Question

A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60 V m^-1, the equation for magnetic field is:

• A. \( B_z = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T} \)
• B. \( B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T} \)
• C. \( B_x = 60 \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{i} \, \text{T} \)
• D. \( B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T} \)

Answer 1

The Correct Option is B

Given the problem of determining the equation for the magnetic field of an electromagnetic wave, we start by analyzing the provided information:

• The electromagnetic wave has a wavelength (\(\lambda\)) of 4 mm or 0.004 meters.
• It propagates along the x-direction, meaning the wave vector \(\mathbf{k}\) is oriented along x.
• The electric field \(\mathbf{E}\) is oriented along the y-direction with an amplitude of 60 V/m.
• The speed of light \(c = 3 \times 10^8 \, \text{m/s}\).

We aim to determine the equation for the magnetic field, \(\mathbf{B}\).

Step-by-Step Solution:

1. Start by determining the wave number \(k\), which is given by the formula:
   • \( k = \frac{2\pi}{\lambda} = \frac{2\pi}{0.004} = 500\pi \, \text{m}^{-1} \).
2. The general equation for a wave in one dimension moving in the positive x-direction is:
   • \( E = E_0 \sin(kx - \omega t) \).
   • Here, the electric field is \(E = 60 \sin(kx - \omega t) \hat{j} \). Given \(E_0 = 60 \, \text{V/m}\).
3. The frequency \(\nu\) of the wave can be calculated using the wave equation:
   • \( c = \nu \lambda \Rightarrow \nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{0.004} = 7.5 \times 10^{10} \, \text{Hz} \).
4. The angular frequency is \(\omega = 2\pi \nu = 2\pi \cdot 7.5 \times 10^{10} = 1.5 \times 10^{11} \, \text{rad/s} \).
5. The relation between electric and magnetic fields in an electromagnetic wave is given by:
   • \( B_0 = \frac{E_0}{c} \), where \( B_0 \) is the magnetic field amplitude.
   • \( B_0 = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \, \text{T} \).
6. The direction of \(\mathbf{B}\) is determined by taking the cross-product of the \(\mathbf{k}\) and \(\mathbf{E}\) using the right-hand rule. Since \(\mathbf{E}\) is in \(\hat{j}\) and propagation is along \(\hat{i}\), \(\mathbf{B}\) is in \(\hat{k}\).
7. Thus, the magnetic field equation is:
   • \( B_z = 2 \times 10^{-7} \sin(kx - \omega t) \hat{k} \).
   • Substitute \(k = 500\pi\) and \(\omega = 1.5 \times 10^{11}\):
   • \( B_z = 2 \times 10^{-7} \sin\left(500\pi \cdot x - 1.5 \times 10^{11} \cdot t\right) \hat{k} \, \text{T} \).
8. After simplifying the wave number \(k\) and frequency in terms relevant to the given options, the equation for the magnetic field becomes:
   • \( B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T} \).
   • This matches the correct option given in the question.

Therefore, the correct option is \( B_z = 2 \times 10^{-7} \sin \left[ \frac{\pi}{2} \times 10^3 \left( x - 3 \times 10^8 t \right) \right] \hat{k} \, \text{T} \).

Answer 2

Step 1: Relation between electric and magnetic fields

The relationship between the electric field \(E\) and magnetic field \(B\) is:

\[ E = cB, \]

where \(c = 3 \times 10^8 \, \text{m/s}\).

Substitute \(E = 60 \, \text{Vm}^{-1}\):

\[ 60 = 3 \times 10^8 \cdot B \implies B = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \, \text{T}. \]

Step 2: Calculate the frequency

The wavelength is given as:

\[ \lambda = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m}. \]

The wave velocity \(c\) is related to the frequency \(f\) as:

\[ c = f\lambda \implies f = \frac{c}{\lambda} = \frac{3 \times 10^8}{4 \times 10^{-3}} = \frac{3}{4} \times 10^{11} \, \text{Hz}. \]

Step 3: Angular frequency

The angular frequency \(\omega\) is given by:

\[ \omega = 2\pi f = 2\pi \cdot \frac{3}{4} \times 10^{11} = \frac{3\pi}{2} \times 10^{11}. \]

Thus:

\[ \omega = \frac{\pi}{2} \times 10^3. \]

Step 4: Determine the direction of the fields

• The electric field is in the \(y\)-direction (\(\hat{\jmath}\)).
• The wave propagates in the \(x\)-direction (\(\hat{\imath}\)).
• The magnetic field must be perpendicular to both \(\hat{\imath}\) and \(\hat{\jmath}\), i.e., in the \(z\)-direction (\(\hat{k}\)).

Step 5: Equation of the magnetic field

The magnetic field \(B_z\) is:

\[ B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3 \left(x - 3 \times 10^8 t\right)\right] \hat{k}. \]

Final Answer: \[ B_z = 2 \times 10^{-7} \sin\left[\frac{\pi}{2} \times 10^3 \left(x - 3 \times 10^8 t\right)\right] \, \hat{k} \, \text{kT}. \]