Problem set 45

1. A plane electromagnetic wave is traveling along the positive z-direction. The maximum electric field along the x-direction is 10 V/m. The approximate maximum values of the power per unit area and the magnetic induction $B$, respectively, are
1. $3.3\times 10^{-7} watts/m^2$ and 10 tesla
2. $3.3\times 10^{-7} watts/m^2$ and $3.3\times 10^{-8}$ tesla
3. $0.265 watts/m^2$ and $10$ tesla
4. $0.265 watts/m^2$ and $3.3\times 10^{-8}$ tesla

The power per unit area transported by the electromagnetic waves is given $$\begin{align*} I&=c\epsilon_0E_0^2\\ &=3\times 10^8\times8.854\times10^{-12}\times10^2\\ &=0.265 watts/m^2 \end{align*}$$

Magnetic induction $$\begin{align*} B&=\frac{1}{c}E_0\\ &=\frac{1}{3\times10^{-8}}\times10\\ &=3.3\times 10^{-8} \: tesla \end{align*}$$

Hence, answer is (D)

2. A particle moves in one dimension in the potential $V=\frac{1}{2}k(t)x^2$, where $k(t)$ is a time dependent parameter. Then $\frac{d}{dt}\left < V\right >$, the rate of change of the expectation value $\left < V\right >$ of the potential energy, is
1. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{k}{2m}\left < xp+px\right >$
2. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right > +\frac{1}{2m}\left < p^2\right >$
3. $\frac{k}{2m}\left < xp+px\right >$
4. $\frac{1}{2}\frac{dk}{dt}\left < x^2\right >$

$$\frac{d\left <\hat A\right >}{dt}=\left < \frac{\partial \hat A}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\hat H,\hat A\right]\right >$$ $$\begin{align*} &\frac{d\left < V\right >}{dt}={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{\hbar}\left < \left[\frac{p^2}{2m}+V, V\right]\right >}\\ &={\textstyle\left < \frac{\partial V}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,V\right]\right > }\\ &={\scriptstyle\left < \frac{\partial \left(\frac{1}{2}k(t)x^2\right)}{\partial t}\right > +\frac{i}{2m\hbar}\left < \left[p^2,\frac{1}{2}k(t)x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < \left[p^2,x^2\right]\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > }\\ &{\scriptstyle+\frac{ik}{4m\hbar}\left < xp\left[p,x\right]+\left[p,x\right]p+p\left[p,x\right]x+\left[p,x\right]xp\right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{ik}{4m\hbar}\left < -2i\hbar xp-2i\hbar px \right > }\\ &={\textstyle\frac{1}{2}\frac{dk}{d t}\left < x^2\right > +\frac{k}{2m}\left < xp+ px \right > } \end{align*}$$

3. Consider three inertial frames of reference A, B and C. The frame B moves with a velocity $c/2$ with respect to $A$, and $C$ moves with velocity $c/10$ with respect to B in the same direction. The velocity of C as measured in A is
1. $\frac{3c}{7}$
2. $\frac{4c}{7}$
3. $\frac{c}{7}$
4. $\frac{\sqrt{3}c}{7}$

Let $V$ be the velocity of frame $B$ with respect to frame A, $u_x$ be the velocity of frame C with respect to frame A, and $u_x'$ be the velocity of frame C with respect to frame B, then we have $$u_x'=\frac{u_x-V}{1-u_xV/c^2}$$ and $$u_x=\frac{u_x'+V}{1+u_x'V/c^2}$$ $$\begin{align*} u_x&=\frac{u_x'+V}{1+u_x'V/c^2}\\ &=\frac{\frac{c}{10}+\frac{c}{2}}{1+\frac{c}{10}\frac{c}{2}\frac{1}{c^2}}=\frac{4}{7}c \end{align*}$$

Hence, answer is (B)

4. Suppose the yz-plane forms a chargeless boundary between two media of permittivities $\epsilon_{left}$ and $\epsilon_{right}$ where $\epsilon_{left}:\epsilon_{right}=1:2$. If the uniform electric field on the left is $\vec E_{left}=c\left(\hat i+\hat j+\hat k\right)$ (where $c$ is constant), then the electric field on the right $\vec E_{right}$ is
1. $c\left(2\hat i+\hat j+\hat k\right)$
2. $c\left(\hat i+2\hat j+2\hat k\right)$
3. $c\left(\frac{1}{2}\hat i+\hat j+\hat k\right)$
4. $c\left(\hat i+\frac{1}{2}\hat j+\frac{1}{2}\hat k\right)$

Using relation $$E_{left}\epsilon_{left}\cos{\theta_1}=E_{right}\epsilon_{right}\cos{\theta_2}$$ where, $\theta_1$ is the angle between $\vec E_{left}$ and the vector normal to the yz-plane and $\theta_2$ is the angle between $\vec E_{right}$ and the vector normal to the yz-plane. The vector normal to yz-plane is $\hat i$

Using $\epsilon_{left}:\epsilon_{right}=1:2$ in above equation $$E_{left}\cos{\theta_1}=2E_{right}\cos{\theta_2}\quad--(1)$$ Magnitude of $\vec E_{left}$ is given by $$E_{left}=\sqrt{c^2+c^2+c^2}=\sqrt{3}c$$ Let us find angle $\cos{\theta_1}$ $$\begin{align*} \cos{\theta}&={\textstyle\frac{a_1b_1+a_2b_2+a_3b_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}}\\ &=\frac{c}{\sqrt{3c^2}\sqrt{1}}=\frac{1}{\sqrt{3}} \end{align*}$$ Hence, equation (1) becomes $$\sqrt{3}c\times\frac{1}{\sqrt{3}}=2E_{right}\cos{\theta_2}$$ $$E_{right}\cos{\theta_2}=\frac{c}{2}$$ One cane verify that $E_{right}=c\left(\hat i+2\hat j+2\hat k\right)$ satisfies the above condition. For this vector find $E_{right}$ and $\cos\theta_2$.

Hence, answer is (B)

5. Which of the following transformations $\left(V,\vec A\right)\rightarrow \left(V',\vec A'\right)$ of electrostatic potential $V$ and the vector potential $\vec A$ is a gauge transformation?
1. $\left(V'=V+ax,\vec A'=\vec A+at\hat k\right)$
2. $\left(V'=V+ax,\vec A'=\vec A-at\hat k\right)$
3. $\left(V'=V+ax,\vec A'=\vec A+at\hat i\right)$
4. $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

The general gauge transformation of electrostatic potential $V$ and the vector potential $\vec A$ is given by the equations $$V'=V-\frac{\partial f}{\partial t}$$ $$\vec A'=\vec A+\nabla f$$ In all options, we have, $V'=V+ax$. This implies $$-\frac{\partial f}{\partial t}=ax$$ Hence, $f=-atx$

Hence, $\nabla f=-at\hat i$

Hence, we have, $\left(V'=V+ax,\vec A'=\vec A-at\hat i\right)$

Hence, answer is (D)