`1micro C` charge moves with the velocity `vecv=4hati+6hatj+3hatk` in uniform magnetic field, ` vecB= 3hati+4hatj-3hatk`xx 10^(-3). Force experience by charged particle in units of 10^(-9) N will be,

Options

A. `-3hati+2.1hatj+0.4hatk`
B. `-30hati+21hatj-2hatk`
C. `-0.03hati+0.21hatj+0.04hatk`
D. `-3hati+0.2hatj+0.4hatk`

Answer

To solve the problem, we need to calculate the magnetic force experienced by a charged particle moving in a magnetic field. The formula for the magnetic force \( \vec{F} \) on a charged particle is given by:

\[ \vec{F} = q (\vec{v} \times \vec{B}) \]

where:
- \( q \) is the charge,
- \( \vec{v} \) is the velocity vector of the charge,
- \( \vec{B} \) is the magnetic field vector,
- \( \times \) denotes the cross product.

### Step 1: Identify the given values
- Charge \( q = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Velocity \( \vec{v} = 4 \hat{i} + 6 \hat{j} + 3 \hat{k} \)
- Magnetic field \( \vec{B} = (3 \hat{i} + 4 \hat{j} - 3 \hat{k}) \times 10^{-3} \)

### Step 2: Factor out the \( 10^{-3} \) from the magnetic field
We can rewrite the magnetic field as:
\[ \vec{B} = 3 \hat{i} + 4 \hat{j} - 3 \hat{k} \quad \text{(in units of } 10^{-3} \text{ T)} \]

### Step 3: Calculate the cross product \( \vec{v} \times \vec{B} \)
We will calculate the cross product using the determinant of a matrix:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 3 \\ 3 & 4 & -3 \end{vmatrix} \]

### Step 4: Expand the determinant
Using the determinant, we have:

\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} 6 & 3 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 3 \\ 3 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 6 \\ 3 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ 6 \cdot (-3) - 3 \cdot 4 = -18 - 12 = -30 \]

2. For \( \hat{j} \):
\[ 4 \cdot (-3) - 3 \cdot 3 = -12 - 9 = -21 \quad \text{(note the negative sign in front)} \]

3. For \( \hat{k} \):
\[ 4 \cdot 4 - 6 \cdot 3 = 16 - 18 = -2 \]

Putting it all together, we have:
\[ \vec{v} \times \vec{B} = -30 \hat{i} + 21 \hat{j} - 2 \hat{k} \]

### Step 5: Multiply by the charge and the factor from the magnetic field
Now we multiply by the charge \( q \) and the factor \( 10^{-3} \):
\[ \vec{F} = q (\vec{v} \times \vec{B}) = 1 \times 10^{-6} \left(-30 \hat{i} + 21 \hat{j} - 2 \hat{k}\right) \times 10^{-3} \]

This simplifies to:
\[ \vec{F} = -30 \times 10^{-9} \hat{i} + 21 \times 10^{-9} \hat{j} - 2 \times 10^{-9} \hat{k} \]

### Step 6: Final expression for the force
Thus, the force experienced by the charged particle is:
\[ \vec{F} = (-30 \hat{i} + 21 \hat{j} - 2 \hat{k}) \times 10^{-9} \, N \]

### Conclusion
The force experienced by the charged particle in units of \( 10^{-9} \, N \) is:
\[ \vec{F} = -30 \hat{i} + 21 \hat{j} - 2 \hat{k} \]