Problem set 59

1. A dipole of moment $\vec p$, oscillating at frequency $\omega$, radiates spherical waves. The vector potential at large distance is $$\vec A(\vec r)=\frac{\mu_0}{4\pi}i\omega\frac{e^{ikr}}{r}\vec p$$. To order $(1/r)$ the magnetic field $\vec B$ at a point $\vec r=r\hat n$ is
1. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\cdot\vec p\right)\frac{e^{ikr}}{r}$
2. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\left(\hat n\times\vec p\right)\frac{e^{ikr}}{r}$
3. $-\frac{\mu_0}{4\pi}\omega^2k\left(\hat n\cdot\vec p\right)\vec p\frac{e^{ikr}}{r}$
4. $-\frac{\mu_0}{4\pi}\frac{\omega^2}{c}\vec p\frac{e^{ikr}}{r}$

Since $\vec B=\vec\nabla\times\vec A$, option (B) is correct

Hence, answer is (B)

2. For a two level system, the population of atoms in the upper and lower levels are $3\times10^{18}$ and $0.7\times10^{18}$, respectively. If the coefficient of stimulated emission is $3\times10^{5}\:m^3/W-s^3$ and the energy density is $9.0 J/m^3Hz$, the rate of stimulated emission will be
1. $6.3\times10^{16}\:s^{-1}$
2. $4.1\times10^{16}\:s^{-1}$
3. $2.7\times10^{16}\:s^{-1}$
4. $1.8\times10^{16}\:s^{-1}$

Rate of stimulated emission is $$R_{stimu}=\rho B$$ $$R_{stimu}= 9.0 \times 3\times10^{5}$$ $$R_{stimu}=2.7\times10^{16}\:s^{-1}$$

Hence, answer is (C)

3. The first ionization potential of K is 4.34 eV, the electron affinity of Cl is 3.82 eV and the equilibrium separation of KCl is 0.3 nm. The energy required to dissociate a KCl molecule into a K and a Cl atom is
1. 8.62 eV
2. 8.16 eV
3. 4.28 eV
4. 4.14 eV

The ionization energy to form K is 4.34 eV, hence we will require 4.34 eV energy to form $K^+$. Now, the electron affinity of Cl is 3.82 eV. Hence, when the electron liberated from potassium is put on Cl to make $Cl^-$ 3.82 eV energy liberated. Hence, net amount of energy to produce $K^+$ and $Cl^-$ is 4.34-3.82=0.52 eV.

The energy required to dissociate KCl into $K^+$ and $Cl^-$ is nothing but the potential energy of attraction between $K^+$ and $Cl^-$ ions, given by \begin{align*} V&=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}\\ &={\scriptstyle-9\times10^9\times\frac{\left(1.6\times10^{-19}\right)^2}{0.3\times10^{-9}}}\\ &=-7.7\times10^{-19}\:J\\ &=-4.79\:eV \end{align*} Thus, the energy required to dissociate KCl into $K^+$ and $Cl^-$ is $4.79\:eV$

Hence, The energy required to dissociate KCl into K and Cl is $4.79-0.52=4.27\:eV$

Hence, answer is (C)

4. Considers circuits as shown in figures (a) and (b) below.
   
   
   If transistors in figures (a) and (b) have current gain ($\beta_{dc}$) of 100 and 10 respectively, then they operate in the
1. active region and saturation region respectively
2. saturation region and active region respectively
3. saturation region in both cases
4. active region in both cases

For figure (a) $I_B=\frac{10.7-0.7}{10k}=1\:mA$. Hence, $I_c=\beta I_B=100\:mA$. Thus, \begin{align*} V_{CB}&=V_C-V_B\\ &=(10-2\times100)-0.7\\ &=-ve \end{align*}

Hence, collector to base junction is forward biased.

As both base to emitter and collector to base junctions are forward biased, the transistor is in saturation region.

For figure (b)$I_B=\frac{5-0.7}{10k}=0.43\:mA$. Hence, $I_c=\beta I_B=4.3\:mA$. Thus, \begin{align*} V_{CB}&=V_C-V_B\\ &=(10-1\times4.3)-0.7\\ &=+ve \end{align*}

Hence, collector to base junction is reverse biased.

As base to emitter junction is forward biased and collector to base junctions is reversed biased, the transistor is in active region.

Hence, answer is (B)

5. A small magnetic needle is kept at $(0,0)$ with its moment along the x-axis. Another small magnetic needle is at the point $(1, 1)$ and is free to rotate in the $xy$-plane. In equilibrium the angle $\theta$ between their magnetic moments is such that
1. $\tan\theta=1/3$
2. $\tan\theta=0$
3. $\tan\theta=3$
4. $\tan\theta=1$

Potential energy of two dipole system is given by \begin{align*} U&=\frac{\mu_0}{4\pi r^3}\left[\vec m_1\cdot\vec m_2-3\vec m_1\cdot\vec r)(\vec m_2\cdot\vec r) \right]\\ &={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\cos\theta -3(\cos{45^0})(\cos{\left(\theta-45^0\right)}) \right]}\\ \end{align*} For equilibrium potential energy is minimum i.e. $\frac{\partial U}{\partial\theta}=0$ \begin{align*} \frac{\partial U}{\partial\theta}&={\scriptstyle\frac{\mu_0m_1m_2}{4\pi r^3}\left[\sin\theta +\frac{\sqrt{3}}{2}\sin{\left(\theta-45^0\right)} \right]}\\ &=0\\ \end{align*} $$\Rightarrow \tan\theta=3$$

Hence, answer is (C)