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

\[ \vec{F} = Q (\vec{V} \times \vec{B}) \]

where:
- \(Q\) is the charge of the particle,
- \(\vec{V}\) is the velocity vector of the particle,
- \(\vec{B}\) is the magnetic field vector.

### Step 1: Identify the given values
- 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 2: Calculate the cross product \(\vec{V} \times \vec{B}\)
To find \(\vec{V} \times \vec{B}\), we can use the determinant of a matrix formed by the unit vectors and the components of \(\vec{V}\) and \(\vec{B}\):

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

### Step 3: Calculate 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 determinants:

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

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

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

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 the charge \(Q\)
Now we can find \(\vec{F}\):
\[ \vec{F} = Q (\vec{V} \times \vec{B}) = (1 \times 10^{-6}) \left((-30 \times 10^{-3}) \hat{i} + (32 \times 10^{-3}) \hat{j} + (-9 \times 10^{-3}) \hat{k}\right) \]

Calculating this gives:
\[ \vec{F} = (-30 \times 10^{-9}) \hat{i} + (32 \times 10^{-9}) \hat{j} + (-9 \times 10^{-9}) \hat{k} \]

Thus, the final expression for \(\vec{F}\) is:
\[ \vec{F} = -30 \hat{i} + 32 \hat{j} - 9 \hat{k} \, \times 10^{-9} \, N \]

### Final Answer
The vector \(\vec{F}\) is:
\[ \vec{F} = (-30 \hat{i} + 32 \hat{j} - 9 \hat{k}) \times 10^{-9} \, N \]