Question

In lossless medium for which \( \eta = 60\pi \), \( \mu_r = 1 \) and \( H = -0.1\cos(\omega t - z) \hat{a}_x + 0.5\sin(\omega t - z) \hat{a}_y \text{ A/m} \), calculate \(\omega\).

• A. \( 1.0 \times 10^6 \) rad/s
• B. \( 1.0 \times 10^8 \) rad/s
• C. \( 1.5 \times 10^6 \) rad/s
• D. \( 1.5 \times 10^8 \) rad/s

Hint:

For a plane wave in a lossless medium: \(\beta = \omega\sqrt{\mu\epsilon}\) and \(\eta = \sqrt{\mu/\epsilon}\).
From these, \(v_p = \omega/\beta = 1/\sqrt{\mu\epsilon}\).
Also, \( \beta = \omega \mu / \eta \) or \( \omega = \beta \eta / \mu \).
In the wave expression \(f(\omega t - \beta z)\), the coefficient of \(z\) is \(\beta\).
\(\mu_0 = 4\pi \times 10^{-7}\) H/m.

Answer 1

The Correct Option is D

For a lossless medium, the intrinsic impedance \( \eta \) is given by \( \eta = \sqrt{\frac{\mu}{\epsilon}} \), where \( \mu = \mu_0 \mu_r \) and \( \epsilon = \epsilon_0 \epsilon_r \). Also, the wave number (phase constant) \( \beta \) is related to \( \omega \), \( \mu \), and \( \epsilon \) by \( \beta = \omega\sqrt{\mu\epsilon} \). 

The wave propagates as \( \cos(\omega t - \beta z) \) or \( \sin(\omega t - \beta z) \). From the given H-field expression, \( H = -0.1\cos(\omega t - z) \hat{a}_x + 0.5\sin(\omega t - z) \hat{a}_y \), we can see that the term multiplying \( z \) is the wave number \( \beta \). So, \( \beta = 1 \) rad/m.

We are given \( \eta = 60\pi \, \Omega \) and \( \mu_r = 1 \). Since \( \mu_r = 1 \), \( \mu = \mu_0 = 4\pi \times 10^{-7} \) H/m. From \( \eta = \sqrt{\frac{\mu}{\epsilon}} \), we have:

\( \eta^2 = \frac{\mu}{\epsilon} \Rightarrow \epsilon = \frac{\mu}{\eta^2} \)

Substituting \( \mu_0 \) and \( \mu_r \), we get:

\( \epsilon = \frac{\mu_0 \mu_r}{\eta^2} = \frac{4\pi \times 10^{-7} \times 1}{(60\pi)^2} = \frac{4\pi \times 10^{-7}}{3600\pi^2} = \frac{1 \times 10^{-7}}{900\pi} \, \text{F/m} \)

Now use \( \beta = \omega\sqrt{\mu\epsilon} \). We know \( \beta = 1 \).

\( 1 = \omega \sqrt{(\mu_0 \mu_r) \left(\frac{\mu_0 \mu_r}{\eta^2}\right)} = \omega \sqrt{\frac{(\mu_0 \mu_r)^2}{\eta^2}} = \omega \frac{\mu_0 \mu_r}{\eta} \)

So, \( \omega = \frac{\eta}{\mu_0 \mu_r} \). Substituting the values:

\( \omega = \frac{60\pi}{4\pi \times 10^{-7} \times 1} = \frac{60}{4 \times 10^{-7}} = \frac{15}{10^{-7}} = 15 \times 10^7 \, \text{rad/s} \)

Thus, \( \omega = 1.5 \times 10^8 \, \text{rad/s} \), which matches option (d).

Alternatively, the phase velocity \(v_p = \frac{\omega}{\beta}\). Also, \(v_p = \frac{1}{\sqrt{\mu\epsilon}}\).

And \( \eta = \sqrt{\frac{\mu}{\epsilon}} \).

From \( \eta = \sqrt{\frac{\mu}{\epsilon}} \) and \(v_p = \frac{1}{\sqrt{\mu\epsilon}}\):

\( \eta v_p = \sqrt{\frac{\mu}{\epsilon}} \frac{1}{\sqrt{\mu\epsilon}} = \frac{1}{\epsilon} \).

\( \frac{v_p}{\eta} = \frac{1}{\sqrt{\mu\epsilon}} \sqrt{\frac{\epsilon}{\mu}} = \frac{1}{\mu} \). So, \(v_p = \eta/\mu\) is not correct.

We have \( \beta = \omega/v_p \). So \(v_p = \omega/\beta\).

Also, \(v_p = 1/\sqrt{\mu\epsilon}\).

\( \eta = \sqrt{\mu/\epsilon} \implies \epsilon = \mu/\eta^2 \).

\(v_p = 1/\sqrt{\mu (\mu/\eta^2)} = 1/\sqrt{\mu^2/\eta^2} = \eta/\mu\).

Given \( \beta = 1 \), then \( \omega = v_p \beta = v_p \).

So, \( \omega = \frac{\eta}{\mu} = \frac{\eta}{\mu_0 \mu_r} = \frac{60\pi}{4\pi \times 10^{-7} \times 1} = 1.5 \times 10^8 \) rad/s.

\( \boxed{1.5 \times 10^8 \, \text{rad/s}} \)