A particle is moving with velocity `vec(v)=hat(i)+3hatj` and it produces an electrostatic field at a point given by `vec(E)=2hat(k)`. It will produce magnetic field at that point equal to (all quantities are in S.I. units and speed of light is `c`)

Options

A. `(6 hat(i) - 2hat(j))/(c^2)`
B. `(6 hat(i) + 2hat(j))/(c^2)`
C. zero
D. data insufficient

Answer

To find the magnetic field produced by a particle moving with a given velocity and creating an electrostatic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Quantities**:
- Velocity of the particle: \(\vec{v} = \hat{i} + 3\hat{j}\)
- Electrostatic field at the point: \(\vec{E} = 2\hat{k}\)

2. **Understand the Relationship**:
- The magnetic field \(\vec{B}\) produced by a moving charge can be calculated using the formula:
\[ \vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{q \cdot \vec{v} \times \hat{r}}{r^2} \]
- Here, \(\hat{r}\) is the unit vector pointing from the charge to the point where the field is measured, and \(r\) is the distance from the charge to that point.

3. **Determine the Unit Vector \(\hat{r}\)**:
- Since the electrostatic field is given as \(2\hat{k}\), we can infer that the point where the electric field is measured is along the z-axis. Thus, the unit vector \(\hat{r}\) can be taken as \(\hat{k}\).

4. **Cross Product Calculation**:
- Calculate the cross product \(\vec{v} \times \hat{r}\):
\[ \vec{v} = \hat{i} + 3\hat{j}, \quad \hat{r} = \hat{k} \]
\[ \vec{v} \times \hat{r} = (\hat{i} + 3\hat{j}) \times \hat{k} \]
- Using the right-hand rule:
\[ \hat{i} \times \hat{k} = -\hat{j}, \quad \hat{j} \times \hat{k} = \hat{i} \]
\[ \vec{v} \times \hat{r} = \hat{i} \times \hat{k} + 3\hat{j} \times \hat{k} = -\hat{j} + 3\hat{i} = 3\hat{i} - \hat{j} \]

5. **Substituting into the Magnetic Field Formula**:
- Substitute \(\vec{v} \times \hat{r}\) into the magnetic field formula:
\[ \vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{q (3\hat{i} - \hat{j})}{r^2} \]

6. **Using the Relationship Between \(\mu_0\), \(\epsilon_0\), and \(c\)**:
- Recall that:
\[ \frac{1}{c^2} = \mu_0 \epsilon_0 \]
- Thus, we can express \(\mu_0\) in terms of \(\epsilon_0\) and \(c\).

7. **Final Expression for Magnetic Field**:
- The magnetic field can be expressed as:
\[ \vec{B} = \frac{\mu_0 q}{4\pi r^2} (3\hat{i} - \hat{j}) \]
- After substituting the values and simplifying, we arrive at:
\[ \vec{B} = \frac{6\hat{i} - 2\hat{j}}{c^2} \]

### Conclusion:
The magnetic field produced at that point is:
\[ \vec{B} = \frac{6\hat{i} - 2\hat{j}}{c^2} \]