A 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. 11 q
B. 5 q
C. 3 q
D. 2 q

Explanation

To find the y-component of the force experienced by a charge \( +q \) moving in an electromagnetic field, we can use the Lorentz force equation:

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

Where:
- \( \vec{F} \) is the force,
- \( q \) is the charge,
- \( \vec{E} \) is the electric field,
- \( \vec{v} \) is the velocity of the charge,
- \( \vec{B} \) is the magnetic field.

### Step 1: Identify the vectors
Given:
- \( \vec{v} = 3\hat{i} + 4\hat{j} + \hat{k} \)
- \( \vec{E} = 3\hat{i} + \hat{j} + 2\hat{k} \)
- \( \vec{B} = \hat{i} + \hat{j} - 3\hat{k} \)

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)
We can calculate the cross product using the determinant of a matrix:

\[ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 1 \\ 1 & 1 & -3 \end{vmatrix} \]

Calculating the determinant:

1. For \( \hat{i} \):
\[ \hat{i} \left( 4 \cdot (-3) - 1 \cdot 1 \right) = \hat{i} (-12 - 1) = -13\hat{i} \]

2. For \( \hat{j} \):
\[ -\hat{j} \left( 3 \cdot (-3) - 1 \cdot 1 \right) = -\hat{j} (-9 - 1) = 10\hat{j} \]

3. For \( \hat{k} \):
\[ \hat{k} \left( 3 \cdot 1 - 4 \cdot 1 \right) = \hat{k} (3 - 4) = -1\hat{k} \]

Combining these, we get:
\[ \vec{v} \times \vec{B} = -13\hat{i} + 10\hat{j} - \hat{k} \]

### Step 3: Calculate the total force \( \vec{F} \)
Using the Lorentz force equation:
\[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) \]

Substituting the values:
\[ \vec{F} = q \left( (3\hat{i} + \hat{j} + 2\hat{k}) + (-13\hat{i} + 10\hat{j} - \hat{k}) \right) \]

Combining the components:
\[ \vec{F} = q \left( (3 - 13)\hat{i} + (1 + 10)\hat{j} + (2 - 1)\hat{k} \right) \]
\[ \vec{F} = q \left( -10\hat{i} + 11\hat{j} + 1\hat{k} \right) \]

### Step 4: Identify the y-component of the force
The y-component of the force \( \vec{F} \) is:
\[ F_y = 11q \]

### Final Answer
The y-component of the force experienced by the charge \( +q \) is \( 11q \).

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