Question

In a non-magnetic medium, the EM-wave equation is given as $\vec{E} = [30 \cos{(10^9t - 8x)}\hat{j} + 40 \sin{(10^9t - 8x)}\hat{k}] V/m$

• A. The value of dielectric constant is 8.
• B. The value of dielectric constant is $\frac{144}{25}$
• C. $\vec{B} = [24 \cos{(10^9t - 8x)}\hat{k} - 32 \sin{(10^9t - 8x)}\hat{j}] \times 10^{-8}T$
• D. $\vec{B} = [24 \sin{(10^9t - 8x)}\hat{k} + 32 \cos{(10^9t - 8x)}\hat{j}] \times 10^{-8}T$

Answer 1

Answer: (B), (C)

B and C

Answer 2

The problem deals with an electromagnetic wave propagating in a non-magnetic medium. We are given the electric field vector and need to find the dielectric constant of the medium and the magnetic field vector.

1. Determine the wave parameters: The given electric field equation is $\vec{E} = [30 \cos{(10^9t - 8x)}\hat{j} + 40 \sin{(10^9t - 8x)}\hat{k}] V/m$. Comparing this with the general form of an electromagnetic wave $E = E_0 \cos(\omega t - kx)$ or $E = E_0 \sin(\omega t - kx)$, we can identify:

• Angular frequency, $\omega = 10^9$ rad/s
• Wave number, $k = 8$ rad/m The wave propagates in the +x direction.

2. Calculate the speed of the wave in the medium: The speed of an electromagnetic wave in a medium is given by $v = \frac{\omega}{k}$. $v = \frac{10^9 \text{ rad/s}}{8 \text{ rad/m}} = 0.125 \times 10^9 \text{ m/s} = 1.25 \times 10^8 \text{ m/s}$.

3. Calculate the dielectric constant (K): For a non-magnetic medium, the permeability is approximately that of free space, i.e., $\mu = \mu_0$. The speed of light in a medium is also given by $v = \frac{1}{\sqrt{\mu\epsilon}}$, where $\epsilon$ is the permittivity of the medium. We know that $\epsilon = K \epsilon_0$, where $K$ is the dielectric constant (relative permittivity) and $\epsilon_0$ is the permittivity of free space. So, $v = \frac{1}{\sqrt{\mu_0 K \epsilon_0}}$. We also know that the speed of light in vacuum is $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3 \times 10^8 \text{ m/s}$. Therefore, $v = \frac{c}{\sqrt{K}}$. Rearranging for $K$: $K = \left(\frac{c}{v}\right)^2$ Substitute the values of $c$ and $v$: $K = \left(\frac{3 \times 10^8 \text{ m/s}}{1.25 \times 10^8 \text{ m/s}}\right)^2 = \left(\frac{3}{1.25}\right)^2 = \left(\frac{3}{5/4}\right)^2 = \left(\frac{12}{5}\right)^2 = \frac{144}{25}$. As a decimal, $K = 5.76$.

4. Calculate the magnetic field vector ($\vec{B}$): For a plane electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and the direction of propagation $\hat{k}_{prop}$ are mutually perpendicular. Their relationship is given by $\vec{B} = \frac{1}{v} (\hat{k}_{prop} \times \vec{E})$. Here, the propagation direction is $\hat{k}_{prop} = \hat{x}$. $\vec{E} = [30 \cos{(10^9t - 8x)}\hat{j} + 40 \sin{(10^9t - 8x)}\hat{k}]$ Now, substitute these into the formula for $\vec{B}$: $\vec{B} = \frac{1}{1.25 \times 10^8} \left( \hat{x} \times [30 \cos{(10^9t - 8x)}\hat{j} + 40 \sin{(10^9t - 8x)}\hat{k}] \right)$ $\vec{B} = \frac{1}{1.25 \times 10^8} \left( 30 \cos{(10^9t - 8x)}(\hat{x} \times \hat{j}) + 40 \sin{(10^9t - 8x)}(\hat{x} \times \hat{k}) \right)$ Using the cross product identities: $\hat{x} \times \hat{j} = \hat{k}$ and $\hat{x} \times \hat{k} = -\hat{j}$. $\vec{B} = \frac{1}{1.25 \times 10^8} \left( 30 \cos{(10^9t - 8x)}\hat{k} - 40 \sin{(10^9t - 8x)}\hat{j} \right)$ Calculate the pre-factor: $\frac{1}{1.25 \times 10^8} = \frac{1}{5/4 \times 10^8} = \frac{4}{5} \times 10^{-8} = 0.8 \times 10^{-8}$. $\vec{B} = 0.8 \times 10^{-8} \left[ 30 \cos{(10^9t - 8x)}\hat{k} - 40 \sin{(10^9t - 8x)}\hat{j} \right]$ $\vec{B} = \left[ (0.8 \times 30) \cos{(10^9t - 8x)}\hat{k} - (0.8 \times 40) \sin{(10^9t - 8x)}\hat{j} \right] \times 10^{-8}$ $\vec{B} = \left[ 24 \cos{(10^9t - 8x)}\hat{k} - 32 \sin{(10^9t - 8x)}\hat{j} \right] \times 10^{-8} \text{ T}$.