The magnetic force action on a charged particle of charge `-2muC` in a magnetic field of 2T acting in y direction , when the particle velocity is `(2hati+3hatj)xx10^6ms^(-1)` , is

Options

A. 4 N in x direction
B. 8 N in x direction
C. 8 N in z direction
D. 8 N in -z direction

Explanation

To solve the problem of finding the magnetic force acting on a charged particle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Charge of the particle, \( q = -2 \, \mu C = -2 \times 10^{-6} \, C \)
- Magnetic field, \( \vec{B} = 2 \, T \, \hat{j} \) (acting in the y-direction)
- Velocity of the particle, \( \vec{v} = (2 \hat{i} + 3 \hat{j}) \times 10^6 \, m/s \)

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

3. **Substitute the Values into the Formula**:
Substitute the values of \( q \), \( \vec{v} \), and \( \vec{B} \):
\[ \vec{F} = -2 \times 10^{-6} \, C \, \left( (2 \hat{i} + 3 \hat{j}) \times 10^6 \, m/s \right) \times (2 \hat{j}) \]

4. **Calculate the Cross Product**:
First, calculate the cross product \( \vec{v} \times \vec{B} \):
\[ \vec{v} = 2 \hat{i} + 3 \hat{j} \]
\[ \vec{B} = 2 \hat{j} \]
Using the determinant method for the cross product:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 0 & 2 & 0 \end{vmatrix} \]
This determinant expands to:
\[ \hat{i}(3 \cdot 0 - 0 \cdot 2) - \hat{j}(2 \cdot 0 - 0 \cdot 2) + \hat{k}(2 \cdot 2 - 3 \cdot 0) \]
Simplifying this gives:
\[ \vec{v} \times \vec{B} = 0 \hat{i} - 0 \hat{j} + 4 \hat{k} = 4 \hat{k} \]

5. **Calculate the Magnetic Force**:
Now substitute back into the force equation:
\[ \vec{F} = -2 \times 10^{-6} \, C \times 4 \hat{k} = -8 \times 10^{-6} \hat{k} \, N \]

6. **Final Result**:
The magnetic force acting on the charged particle is:
\[ \vec{F} = -8 \times 10^{-6} \hat{k} \, N \]
This means the magnitude of the force is \( 8 \, \mu N \) and it acts in the negative z-direction.