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Step-by-Step Solution
Step 1: Identify the Given Parameters
• Frequency of light: $v = \frac{3}{2\pi} \times 10^{12}\,\text{Hz}$.
• Direction of propagation of wave: $ \displaystyle \frac{\widehat{i} + \widehat{j}}{\sqrt{2}} $.
• Polarization (electric field) along $ \widehat{k} $ direction.
Step 2: Determine the Angular Frequency
The angular frequency $ \omega $ is related to frequency $ f $ by
$ \omega = 2\pi f $.
Thus,
\[
\omega = 2\pi \times \frac{3}{2\pi}\times 10^{12} = 3\times 10^{12}\,\text{rad s}^{-1}.
\]
Step 3: Find the Direction of the Magnetic Field
For an electromagnetic wave, the magnetic field $ \vec{B} $ is perpendicular to both the electric field $ \vec{E} $ (given along $ \widehat{k} $) and the direction of propagation $ \vec{k} $ (which is along $ \frac{\widehat{i} + \widehat{j}}{\sqrt{2}} $). The direction of $ \vec{B} $ is given by
\[
\vec{B} \propto \left( \frac{\widehat{i} + \widehat{j}}{\sqrt{2}} \times \widehat{k} \right).
\]
Cross Product Calculation
Using the right-hand rule for the cross product:
\[
\frac{\widehat{i} + \widehat{j}}{\sqrt{2}} \times \widehat{k}
= \frac{1}{\sqrt{2}}
\left( \widehat{i} \times \widehat{k} + \widehat{j} \times \widehat{k} \right).
\]
Now,
\[
\widehat{i} \times \widehat{k} = -\widehat{j},
\quad
\widehat{j} \times \widehat{k} = \widehat{i}.
\]
So,
\[
\frac{1}{\sqrt{2}}
\left( -\widehat{j} + \widehat{i} \right)
= \frac{\widehat{i} - \widehat{j}}{\sqrt{2}}.
\]
Hence, the magnetic field direction is
\[
\frac{\widehat{i} - \widehat{j}}{\sqrt{2}}.
\]
Step 4: Express the Magnetic Field as a Function of Space and Time
A standard electromagnetic wave traveling in the direction
$ \vec{k} = \frac{\widehat{i} + \widehat{j}}{\sqrt{2}} $
can be written in the form:
\[
\vec{B} = \vec{B}_0 \cos \bigl(\vec{k}\cdot \vec{r} - \omega t\bigr).
\]
Given that $ \vec{B}_0 $ points along
$ \frac{\widehat{i} - \widehat{j}}{\sqrt{2}} $,
we have
\[
\vec{B}_0 = B_0\,\frac{\widehat{i} - \widehat{j}}{\sqrt{2}}.
\]
In an electromagnetic wave,
$ B_0 = \frac{E_0}{c}
\quad (\text{where } c \text{ is speed of light}).
\]
Thus,
\[
\vec{B}_0 = \frac{E_0}{c} \cdot \frac{\widehat{i} - \widehat{j}}{\sqrt{2}}.
\]
Step 5: Final Expression for the Magnetic Field
Combining all these factors, the magnetic field is:
\[
\vec{B}
= \frac{E_0}{c}\,\frac{\widehat{i} - \widehat{j}}{\sqrt{2}}
\cos \Bigl[\Bigl(\frac{\widehat{i} + \widehat{j}}{\sqrt{2}}\Bigr)\cdot \vec{r} - \omega t\Bigr].
\]
Step 6: Match with the Correct Answer Option
From the given options, the expression that aligns with the above form is:
