Question

The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:

• A. \( 0.5 \times 10^3 \, \text{m}^{-1} \)
• B. \( 6.0 \times 10^2 \, \text{m}^{-1} \)
• C. \( 7.5 \times 10^2 \, \text{m}^{-1} \)
• D. \( 1.5 \times 10^3 \, \text{m}^{-1} \)

Hint:

For EM waves in a medium: \( k = \frac{\omega}{v} \), and \( v = \frac{c}{n} \). Always find speed first, then wave number.

Answer 1

The Correct Option is C

To determine the value of \(\alpha\) in the given electromagnetic wave equation, we start by analyzing the properties of the wave provided in the question:

The magnetic field component of the electromagnetic wave is given by:

\(B_y = (2 \times 10^{-7} \, \text{T}) \sin(\alpha x + 1.5 \times 10^{11} t)\)

Here, the wave is traveling in glass, which has a refractive index \(n = 1.5\). The speed of light in a medium is given by:

\(v = \frac{c}{n}\)

where \(c = 3 \times 10^8 \, \text{m/s}\) is the speed of light in a vacuum. Thus, the speed of the electromagnetic wave in glass is:

\(v = \frac{3 \times 10^8 \, \text{m/s}}{1.5} = 2 \times 10^8 \, \text{m/s}\)

The wave number \(\alpha\) and the angular frequency \(\omega\) are related to the wave's speed by the formula:

\(v = \frac{\omega}{\alpha}\)

From the given equation, the angular frequency \(\omega\) is:

\(\omega = 1.5 \times 10^{11} \, \text{rad/s}\)

Rearranging the formula for speed:

\(\alpha = \frac{\omega}{v}\)

Substituting the known values:

\(\alpha = \frac{1.5 \times 10^{11} \, \text{rad/s}}{2 \times 10^8 \, \text{m/s}} = 7.5 \times 10^2 \, \text{m}^{-1}\)

Therefore, the value of \(\alpha\) is \(7.5 \times 10^2 \, \text{m}^{-1}\), which corresponds to the correct option.

Answer 2

There's a direct shortcut formula that skips finding the medium's speed as a separate step: since a wave's speed in a medium of refractive index \( n \) is \( c/n \), and wave number is always angular frequency divided by speed, the two facts combine into a single formula, \( \alpha = \dfrac{n\omega}{c} \).

Reading the known values straight off the given wave equation and the problem statement: the angular frequency is \( \omega = 1.5 \times 10^{11}\,\text{rad/s} \), the refractive index of glass is \( n = 1.5 \), and the speed of light in vacuum is \( c = 3 \times 10^8\,\text{m/s} \).

Substituting all three directly into the single-step formula:

\[ \alpha = \frac{n\omega}{c} = \frac{1.5 \times (1.5 \times 10^{11})}{3 \times 10^8} = \frac{2.25 \times 10^{11}}{3 \times 10^8} = 7.5 \times 10^2 \, \text{m}^{-1} \]

As a quick sanity check, this value is 1.5 times larger than the vacuum wave number would be for the same frequency (\( 5\times10^2\,\text{m}^{-1} \)), which makes sense since entering a denser medium always shortens the wavelength, and hence raises the wave number, by exactly the refractive index factor.

Therefore, the value of \( \alpha \) is \( 7.5 \times 10^2 \, \text{m}^{-1} \).