Question

For an electromagnetic wave, consider an electric field \(E = E_0e^{-i[a(x+y)-\omega t]\hat{k}\). The corresponding magnetic field \(B\) is (\(E_0, a, \omega\) are constants of appropriate dimensions and c is the speed of light)}

• A. \(\frac{1}{c\sqrt{2}}E_0e^{-i[a(x+y)-\omega t]}(\hat{i} - \hat{j})\)
• B. \(\frac{1}{c\sqrt{2}}E_0e^{-i[a(x+y)-\omega t]}(\hat{i} + \hat{j})\)
• C. \(\frac{1}{c\sqrt{2}}E_0e^{-i[a(x+y)-\omega t]}(-\hat{i} - \hat{j})\)
• D. \(\frac{1}{c\sqrt{2}}E_0e^{-i[a(x+y)-\omega t]}(-\hat{i} + \hat{j})\)

Hint:

A faster method for plane waves is to use the relation \(\vec{B} = \frac{1}{c}(\hat{k}_{prop} \times \vec{E})\), where \(\hat{k}_{prop}\) is the unit vector in the direction of propagation. From the phase \(a(x+y)-\omega t\), the direction of propagation is \((\hat{i}+\hat{j})\), so \(\hat{k}_{prop} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}\). \(\vec{E}\) is in the \(\hat{k}\) direction. The cross product \((\hat{i}+\hat{j}) \times \hat{k} = (\hat{i}\times\hat{k}) + (\hat{j}\times\hat{k}) = -\hat{j} + \hat{i} = \hat{i} - \hat{j}\). This quickly gives the direction of \(\vec{B}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The relationship between the electric field (\(\vec{E}\)) and magnetic field (\(\vec{B}\)) of a plane electromagnetic wave is governed by Maxwell's equations. Specifically, Faraday's law of induction, \(\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}\), connects the spatial variation of \(\vec{E}\) to the temporal variation of \(\vec{B}\).
Step 2: Key Formula or Approach
We will use Faraday's Law in differential form: \[ \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \] From this, we can find \(\vec{B}\) by integrating \(-(\nabla \times \vec{E})\) with respect to time.
For a plane wave, this relationship can be written as \(\vec{k} \times \vec{E} = \omega \vec{B}\), where \(\vec{k}\) is the wave vector.
Step 3: Detailed Explanation
Given the electric field: \(\vec{E} = E_0 e^{-i[a(x+y)-\omega t]} \hat{k}\). This can be written as \(\vec{E} = E_z \hat{k}\), where \(E_z = E_0 e^{-i[a(x+y)-\omega t]}\).
First, let's compute the curl of \(\vec{E}\): \[ \nabla \times \vec{E} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
0 & 0 & E_z \end{vmatrix} = \hat{i}\left(\frac{\partial E_z}{\partial y} - 0\right) - \hat{j}\left(\frac{\partial E_z}{\partial x} - 0\right) + \hat{k}(0-0) \] \[ \nabla \times \vec{E} = \hat{i}\frac{\partial E_z}{\partial y} - \hat{j}\frac{\partial E_z}{\partial x} \] Now, calculate the partial derivatives: \[ \frac{\partial E_z}{\partial y} = \frac{\partial}{\partial y} \left(E_0 e^{-i[a(x+y)-\omega t]}\right) = E_z \cdot (-ia) \] \[ \frac{\partial E_z}{\partial x} = \frac{\partial}{\partial x} \left(E_0 e^{-i[a(x+y)-\omega t]}\right) = E_z \cdot (-ia) \] Substitute these back into the curl expression: \[ \nabla \times \vec{E} = \hat{i}(-ia E_z) - \hat{j}(-ia E_z) = -ia E_z (\hat{i} - \hat{j}) \] Now, use Faraday's Law: \[ -\frac{\partial \vec{B}}{\partial t} = \nabla \times \vec{E} = -ia E_z (\hat{i} - \hat{j}) \] \[ \frac{\partial \vec{B}}{\partial t} = ia E_z (\hat{i} - \hat{j}) = ia E_0 e^{-i[a(x+y)-\omega t]} (\hat{i} - \hat{j}) \] To find \(\vec{B}\), we integrate with respect to time \(t\): \[ \vec{B} = \int ia E_0 e^{-ia(x+y)} e^{i\omega t} (\hat{i} - \hat{j}) dt \] \[ \vec{B} = ia E_0 e^{-ia(x+y)} (\hat{i} - \hat{j}) \int e^{i\omega t} dt \] \[ \vec{B} = ia E_0 e^{-ia(x+y)} (\hat{i} - \hat{j}) \left( \frac{e^{i\omega t}}{i\omega} \right) \] \[ \vec{B} = \frac{a}{\omega} E_0 e^{-i[a(x+y)-\omega t]} (\hat{i} - \hat{j}) \] Finally, we need to find the relationship between \(a\) and \(\omega\). The phase of the wave is \(a(x+y) - \omega t\), which is \(\vec{k} \cdot \vec{r} - \omega t\).
So, the wave vector is \(\vec{k} = a\hat{i} + a\hat{j}\).
The magnitude of the wave vector is \(k = |\vec{k}| = \sqrt{a^2 + a^2} = a\sqrt{2}\).
For an electromagnetic wave in vacuum, the dispersion relation is \(\omega = ck\).
\[ \omega = c(a\sqrt{2}) \implies \frac{a}{\omega} = \frac{1}{c\sqrt{2}} \] Step 4: Final Answer
Substitute the value of \(\frac{a}{\omega}\) back into the expression for \(\vec{B}\): \[ \vec{B} = \frac{1}{c\sqrt{2}} E_0 e^{-i[a(x+y)-\omega t]} (\hat{i} - \hat{j}) \] This matches option (A).