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Step-by-Step Detailed Solution
Step 1: Identify the given fields at t = 0, origin
At time t = 0 and position at the origin (x = 0, y = 0), the cosine terms become
$ \cos(\omega \times 0 - k \times 0) = \cos(0) = 1.$
Thus the electric fields reduce to:
$\overrightarrow{E_1} = E_0 \,\hat{j}\,(\cos(0)) = E_0 \,\hat{j}$.
$\overrightarrow{E_2} = E_0 \,\hat{k}\,(\cos(0)) = E_0 \,\hat{k}$.
Similarly, the corresponding magnetic fields at t = 0, origin are:
$\overrightarrow{B_1} = \frac{E_0}{c}\,\hat{k}\,\bigl(\cos(0)\bigr) = \frac{E_0}{c}\,\hat{k}.$
$\overrightarrow{B_2} = \frac{E_0}{c}\,\hat{i}\,\bigl(\cos(0)\bigr) = \frac{E_0}{c}\,\hat{i}.$
Step 2: Write the total electric field and total magnetic field
The net electric field is
$ \overrightarrow{E}_{\text{net}} = \overrightarrow{E_1} + \overrightarrow{E_2}
= E_0 \,\hat{j} + E_0 \,\hat{k} = E_0(\hat{j} + \hat{k}). $
The net magnetic field is
$ \overrightarrow{B}_{\text{net}} = \overrightarrow{B_1} + \overrightarrow{B_2}
= \frac{E_0}{c}\,\hat{k} + \frac{E_0}{c}\,\hat{i}
= \frac{E_0}{c}(\hat{k} + \hat{i}). $
Step 3: Write down the Lorentz force expression
The total force on the charged particle (charge = q) moving with velocity
$ \overrightarrow{v} $ is given by the Lorentz force law:
$ \overrightarrow{F} = q \,\bigl(\overrightarrow{E}_{\text{net}} + \overrightarrow{v} \times \overrightarrow{B}_{\text{net}}\bigr). $
Given:
$ \overrightarrow{v} = 0.8\,c\,\hat{j}. $
Step 4: Compute the cross product $ \overrightarrow{v} \times \overrightarrow{B}_{\text{net}} $
Substitute $ \overrightarrow{v} = 0.8\,c\,\hat{j} $ and
$ \overrightarrow{B}_{\text{net}} = \frac{E_0}{c}(\hat{k} + \hat{i}) $:
$
\overrightarrow{v} \times \overrightarrow{B}_{\text{net}}
= 0.8\,c\,\hat{j} \;\times\; \frac{E_0}{c}(\hat{k} + \hat{i})
= 0.8\,E_0\,\Bigl[\hat{j} \times (\hat{k} + \hat{i})\Bigr].
$
Now, handle the vector cross products individually:
$ \hat{j} \times \hat{k} = \hat{i}. $
$ \hat{j} \times \hat{i} = -\,\hat{k}. $
Hence,
$
\hat{j} \times (\hat{k} + \hat{i})
= \hat{j} \times \hat{k} + \hat{j} \times \hat{i}
= \hat{i} - \hat{k}.
$
Therefore,
$
\overrightarrow{v} \times \overrightarrow{B}_{\text{net}}
= 0.8\,E_0\,(\hat{i} - \hat{k}).
$
Step 5: Sum the electric field and magnetic force contributions
The net force is:
$
\overrightarrow{F}
= q\,\Bigl[\overrightarrow{E}_{\text{net}} + \bigl(\overrightarrow{v} \times \overrightarrow{B}_{\text{net}}\bigr)\Bigr]
= q\,\Bigl[E_0(\hat{j} + \hat{k}) + 0.8\,E_0\,(\hat{i} - \hat{k})\Bigr].
$
Combine like terms:
$
= q\,E_0\,\Bigl[0.8\,\hat{i} + \hat{j} + \hat{k} - 0.8\,\hat{k}\Bigr]
= q\,E_0\,\Bigl[0.8\,\hat{i} + \hat{j} + (1 - 0.8)\,\hat{k}\Bigr]
= q\,E_0\,\bigl[0.8\,\hat{i} + \hat{j} + 0.2\,\hat{k}\bigr].
$
Step 6: State the final result
Thus, the instantaneous force experienced by the particle is:
$ \overrightarrow{F} = E_0\,q\,\Bigl(0.8\,\hat{i} + \hat{j} + 0.2\,\hat{k}\Bigr). $
