A charge particle is moving with a velocity `3hat(i) + 4 hat(j) m//sec` and it has electric field `E = 10hat(k) N//C` at a given point. Find the magnitude of magneitc field at the same point due to the motion of the charge particle.

Options

A. `100(mu_0 varepsilon_0)`
B. `25(mu_0varepsilon_0)`
C. `12.5(mu_0varepsilon_0)`
D. `50(mu_0varepsilon_0)`

Explanation

To solve the problem, we need to find the magnitude of the magnetic field at a point due to a charged particle moving in an electric field. The steps are as follows:

### Step 1: Identify the Given Values
- Velocity of the charged particle, **v** = \(3\hat{i} + 4\hat{j}\) m/s
- Electric field, **E** = \(10\hat{k}\) N/C

### Step 2: Use the Relationship Between Electric Field and Magnetic Field
The magnetic field **B** due to a moving charge can be calculated using the formula:
\[ \mathbf{B} = \frac{\mu_0}{4\pi} \frac{q \mathbf{v} \times \mathbf{r}}{r^2} \]
However, we can also relate the electric field to the magnetic field using:
\[ \mathbf{E} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \hat{r} \]
From this, we can express \( \frac{q}{r^2} \) in terms of the electric field:
\[ \frac{q}{r^2} = 10 \quad \text{(since } E = 10\hat{k} \text{)} \]

### Step 3: Calculate the Cross Product
We need to find \( \mathbf{v} \times \mathbf{r} \). Here, we can assume the position vector \( \mathbf{r} \) is along the z-axis, thus:
\[ \mathbf{r} = \hat{k} \]
Now, calculate the cross product:
\[ \mathbf{v} \times \mathbf{r} = (3\hat{i} + 4\hat{j}) \times \hat{k} \]
Using the right-hand rule:
\[ \hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i} \]
Thus,
\[ \mathbf{v} \times \mathbf{r} = 3(-\hat{j}) + 4\hat{i} = 4\hat{i} - 3\hat{j} \]

### Step 4: Substitute Back to Find Magnetic Field
Now we can substitute this back into the equation for the magnetic field:
\[ \mathbf{B} = 10 \mu_0 \epsilon_0 (4\hat{i} - 3\hat{j}) \]

### Step 5: Calculate the Magnitude of the Magnetic Field
To find the magnitude of **B**:
\[ |\mathbf{B}| = 10 \mu_0 \epsilon_0 \sqrt{(4^2 + (-3)^2)} = 10 \mu_0 \epsilon_0 \sqrt{16 + 9} = 10 \mu_0 \epsilon_0 \sqrt{25} = 10 \mu_0 \epsilon_0 \times 5 \]
Thus,
\[ |\mathbf{B}| = 50 \mu_0 \epsilon_0 \]

### Final Answer
The magnitude of the magnetic field at the point due to the motion of the charged particle is:
\[ \boxed{50 \mu_0 \epsilon_0} \]