The magnetic force acting on a charged particle of charge `-2 muC` in a magnetic field of `2 T` acting `y` direction, when the particle velocity is `(2i + 3 hat(j)) xx 10^(6) ms^(-1)`, is

Options

A. 8 N in z-direction
B. 4 N in z-direction
C. 8 N in y-direction
D. 8 N in z-direction

Answer

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

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

where:
- \( q \) is the charge of the particle,
- \( \mathbf{v} \) is the velocity vector of the particle,
- \( \mathbf{B} \) is the magnetic field vector,
- \( \times \) denotes the cross product.

### Step 1: Identify the given values
- Charge \( q = -2 \, \mu C = -2 \times 10^{-6} \, C \)
- Magnetic field \( \mathbf{B} = 2 \, T \, \hat{j} \)
- Velocity \( \mathbf{v} = (2 \hat{i} + 3 \hat{j}) \times 10^6 \, m/s \)

### Step 2: Write the velocity and magnetic field vectors
The velocity vector can be expressed as:
\[ \mathbf{v} = 2 \times 10^6 \hat{i} + 3 \times 10^6 \hat{j} \]
The magnetic field vector is:
\[ \mathbf{B} = 2 \hat{j} \]

### Step 3: Calculate the cross product \( \mathbf{v} \times \mathbf{B} \)
To find the cross product, we can use the determinant of a matrix:
\[ \mathbf{v} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 \times 10^6 & 3 \times 10^6 & 0 \\ 0 & 2 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{v} \times \mathbf{B} = \hat{i} \left(3 \times 10^6 \cdot 0 - 0 \cdot 2\right) - \hat{j} \left(2 \times 10^6 \cdot 0 - 0 \cdot 0\right) + \hat{k} \left(2 \times 10^6 \cdot 2 - 3 \times 10^6 \cdot 0\right) \]
\[ = 0 \hat{i} - 0 \hat{j} + (4 \times 10^6) \hat{k} \]
\[ = 4 \times 10^6 \hat{k} \]

### Step 4: Multiply by the charge to find the force
Now, we can substitute this result back into the force equation:
\[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) = -2 \times 10^{-6} (4 \times 10^6 \hat{k}) \]
\[ = -8 \hat{k} \, N \]

### Step 5: Interpret the result
The negative sign indicates that the force is in the negative \( z \) direction. Therefore, the magnetic force acting on the charged particle is:
\[ \mathbf{F} = -8 \, N \, \hat{k} \]

### Final Answer
The magnetic force acting on the charged particle is \( 8 \, N \) in the negative \( z \) direction.

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