In a magnetic field `vecB=hati+yhatj+3hatk` a charge particle (q, m) is moving with velocity `vecV=2hati+3hatj+zhatk` experiences a force `vecF=-hati+2hatj+hatk`. The value of y and z may be

Options

A. y=4, z=2
B. y=2, z=4
C. y=3, z=6
D. y=6, z=3

Answer

To solve the problem, we need to find the values of \(y\) and \(z\) given the magnetic field \(\vec{B}\), the velocity of the charged particle \(\vec{V}\), and the force \(\vec{F}\) experienced by the particle. We will use the formula for the magnetic force on a charged particle, which is given by:

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

### Step 1: Write down the vectors
Given:
- Magnetic field: \(\vec{B} = \hat{i} + y \hat{j} + 3 \hat{k}\)
- Velocity: \(\vec{V} = 2 \hat{i} + 3 \hat{j} + z \hat{k}\)
- Force: \(\vec{F} = -\hat{i} + 2 \hat{j} + \hat{k}\)

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

\[ \vec{V} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & z \\ 1 & y & 3 \end{vmatrix} \]

### Step 3: Expand the determinant
Calculating the determinant, we have:

\[ \vec{V} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & z \\ y & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & z \\ 1 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 1 & y \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} 3 & z \\ y & 3 \end{vmatrix} = 9 - zy\)
2. \(\begin{vmatrix} 2 & z \\ 1 & 3 \end{vmatrix} = 6 - z\)
3. \(\begin{vmatrix} 2 & 3 \\ 1 & y \end{vmatrix} = 2y - 3\)

Thus, we have:

\[ \vec{V} \times \vec{B} = (9 - zy) \hat{i} - (6 - z) \hat{j} + (2y - 3) \hat{k} \]

### Step 4: Set the cross product equal to the force vector
From the equation \(\vec{F} = q (\vec{V} \times \vec{B})\), we can equate components:

1. For \(\hat{i}\): \(9 - zy = -1\)
2. For \(\hat{j}\): \(- (6 - z) = 2\)
3. For \(\hat{k}\): \(2y - 3 = 1\)

### Step 5: Solve the equations
1. From the first equation:
\[ 9 - zy = -1 \implies zy = 10 \quad (1) \]

2. From the second equation:
\[ -6 + z = 2 \implies z = 8 \quad (2) \]

3. From the third equation:
\[ 2y - 3 = 1 \implies 2y = 4 \implies y = 2 \quad (3) \]

### Step 6: Substitute \(z\) back into equation (1)
Substituting \(z = 8\) into equation (1):

\[ 8y = 10 \implies y = \frac{10}{8} = 1.25 \]

However, from the third equation, we found \(y = 2\). Thus, we have:

- \(y = 2\)
- \(z = 8\)

### Conclusion
The values of \(y\) and \(z\) are:
- \(y = 2\)
- \(z = 8\)