Problem set 71

1. A rod of length $L$ carries a total charge $Q$ distributed uniformly. If this is observed in a frame moving with a speed $v$ along the rod, the change per unit length (as measured by the moving observer) is
1. $\frac{Q}{L}\left(1-\frac{v^2}{c^2}\right)$
2. $\frac{Q}{L}\sqrt{1-\frac{v^2}{c^2}}$
3. $\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$
4. $\frac{Q}{L\left(1-\frac{v^2}{c^2}\right)}$

Length contraction is given by $$L'=L \sqrt{1-\frac{v^2}{c^2}}$$ Hence, charge per unit length is given by $$\frac{Q}{L\sqrt{1-\frac{v^2}{c^2}}}$$

Hence, answer is (C)

2. The electric and magnetic fields in the charge free region $z > 0$ are given by $$\vec E(\vec r,t)=E_0e^{-k_1z}\cos{(k_2x-\omega t)}\hat j$$ $${\scriptstyle\vec B(\vec r,t)=\frac{E_0}{\omega}e^{-k_1z}\left[k_1\!\sin{(k_2x-\omega t)}\hat i+k_2\!\cos{(k_2x-\omega t)}\hat k\right]}$$ where $\omega$, $k_1$, and $k_2$ are positive constants. The average energy flow in $x$-direction is
1. $\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$
2. $\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}$
3. $\frac{E_0^2k_1}{2\mu_0\omega}e^{-2k_1z}$
4. $\frac{1}{2}c\epsilon_0E_0^2e^{-2k_1z}$

According to Poynting theorem the energy flux is given by $$\vec S=\frac{1}{\mu_0}\vec E\times \vec B$$ Hence, $$\begin{align*} S_x\hat i&=\frac{1}{\mu_0}\left(E_0e^{-k_1z}\cos{(k_2x-\omega t)}\right)\hat j\times\\ &~~~~\left(\frac{E_0}{\omega}e^{-k_1z}k_2\cos{(k_2x-\omega t)} \right)\hat k\\ &=\frac{E_0^2k_2}{\mu_0\omega}e^{-2k_1z}\cos^2{(k_2x-\omega t)}\hat i\\ \end{align*}$$ $$\left < S_x \right > =\frac{E_0^2k_2}{2\mu_0\omega}e^{-2k_1z}$$

Hence, answer is (A)

3. A uniform magnetic field in the positive $z$-direction passes through a circular wire loop of radius 1 cm and resistance 1 $\omega$ lying in $xy$-plane. The field strength is reduced from 10 tesla to 9 tesla in 1 s. The charge transferred across any point in the wire is approximately
1. $3.1\times10^{-4}$ coulomb
2. $3.4\times10^{-4}$ coulomb
3. $4.2\times10^{-4}$ coulomb
4. $5.2\times10^{-4}$ coulomb

Magnetic flux through circular wire is $$\begin{align*} \phi&=\int B\:dA\\ &=B\pi r^2\\ &=B\times3.14\times1\times10^{-4}\\ &=3.14\times10^{-4}B\\ \end{align*}$$ Change in magnetic flux per unit time is $$\begin{align*} \frac{d\phi}{dt}&=3.14\times10^{-4}(B_2-B_1)\\ &=3.14\times10^{-4}(10-9)\\ &=3.14\times10^{-4} \end{align*}$$ Faraday's law of induction can now be used to determine the induced emf: $$\epsilon=-\frac{d\phi}{dt}=-3.14\times10^{-4}\:volts$$ Current $$I=\frac{\epsilon}{R}=-3.14\times10^{-4}\:Amp$$ Hence, charge transferred in one second at point is $3.14\times10^{-4}$.

Hence, answer is (A)

4. A particle of mass $m$ is in a potential $V=\frac{1}{2}m\omega^2 x^2$, where $\omega$ is a constant. Let $\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i\hat p}{m\omega}\right)$. In the Heisenberg picture $\frac{d\hat a}{dt}$ is given by
1. $\omega\hat a$
2. $-i\omega\hat a$
3. $\omega\hat a^\dagger$
4. $i\omega\hat a^\dagger$

In Heisenberg picture the time evolution of operator is given by $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]+\frac{\partial\hat A_H(t)}{\partial t}$$ As $\hat A_H(t)$ does not explicitly depend on time $$\frac{d\hat A_H(t)}{dt}=\frac{i}{\hbar}\left[\hat H,\hat A_H(t)\right]$$ For harmonic oscillator $$\hat H=\frac{\hat p^2}{2m}+\frac{1}{2}m\omega^2 x^2$$ $$\begin{align*} \frac{d\hat p}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat p\right]\\ &=-m\omega^2\hat x \end{align*}$$ $$\begin{align*} \frac{d\hat x}{dt}&=\frac{i}{\hbar}\left[\hat H,\hat x\right]\\ &=\frac{\hat p}{m} \end{align*}$$ $$\begin{align*} \frac{d\hat a}{dt}&=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{d\hat x}{dt}+\frac{i}{m\omega}\frac{d\hat p}{dt}\right)\\ &=\sqrt{\frac{m\omega}{2\hbar}}\left(\frac{\hat p}{m}-\frac{i}{m\omega}m\omega^2\hat x\right)\\ &=-i\omega\hat a \end{align*}$$

Hence, answer is (B)

5. Two different sets of orthogonal basis vectors $\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\}$ and $\left\{\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix},\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\right\}$ are given for a two-dimensional real vector space. The matrix representation of a linear operator $\hat A$ in these bases are related by a unitary transformation. The unitary matrix may be chosen to be
1. $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$
2. $\begin{pmatrix}0&1\\1&0\end{pmatrix}$
3. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$
4. $\frac{1}{\sqrt{2}}\begin{pmatrix}1&0\\1&1\end{pmatrix}$

$$U_{ij}=< \psi_{new_i}|\psi_{old_j} >$$ $$\begin{align*} U_{11}&=< \psi_{new_1}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*}$$ $$\begin{align*} U_{12}&=< \psi_{new_1}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*}$$ $$\begin{align*} U_{21}&=< \psi_{new_2}|\psi_{old_1} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\ &=\frac{1}{\sqrt{2}} \end{align*}$$ $$\begin{align*} U_{22}&=< \psi_{new_2}|\psi_{old_2} > \\ &=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}\\ &=-\frac{1}{\sqrt{2}} \end{align*}$$ $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$

Hence, answer is (C)