Question

The magnetic field in a plane electromagnetic wave travelling in glass \((n = 1.5)\) is given by \[ B_y=(2\times10^{-7}\,\text{T})\sin(\alpha x+1.5\times10^{11}t) \] where \(x\) is in metres and \(t\) is in seconds. The value of \(\alpha\) is:

• A. \(0.5\times10^3\ \text{m}^{-1}\)
• B. \(6.0\times10^2\ \text{m}^{-1}\)
• C. \(7.5\times10^2\ \text{m}^{-1}\)
• D. \(1.5\times10^3\ \text{m}^{-1}\)

Hint:

For an electromagnetic wave, \[ v=\frac{\omega}{k} \] and \[ v=\frac{c}{n}. \] Therefore, \[ k=\frac{\omega n}{c}. \] This direct formula is often the fastest way to solve such questions.

Answer 1

The Correct Option is C

Concept: The general equation of a plane electromagnetic wave is \[ B=B_0\sin(kx\pm\omega t), \] where \[ k=\frac{2\pi}{\lambda} \] is called the wave number and \[ \omega=2\pi\nu \] is the angular frequency. The speed of an electromagnetic wave in a medium of refractive index \(n\) is \[ v=\frac{c}{n}, \] where \(c\) is the speed of light in vacuum. Also, \[ v=\frac{\omega}{k}. \] Therefore, \[ k=\frac{\omega}{v}. \] Since \(\alpha\) is the coefficient of \(x\), we identify \[ \alpha=k. \]

Step 1: Determine the speed of the wave in glass. Given, \[ n=1.5. \] Using \[ v=\frac{c}{n}, \] we obtain \[ v=\frac{3\times10^8}{1.5} =2\times10^8\ \text{m s}^{-1}. \]

Step 2: Identify the angular frequency from the given wave equation. Comparing \[ B_y=(2\times10^{-7})\sin(\alpha x+1.5\times10^{11}t) \] with \[ B=B_0\sin(kx+\omega t), \] we get \[ \omega=1.5\times10^{11}\ \text{rad s}^{-1}. \]

Step 3: Calculate the wave number. Using \[ k=\frac{\omega}{v}, \] we have \[ k=\frac{1.5\times10^{11}} {2\times10^8}. \] Therefore, \[ k=0.75\times10^3. \] Thus, \[ k=7.5\times10^2\ \text{m}^{-1}. \] Since \[ \alpha=k, \] we obtain \[ \boxed{\alpha=7.5\times10^2\ \text{m}^{-1}}. \] Hence, the correct answer is \[ \boxed{\text{(C)}} \]