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Step-by-Step Solution
Step 1: Identify the Given Magnetic Field
The problem provides the magnetic field of a plane electromagnetic wave as
$ \overrightarrow{B} = 3 \times 10^{-8} \sin\bigl(1.6 \times 10^{3}x \;+\; 4.8 \times 10^{11}t\bigr)\,\hat{j}\,\text{T} $.
Step 2: Recall the Relationship Between Electric and Magnetic Field Amplitudes
For an electromagnetic wave in free space, the magnitudes of the electric field $E_0$ and the magnetic field $B_0$ are related by the speed of light $c$:
$$
\frac{E_0}{B_0} = c.
$$
Here, $ c \approx 3 \times 10^8 \,\text{m/s}$.
Step 3: Compute the Electric Field Amplitude
We have $B_0 = 3 \times 10^{-8}\,\text{T}$. Using the relation $E_0 = B_0\,c$, we get:
$$
E_0 = \left(3 \times 10^{-8}\,\text{T}\right) \times \left(3 \times 10^{8}\,\text{m/s}\right) = 9\,\text{V/m}.
$$
Step 4: Determine the Direction of the Electric Field
In an electromagnetic wave traveling in the $x$-direction, if the magnetic field is along $\hat{j}$, then the electric field is perpendicular to both $\hat{i}$ (the direction of propagation) and $\hat{j}$ (the magnetic field), which means it must lie along $\hat{k}$.
Step 5: Write the Final Expression for the Electric Field
The electric field has the same angular frequency and wave number as the magnetic field. Thus, its expression is:
$$
\overrightarrow{E} = 9\,\sin\bigl(1.6 \times 10^3 x + 4.8 \times 10^{10}t\bigr)\,\hat{k}\,\text{V/m}.
$$
This matches the form given in the correct answer.
