An electric charge `+q` moves with velocity `vecv=3hati+4hatj+hatk`,in an electromagnetic field given by:
`vecE=3hati+hatj+2hatk` and `vecB=hati+hatj+3hatk`.The `y`-component of the force experienced by `+q` is:

Options

A. -7q
B. 5q
C. -3q
D. 2q

Answer

To find the y-component of the force experienced by a charge \( +q \) moving in an electromagnetic field, we need to consider both the electric force and the magnetic force acting on the charge.

### Step-by-Step Solution:

1. **Identify the Electric Force**:
The electric force \( \vec{F}_E \) on a charge in an electric field is given by:
\[ \vec{F}_E = q \vec{E} \]
Given the electric field \( \vec{E} = 3 \hat{i} + \hat{j} + 2 \hat{k} \), the y-component of the electric force \( F_{Ey} \) is:
\[ F_{Ey} = q E_y = q \cdot 1 = q \]

2. **Identify the Magnetic Force**:
The magnetic force \( \vec{F}_B \) on a charge moving in a magnetic field is given by:
\[ \vec{F}_B = q (\vec{v} \times \vec{B}) \]
where \( \vec{v} = 3 \hat{i} + 4 \hat{j} + \hat{k} \) and \( \vec{B} = \hat{i} + \hat{j} + 3 \hat{k} \).

3. **Calculate the Cross Product \( \vec{v} \times \vec{B} \)**:
We can calculate the cross product using the determinant of a matrix:
\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 1 & 1 & 3 \end{vmatrix} \]
Expanding this determinant, we have:
\[ \vec{v} \times \vec{B} = \hat{i} \begin{vmatrix} 4 & 1 \\ 1 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} \]
Calculating the determinants:
- For \( \hat{i} \):
\[ 4 \cdot 3 - 1 \cdot 1 = 12 - 1 = 11 \]
- For \( \hat{j} \):
\[ 3 \cdot 3 - 1 \cdot 1 = 9 - 1 = 8 \]
- For \( \hat{k} \):
\[ 3 \cdot 1 - 4 \cdot 1 = 3 - 4 = -1 \]
Thus, we have:
\[ \vec{v} \times \vec{B} = 11 \hat{i} - 8 \hat{j} - 1 \hat{k} \]

4. **Calculate the y-component of the Magnetic Force**:
The y-component of the magnetic force \( F_{By} \) is:
\[ F_{By} = q (v \times B)_y = q \cdot (-8) = -8q \]

5. **Combine the Forces**:
The total y-component of the force \( F_y \) is the sum of the electric and magnetic y-components:
\[ F_y = F_{Ey} + F_{By} = q + (-8q) = q - 8q = -7q \]

### Final Answer:
The y-component of the force experienced by the charge \( +q \) is:
\[ F_y = -7q \hat{j} \]