Problem set 90

1. A plane polarized EM wave of frequency $\omega$ is incident at an angle $\theta$ in a rectangular waveguide of resonant frequency $\omega_{mn}$. Then energy carried by the wave propagating inside the cavity will propagate with the group velocity of :
1. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}}$
2. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}$
3. $\frac{c}{\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)}}$
4. $c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$

$$v_g=c\sqrt{1-\left(\frac{\omega_{mn}}{\omega}\right)^2}$$

Hence, answer is (D)

2. The electric field of an electromagnetic wave propagating in the free space is given by : $\vec E(r,t)=E_0\hat z\cos{\left[200\sqrt{3}\pi x-200\pi y-\omega t\right]}$. Then the wave vector $\vec k$ is given by
1. $200\frac{\sqrt{3}}{2}\pi\hat x-200\pi\hat y$
2. $400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$
3. $200\sqrt{3}\pi\hat x$
4. $-200\pi\hat y$

General format of electromagnetic wave equation is $$\vec E(r,t)=E_0\hat z\cos{(\vec k\cdot\vec r-\omega t)}$$ $$\vec k\cdot\vec r= 200\sqrt{3}\pi x-200\pi y$$ $$\Rightarrow \vec k=400\pi\left[\frac{\sqrt{3}}{2}\hat x-\frac{1}{2}\hat y\right]$$

Hence, answer is (B)

3. The Ampere's law in the free space takes the form:
1. $\vec\nabla\times\vec B=\mu_0\vec J$
2. $\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
3. $\vec\nabla\times\vec B=\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$
4. $\vec\nabla\times\vec B=\mu_0\vec J-\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

$\vec\nabla\times\vec B=\mu_0\vec J+\epsilon_0\mu_0\frac{\partial\vec E}{\partial t}$

Hence, answer is (B)

4. An electric charge $+Q$ is placed at the center of a cube of sides 10 cm. The electric flux emanating from each of the face of the cube is :
1. $\frac{Q}{\epsilon_0}$
2. $\frac{Q}{10\epsilon_0}$
3. $\frac{Q}{6\epsilon_0}$
4. $\frac{10Q}{\epsilon_0}$

According to Gauss’ Theorem, the total electric flux through all the six faces of the cube is, $$\Phi = \frac{Q}{\epsilon_0}$$ Each face of the cube comprises $1/6$ of the total surface of the cube. Therefore, the flux passing through each face is, $$\Phi = \frac{Q}{6\epsilon_0}$$

Hence, answer is (C)

5. A field at certain point in the space is expressed as the potential function $V=3x^2z-xy^3+z$. Then the potential $V$ at point $(2,-1,1)$ is :
1. 15 V
2. 13 V
3. 0 V
4. 8 V

$$V=32^2\times1-2\times(-1)^3+1=15 V$$

Hence, answer is (A)