Two electromagnetic waves are moving in free space whose electric field vectors are given by `vecE_1 =vecE_0 hatjcos(kx -omegat) & vecE_2 = vecE_0 hatk cos(ky -omegat)`. A charge q is moving with velocity `vecv = 0.8 c hatj`. Find the net Lorentz force on this charge at `t = 0` and when it is at origin

Options

A. Option1 `E_(0)q(0.4hati-3hatj+0.8hatk)`
B. Option2 `E_(0)q(0.8hati-hatj + 0.4hatk)`
C. Option3 `E_(0)q(-0.8hati+hatj+hatk)`
D. Option4 `E_(0)q(0.8hati+hatj+0.2hatk)`

Answer

To find the net Lorentz force on the charge \( q \) moving with velocity \( \vec{v} = 0.8c \hat{j} \) in the presence of two electromagnetic waves, we will follow these steps:

### Step 1: Identify the Electric Fields
The electric fields of the two electromagnetic waves are given as:
\[ \vec{E_1} = \vec{E_0} \hat{j} \cos(kx - \omega t) \]
\[ \vec{E_2} = \vec{E_0} \hat{k} \cos(ky - \omega t) \]

At \( t = 0 \), these fields become:
\[ \vec{E_1} = \vec{E_0} \hat{j} \cos(kx) \]
\[ \vec{E_2} = \vec{E_0} \hat{k} \cos(ky) \]

### Step 2: Calculate the Electric Field at the Origin
At the origin \( (x = 0, y = 0) \):
\[ \vec{E_1}(0, 0) = \vec{E_0} \hat{j} \cos(0) = \vec{E_0} \hat{j} \]
\[ \vec{E_2}(0, 0) = \vec{E_0} \hat{k} \cos(0) = \vec{E_0} \hat{k} \]

### Step 3: Calculate the Total Electric Field
The total electric field \( \vec{E} \) at \( t = 0 \) is:
\[ \vec{E} = \vec{E_1} + \vec{E_2} = \vec{E_0} \hat{j} + \vec{E_0} \hat{k} = \vec{E_0} (\hat{j} + \hat{k}) \]

### Step 4: Calculate the Magnetic Fields
Using the relationship between electric and magnetic fields in electromagnetic waves, we find the magnetic fields \( \vec{B_1} \) and \( \vec{B_2} \).

For \( \vec{E_1} \):
- The direction of propagation is along \( \hat{i} \) (x-direction).
- The magnetic field \( \vec{B_1} \) is given by:
\[ \vec{B_1} = \frac{\vec{E_0}}{c} \hat{k} \cos(kx - \omega t) \]
At \( t = 0 \):
\[ \vec{B_1}(0) = \frac{\vec{E_0}}{c} \hat{k} \cos(k \cdot 0) = \frac{\vec{E_0}}{c} \hat{k} \]

For \( \vec{E_2} \):
- The direction of propagation is along \( \hat{j} \) (y-direction).
- The magnetic field \( \vec{B_2} \) is given by:
\[ \vec{B_2} = \frac{\vec{E_0}}{c} \hat{i} \cos(ky - \omega t) \]
At \( t = 0 \):
\[ \vec{B_2}(0) = \frac{\vec{E_0}}{c} \hat{i} \cos(k \cdot 0) = \frac{\vec{E_0}}{c} \hat{i} \]

### Step 5: Calculate the Total Magnetic Field
The total magnetic field \( \vec{B} \) at \( t = 0 \) is:
\[ \vec{B} = \vec{B_1} + \vec{B_2} = \frac{\vec{E_0}}{c} \hat{k} + \frac{\vec{E_0}}{c} \hat{i} = \frac{\vec{E_0}}{c} (\hat{i} + \hat{k}) \]

### Step 6: Calculate the Lorentz Force
The Lorentz force \( \vec{F} \) on the charge \( q \) is given by:
\[ \vec{F} = q (\vec{E} + \vec{v} \times \vec{B}) \]
Substituting the values:
- \( \vec{E} = \vec{E_0} (\hat{j} + \hat{k}) \)
- \( \vec{v} = 0.8c \hat{j} \)
- \( \vec{B} = \frac{\vec{E_0}}{c} (\hat{i} + \hat{k}) \)

Now, calculate \( \vec{v} \times \vec{B} \):
\[ \vec{v} \times \vec{B} = (0.8c \hat{j}) \times \left(\frac{\vec{E_0}}{c} (\hat{i} + \hat{k})\right) \]
Using the cross product:
\[ = 0.8 \vec{E_0} (\hat{j} \times \hat{i} + \hat{j} \times \hat{k}) = 0.8 \vec{E_0} (-\hat{k} + \hat{i}) \]

### Step 7: Combine Forces
Now, substitute back into the Lorentz force equation:
\[ \vec{F} = q \left( \vec{E_0} (\hat{j} + \hat{k}) + 0.8 \vec{E_0} (-\hat{k} + \hat{i}) \right) \]
\[ = q \left( \vec{E_0} \hat{j} + \vec{E_0} \hat{k} + 0.8 \vec{E_0} (-\hat{k}) + 0.8 \vec{E_0} \hat{i} \right) \]
\[ = q \left( 0.8 \vec{E_0} \hat{i} + \vec{E_0} \hat{j} + (1 - 0.8) \vec{E_0} \hat{k} \right) \]
\[ = q \left( 0.8 \vec{E_0} \hat{i} + \vec{E_0} \hat{j} + 0.2 \vec{E_0} \hat{k} \right) \]

### Final Result
Thus, the net Lorentz force on the charge \( q \) at \( t = 0 \) is:
\[ \vec{F} = q \vec{E_0} (0.8 \hat{i} + \hat{j} + 0.2 \hat{k}) \]