$$

\begin{align} &\int_S\textbf{E}\cdot\mathrm{d}\textbf{S}=\dfrac{1}{\epsilon_0}\int_V\rho\ \mathrm dV\\ \Rightarrow \qquad &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 \qquad &E_r(\textbf{r})=\dfrac{2\pi CR^2}{\epsilon_0}\div 4\pi r^2\\ \end{align} $$ So the electric field intensity is

$$E=\dfrac{CR^2}{2\epsilon_0r^2}$$

Part (b)
The electrostatic potential energy needed to take a unit test charge from infinity (

$$V=0$$ ) to a point outside the region at

$$r>R$$ is:


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

The electrostatic potential inside the sphere,

$$V_{in}(r)$$ , equals:
 * the potential energy to take a test charge from infinity to the region's surface$$V_{out}(R)$$, added to
 * the potential energy to take it from the surface to a point inside the region$$V_{into}(r)$$.

Now, the potential to take it from the surface into the region is
 * $$V_{into}(r)=-\int_{\textbf{r}_1}^{\textbf{r}_2}\textbf{E}\cdot\mathrm d\textbf{l}=-\int_R^r\dfrac{C}{2\epsilon_0}\ \mathrm dr=-\dfrac{C}{2\epsilon_0}\left[r\right]_R^r=\dfrac{C}{2\epsilon_0}(R-r)$$

Hence
 * $$V_{in}=V_{out}(R)+V_{into}(r)=\dfrac{CR^2}{2\epsilon_0r}+\dfrac{C}{2\epsilon_0}(R-r)=\dfrac{C}{2\epsilon_0}\left(\dfrac{R^2}{r}+R-r\right)$$

So the electric potential is:

\begin{align} &V_{out}=\dfrac{CR^2}{2\epsilon_0r} \text{, for } r\ge R\\ &V_{in}=\dfrac{C}{2\epsilon_0}\left(\dfrac{R^2}{r}+R-r\right) \text{, for } 0R$$ , we evaluate

$$\nabla^2V_{out}$$ using the formula from the Equations booklet,
 * but since$$V_{out}=\dfrac{CR^2}{2\epsilon_0r}$$ does not depend on$$\theta$$ or$$\phi$$, we only need the section with$$\dfrac{\partial}{\partial r}$$ terms:



\begin{align} \nabla^2V_{out}&=\dfrac{1}{r^2}\dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\left(\dfrac{CR^2}{2\epsilon_0r}\right)\right)\\ &=\dfrac{1}{r^2}\dfrac{\partial}{\partial r}\left(-\dfrac{CR^2}{\epsilon_0}\right)\\ &=0 \end{align} $$ Since

$$\rho=0$$ outside the spherical region then

$$-\rho/\epsilon_0=0$$ , so
 * $$\nabla^2V_{out}=-\rho/\epsilon_0=0$$, as required.

Part (ii)
For

$$0<r<R$$ , we evaluate

$$\nabla^2V_{in}$$ ,
 * with$$V_{in}=\dfrac{C}{2\epsilon_0}\left(\dfrac{R^2}{r}+R-r\right)$$ not depending on$$\theta$$ or$$\phi$$, so



\begin{align} \nabla^2V_{in}&=\dfrac{1}{r^2}\dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\left(\dfrac{C}{2\epsilon_0}\left(\dfrac{R^2}{r}+R-r\right)\right)\right)\\ &=\dfrac{C}{2\epsilon_0r^2}\dfrac{\partial}{\partial r}\left(r^2\dfrac{\partial}{\partial r}\left(\dfrac{R^2}{r}+R-r\right)\right)\\ &=\dfrac{C}{2\epsilon_0r^2}\dfrac{\partial}{\partial r}\left(r^2\left(-\dfrac{R^2}{r^2}-1\right)\right)\\ &=\dfrac{C}{2\epsilon_0r^2}\dfrac{\partial}{\partial r}\left(-R^2-r^2\right)\\ &=-\dfrac{C}{\epsilon_0r} \end{align} $$ And since

$$\rho=C/r$$ inside the spherical region then
 * $$\nabla^2V_{in}=-\dfrac{C}{\epsilon_0r}=-\rho/\epsilon_0$$, as required.