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Step-by-Step Solution
Step 1: Identify the Electric Field and Its Direction
The given electric field at time t = 0 is
$ \vec{E}(x,z) = 10 \,\hat{j}\,\cos\bigl(6x + 8z\bigr). $
This shows that the amplitude of the electric field is 10 and its direction is along $ \hat{j}. $
Step 2: Determine the Wave Vector and Direction of Propagation
The argument of the cosine function, $6x + 8z,$ reveals that the wave is propagating in the direction of the vector
$ \vec{K} = 6\,\hat{i} + 8\,\hat{k}. $
The magnitude of this wave vector is
$ |\vec{K}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10. $
Hence, the direction of propagation (unit vector) is
$ \hat{c} = \dfrac{\vec{K}}{|\vec{K}|} = \dfrac{6\,\hat{i} + 8\,\hat{k}}{10}. $
Step 3: Relation Between $\vec{E}$, $\vec{B}$, and $\vec{c}$
In an electromagnetic wave, the electric field $\vec{E}$, the magnetic field $\vec{B}$, and the direction of wave propagation $\vec{c}$ (or $\vec{k}$) are mutually perpendicular. The magnetic field direction is given by
$ \hat{B} = \hat{c} \times \hat{E}. $
Here, $ \hat{E} $ is along $ \hat{j} $.
Step 4: Compute the Direction of the Magnetic Field
First, note that
$ \hat{c} = \dfrac{6\,\hat{i} + 8\,\hat{k}}{10}, \quad \hat{E} = \hat{j}. $
So,
$$
\hat{c} \times \hat{E}
= \biggl(\dfrac{6\,\hat{i} + 8\,\hat{k}}{10}\biggr) \times \hat{j}.
$$
Recall:
$ \hat{i} \times \hat{j} = \hat{k}, \quad \hat{k} \times \hat{j} = -\hat{i}. $
Therefore,
$$
\hat{c} \times \hat{E}
= \dfrac{1}{10}\bigl(6\,\hat{k} - 8\,\hat{i}\bigr).
$$
Step 5: Find the Magnitude of $\vec{B}$
For an electromagnetic wave in free space, the magnitudes of $\vec{E}$ and $\vec{B}$ satisfy
$E = c\,B.$
Hence,
$ B = \dfrac{E}{c} = \dfrac{10}{c}. $
Step 6: Construct $\vec{B}$ With Magnitude and Direction
Combining the direction from Step 4 and the magnitude from Step 5, we get
$$
\vec{B} = \dfrac{10}{c} \times \dfrac{1}{10}\bigl(6\,\hat{k} - 8\,\hat{i}\bigr)
= \dfrac{6\,\hat{k} - 8\,\hat{i}}{c}.
$$
Step 7: Include the Time Dependent Phase
The angular wave number is $k = |\vec{K}| = 10$ and the wave travels as
$ \cos\bigl(6x + 8z - 10ct\bigr). $
Therefore, the magnetic field is
$$
\vec{B}(x,z,t)
= \dfrac{6\,\hat{k} - 8\,\hat{i}}{c}\,
\cos\bigl(6x + 8z - 10ct\bigr).
$$
Final Answer
The correct expression for the magnetic field is:
$$
\boxed{
\dfrac{1}{c}\bigl(6\,\hat{k} - 8\,\hat{i}\bigr)\,
\cos\bigl(6x + 8z - 10ct\bigr).
}
$$
