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

Concept: A plane electromagnetic wave is represented as: \[ \sin(kx \pm \omega t) \] Where:

\( k \) = wave number \( = \frac{\omega}{v} \)
\( \omega \) = angular frequency
\( v \) = speed of wave in medium
Speed of EM wave in medium: \[ v = \frac{c}{n} \]
Step 1: Identify given quantities. From the wave equation: \[ \omega = 1.5 \times 10^{11} \, \text{rad/s} \] Refractive index: \[ n = 1.5 \]
Step 2: Speed of wave in glass. \[ v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \, \text{m/s} \]
Step 3: Calculate wave number \( k \). \[ k = \frac{\omega}{v} \] \[ k = \frac{1.5 \times 10^{11}}{2 \times 10^8} = 0.75 \times 10^3 = 7.5 \times 10^2 \, \text{m}^{-1} \]
Step 4: Identify \( \alpha \). Comparing with \( \sin(\alpha x + \omega t) \), we get: \[ \alpha = k = 7.5 \times 10^2 \, \text{m}^{-1} \]

Answer 2

Instead of computing the wave speed and dividing directly, this method goes through the wavelength, using the relation between angular frequency and vacuum wavelength, then adjusting for the medium.

The wave's angular frequency, read off the given equation \( B_y = (2\times10^{-7}\,\text{T})\sin(\alpha x + 1.5\times10^{11}t) \), is \( \omega = 1.5\times10^{11}\,\text{rad/s} \).

In vacuum, an electromagnetic wave of this angular frequency would have wave number:

\[ k_{\text{vac}} = \frac{\omega}{c} = \frac{1.5 \times 10^{11}}{3 \times 10^8} = 5 \times 10^2 \, \text{m}^{-1} \]

When the same wave enters a medium of refractive index \( n \), its frequency stays the same but its speed drops to \( c/n \), which means its wavelength shrinks by the same factor \( n \), and correspondingly its wave number grows by a factor of \( n \):

\[ \alpha = k_{\text{medium}} = n \times k_{\text{vac}} = 1.5 \times (5 \times 10^2) = 7.5 \times 10^2 \, \text{m}^{-1} \]

This confirms the same result through the wavelength-scaling relationship rather than computing the medium's speed directly.

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