A charged particle carrying charge `1 mu C` is moving with velocity `(2 hati + 3hati + 4hatk) ms^(-1)`. If an external magnetic field of `(5 hati + 3hatj - 6hatk) xx 10^(-3)T` exists in the region where the particle is moving then the force on the particle is `vecF xx 10^(-9)N`. The vector `vecF` is :

Options

A. `-0.30 hati + 0.32 hatj - 0.9 hatk`
B. `-3.0 hati + 3.2hatj -0.9 hatk`
C. `-30hati + 32hatj - 9hatk`
D. `-300 hati + 320 hatj - 90hatk`

Explanation

To find the force on a charged particle moving in a magnetic field, we use the formula:

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

where:
- \( \vec{F} \) is the force,
- \( q \) is the charge,
- \( \vec{v} \) is the velocity vector, and
- \( \vec{B} \) is the magnetic field vector.

Given:
- Charge \( q = 1 \mu C = 1 \times 10^{-6} C \)
- Velocity \( \vec{v} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \, m/s \)
- Magnetic field \( \vec{B} = (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) \times 10^{-3} \, T \)

### Step 1: Calculate the cross product \( \vec{v} \times \vec{B} \)

We will set up the determinant for the cross product:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 \times 10^{-3} & 3 \times 10^{-3} & -6 \times 10^{-3} \end{vmatrix} \]

### Step 2: Expand the determinant

Calculating the determinant, we have:

\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 4 \\ 3 \times 10^{-3} & -6 \times 10^{-3} \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 5 \times 10^{-3} & -6 \times 10^{-3} \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 5 \times 10^{-3} & 3 \times 10^{-3} \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ 3 \times (-6 \times 10^{-3}) - 4 \times (3 \times 10^{-3}) = -18 \times 10^{-3} - 12 \times 10^{-3} = -30 \times 10^{-3} \]

2. For \( -\hat{j} \):
\[ 2 \times (-6 \times 10^{-3}) - 4 \times (5 \times 10^{-3}) = -12 \times 10^{-3} - 20 \times 10^{-3} = -32 \times 10^{-3} \]
Thus, it becomes \( +32 \times 10^{-3} \hat{j} \).

3. For \( \hat{k} \):
\[ 2 \times (3 \times 10^{-3}) - 3 \times (5 \times 10^{-3}) = 6 \times 10^{-3} - 15 \times 10^{-3} = -9 \times 10^{-3} \]

### Step 3: Combine the results

Putting it all together, we have:

\[ \vec{v} \times \vec{B} = (-30 \times 10^{-3}) \hat{i} + (32 \times 10^{-3}) \hat{j} + (-9 \times 10^{-3}) \hat{k} \]

### Step 4: Multiply by charge \( q \)

Now, we multiply by the charge \( q \):

\[ \vec{F} = q (\vec{v} \times \vec{B}) = (1 \times 10^{-6}) \times (-30 \times 10^{-3} \hat{i} + 32 \times 10^{-3} \hat{j} - 9 \times 10^{-3} \hat{k}) \]

Calculating this gives:

\[ \vec{F} = (-30 \times 10^{-9} \hat{i} + 32 \times 10^{-9} \hat{j} - 9 \times 10^{-9} \hat{k}) \, N \]

### Final Result

Thus, the force on the particle is:

\[ \vec{F} = -30 \hat{i} + 32 \hat{j} - 9 \hat{k} \times 10^{-9} \, N \]