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Step-by-Step Solution
Step 1: Understand the Given Electric Field
The electric field of the plane electromagnetic wave is given by
$ \overrightarrow{E} = E_0 \,\widehat{j}\,\cos\bigl(\omega t - kx\bigr). $
Here, the wave is propagating along the x-direction, and the electric field oscillates along the y-axis (denoted by $ \widehat{j} $).
Step 2: Recall the Relationship Between $\overrightarrow{E}$ and $\overrightarrow{B}$
For an electromagnetic wave traveling in vacuum:
1) The magnitudes of the electric and magnetic fields are related by
$ E_0 = c\, B_0, $
where $ c $ is the speed of light in vacuum given by
$ c = \frac{1}{\sqrt{\mu_0 \,\epsilon_0}}. $
2) The electric field, magnetic field, and the direction of propagation are mutually perpendicular.
Step 3: Find the Magnitude of $\overrightarrow{B}$
From $ E_0 = c\,B_0 $ and $ c = \frac{1}{\sqrt{\mu_0 \,\epsilon_0}}, $ we get:
$ B_0 = \frac{E_0}{c} = E_0 \,\sqrt{\mu_0 \,\epsilon_0}. $
Step 4: Determine the Direction of $\overrightarrow{B}$
Since the wave propagates along the x-axis ( $ \widehat{i} $ ) and the electric field is along $ \widehat{j} $, the magnetic field must be perpendicular to both $ \widehat{i} $ and $ \widehat{j} $, which means $ \overrightarrow{B} $ is along $ \widehat{k} $.
Step 5: Write the Final Expression for $\overrightarrow{B}$
Plugging the magnitude and direction back into the standard cosine form:
$ \overrightarrow{B} = B_0 \cos(\omega t - kx)\,\widehat{k}
= E_0 \sqrt{\mu_0\,\epsilon_0} \,\cos(\omega t - kx)\,\widehat{k}. $
At time $ t = 0 $, this becomes:
$ \overrightarrow{B} = E_0 \sqrt{\mu_0\,\epsilon_0} \,\cos(kx)\,\widehat{k}. $
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