`1 micro 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 find the force experienced by a charged particle moving in a magnetic field, we can use the formula for the magnetic force:

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

where:
- \( q \) is the charge,
- \( \vec{v} \) is the velocity vector,
- \( \vec{B} \) 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, \( \vec{B} = (3 \hat{i} + 4 \hat{j} - 3 \hat{k}) \times 10^{-3} \, T \)

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)

We will calculate the cross product using the determinant of a matrix:

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

### Step 3: Expand the determinant

Calculating the determinant, we have:

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

Calculating each of these 2x2 determinants:

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

2. For \( \hat{j} \):
\[ 4 \cdot (-3 \times 10^{-3}) - 3 \cdot (3 \times 10^{-3}) = -12 \times 10^{-3} - 9 \times 10^{-3} = -21 \times 10^{-3} \]

3. For \( \hat{k} \):
\[ 4 \cdot (4 \times 10^{-3}) - 6 \cdot (3 \times 10^{-3}) = 16 \times 10^{-3} - 18 \times 10^{-3} = -2 \times 10^{-3} \]

### Step 4: Combine the results

Thus, we have:

\[ \vec{v} \times \vec{B} = (-30 \times 10^{-3}) \hat{i} + (21 \times 10^{-3}) \hat{j} + (-2 \times 10^{-3}) \hat{k} \]

### Step 5: Multiply by the charge

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

\[ \vec{F} = q (\vec{v} \times \vec{B}) = (1 \times 10^{-6}) \cdot (-30 \times 10^{-3} \hat{i} + 21 \times 10^{-3} \hat{j} - 2 \times 10^{-3} \hat{k}) \]

Calculating this gives:

\[ \vec{F} = (-30 \times 10^{-9} \hat{i} + 21 \times 10^{-9} \hat{j} - 2 \times 10^{-9} \hat{k}) \, N \]

### Step 6: Find the magnitude of the force

The magnitude of the force vector is:

\[ |\vec{F}| = \sqrt{(-30 \times 10^{-9})^2 + (21 \times 10^{-9})^2 + (-2 \times 10^{-9})^2} \]

Calculating this:

\[ |\vec{F}| = \sqrt{900 + 441 + 4} \times 10^{-18} = \sqrt{1345} \times 10^{-9} \, N \]

### Step 7: Final answer

Thus, the force experienced by the charged particle in units of \( 10^{-9} \, N \) is approximately:

\[ |\vec{F}| \approx 36.7 \times 10^{-9} \, N \]