Problem set 24

1. The Hamiltonian of a simple pendulum consisting of a mass $m$ attached to a massless string of length $l$ is $H = \frac{p_\theta^2}{2ml^2}+ mgl(1- \cos\theta)$. If $L$ denotes the Lagrangian, the value $\frac{dL}{dt}$ is:
1. $-\frac{2g}{l}p_\theta\sin\theta$
2. $-\frac{g}{l}p_\theta\sin{2\theta}$
3. $\frac{g}{l}p_\theta\cos{\theta}$
4. $lp_\theta^2\cos{\theta}$
$$L=\sum\limits_ip_iq_i-H$$ $$L=p_\theta\dot\theta-\frac{p_\theta^2}{2ml^2}- mgl(1- \cos\theta)$$ As $L\equiv L(q,\dot q,t)$ $$\frac{dL}{dt}=\frac{\partial L}{\partial q}\frac{\partial q}{\partial t}+\frac{\partial L}{\partial\dot q}\frac{\partial\dot q}{\partial t}+\frac{\partial L}{\partial t}$$ In our case$L\equiv L(\theta,\dot \theta)$ $$\frac{dL}{dt}=\frac{\partial L}{\partial \theta}\frac{\partial \theta}{\partial t}+\frac{\partial L}{\partial\dot\theta}\frac{\partial\dot\theta}{\partial t}$$ $$\frac{\partial L}{\partial \theta}=-mgl\sin{\theta}$$ $$\frac{\partial L}{\partial\dot \theta}=p_\theta$$ $$\frac{dL}{dt}=-mgl\sin{\theta}\: \dot\theta+p_\theta\:\ddot\theta$$ $$\dot\theta=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{ml^2}$$ $$\ddot\theta=\frac{\dot p_\theta}{ml^2}$$ But $$\dot p_\theta=-\frac{\partial H}{\partial \theta}=-mgl\sin{\theta}$$ $$\ddot\theta=-\frac{g}{l}\sin{\theta}$$ $$\frac{dL}{dt}=-\frac{2g}{l}p_\theta\sin{\theta}$$ Hence, answer is (A)
2. Two bodies of equal mass $m$ are connected by a massless rigid rod of length $l$ lying in the XY-plane with the centre of the rod at the origin. If this system is rotating about the $z$-axis with a frequency $\omega$, its angular momentum is
1. $ml^2\omega/4$
2. $ml^2\omega/2$
3. $ml^2\omega$
4. $2ml^2\omega$
Angular momentum, $L=I\omega$ $$I=\sum_im_ir_i^2$$ Here, $r_i=\frac{l}{2}$ $$I=m\frac{l^2}{4}+m\frac{l^2}{4}=m\frac{l^2}{2}$$ $$L=\frac{ml^2\omega}{2}$$ Hence, answer is (B)
3. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current $I$. The vector potential $\vec A$ at a distance $R$ from the axis
1. is constant inside and varies as $R$ outside the solenoid
2. varies as $R$ inside and is constant outside the solenoid
3. varies as $\frac{1}{R}$ inside and as $R$ outside the solenoid
4. varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Let $r$ is the radius of solenoid.
Magnetic filed inside the solenoid is $B=\mu_0 n I$, where $n$ is number of turns per unit length, and, $I$ is current flowing through the wire. Magnetic filed outside the solenoid is zero.

Vector potential inside the solenoid:
Now, according to Stoke's law, we have $$\oint \vec A\cdot d\vec l=\int_S\left(\nabla\times\vec A\right)\cdot d\vec S$$ But $\left(\nabla\times\vec A\right)=\vec B$ $$\oint \vec A\cdot d\vec l=\int_S\vec B\cdot d\vec S=\Phi_B=flux$$ Let us calculate flux $\Phi_B$. For this let us consider a circular path of radius $R$ centered along its axis. The flux through the circular area is $$\Phi_B=\pi R^2B=\pi R^2\mu_0 n I$$ Hence, $$\oint \vec A\cdot d\vec l=\pi R^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$, this is because $\vec A$ is along the direction of current which is circumferential. $$2\pi R A_\Phi=\pi R^2\mu_0 n I$$ $$A_\Phi=\mu_0 n I\frac{R}{2}$$

Vector potential outside the solenoid:
Let us consider the point $R>r$. The flux through circular area is $$\Phi_B=\pi r^2B=\pi r^2\mu_0 n I$$ $$\oint \vec A\cdot d\vec l=\pi r^2\mu_0 n I$$ But, $\oint \vec A\cdot d\vec l=2\pi R A_\Phi$ $$2\pi R A_\Phi=\pi r^2\mu_0 n I$$ $$A=\mu_0 n I\frac{r^2}{2R}$$
Hence, $\vec A$ varies as $R$ inside and as $\frac{1}{R}$ outside the solenoid
Hence, answer is (D)
4. Consider an infinite conducting sheet in the xy-plane with a time dependent cunent density $Kt\hat i$, where $K$ is a constant. The vector potential at $(x, y,z)$ is given by $\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$. The magnetic field $\vec B$ is
1. $\frac{\mu_0Kt}{2}\hat j$
2. $-\frac{\mu_0Kz}{2c}\hat j$
3. $-\frac{\mu_0K}{2c}(ct-z)\hat i$
4. $-\frac{\mu_0K}{2c}(ct-z)\hat j$
$$\vec A=\frac{\mu_0K}{4c}(ct-z)^2\hat i$$ $$\begin{align*} \vec B&=\vec\nabla\times\vec A\\ &=\begin{vmatrix} \hat i&\hat j&\hat k\\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ \frac{\mu_0K}{4c}(ct-z)^2&0&0 \end{vmatrix} \end{align*}$$ $$\vec B=-\frac{\mu_0K}{2c}(ct-z)\hat j$$ Hence, answer is (D)
5. When a charged particle emits electromagnetic radiation, the electric field $\vec E$ and the Poynting vector $\vec S = \frac{1}{\mu_0} \vec E\times\vec B$ at a large distance $r$ from the emitter vary as $\frac{1}{r^n}$ and $\frac{1}{r^m}$ respectively. Which of the following choices for $n$ and $m$ are correct?
1. $n = 1$ and $m=1$
2. $n = 2$ and $m=2$
3. $n = 1$ and $m=2$
4. $n = 2$ and $m=4$
When a charged particle emits electromagnetic radiation $E\propto\frac{1}{r}$ and $B\propto\frac{1}{r}$. As $\vec S=\frac{\vec E\times \vec B}{\mu_0}$, $S\propto\frac{1}{r^2}$
Hence, answer is (C)