A charge q moves with a velocity `2m//s` along x- axis in a unifrom magnetic field `B=(hati+2hatj+3hatk)` tesla.

Options

A. Charge will experience a force in z-y plane
B. Charge will experience a force along- y aixs
C. Charge will experience a force + z aixs
D. Charge will experience a force along - z axis

Answer

To solve the problem, we need to find the force acting on a charge \( q \) moving with a velocity \( \mathbf{v} = 2 \, \hat{i} \, \text{m/s} \) in a uniform magnetic field \( \mathbf{B} = \hat{i} + 2\hat{j} + 3\hat{k} \, \text{T} \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Velocity of the charge: \( \mathbf{v} = 2 \hat{i} \, \text{m/s} \)
- Magnetic field: \( \mathbf{B} = \hat{i} + 2\hat{j} + 3\hat{k} \, \text{T} \)

2. **Use the Formula for Magnetic Force:**
The force \( \mathbf{F} \) on a charge \( q \) moving in a magnetic field is given by:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \]

3. **Calculate the Cross Product \( \mathbf{v} \times \mathbf{B} \):**
- Write the vectors in component form:
\[ \mathbf{v} = 2 \hat{i} = (2, 0, 0) \]
\[ \mathbf{B} = \hat{i} + 2\hat{j} + 3\hat{k} = (1, 2, 3) \]
- The cross product can be calculated using the determinant of a matrix:
\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 1 & 2 & 3 \end{vmatrix} \]

4. **Calculate the Determinant:**
\[ \mathbf{v} \times \mathbf{B} = \hat{i} \begin{vmatrix} 0 & 0 \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 0 \\ 1 & 2 \end{vmatrix} \]
- Calculate each of the determinants:
- For \( \hat{i} \): \( 0 \cdot 3 - 0 \cdot 2 = 0 \)
- For \( \hat{j} \): \( 2 \cdot 3 - 0 \cdot 1 = 6 \)
- For \( \hat{k} \): \( 2 \cdot 2 - 0 \cdot 1 = 4 \)

Putting it all together:
\[ \mathbf{v} \times \mathbf{B} = 0 \hat{i} - 6 \hat{j} + 4 \hat{k} = -6 \hat{j} + 4 \hat{k} \]

5. **Calculate the Force:**
\[ \mathbf{F} = q (-6 \hat{j} + 4 \hat{k}) = -6q \hat{j} + 4q \hat{k} \]

6. **Determine the Plane of the Force:**
The force has components in the \( \hat{j} \) (y-axis) and \( \hat{k} \) (z-axis) directions. Therefore, the force acts in the \( yz \)-plane.

### Conclusion:
The charge will experience a force in the \( yz \) or \( zy \) plane.