An electron having a charged e moves with a velcoity v in postive x direction. A magnetic field acts on it in positive y direction. The force on the electron acts in (where outward direction is taken as positive z-axis)

Options

A. Negative direction of y-axis
B. Positive direction on y-axis
C. Positive direftion of Z-axis
D. Negative direction of z-axis

Answer

To solve the problem, we will use the concept of the Lorentz force, which describes the force acting on a charged particle moving in a magnetic field. The force is given by the equation:

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

Where:
- \( \mathbf{F} \) is the force on the charge,
- \( Q \) is the charge of the particle,
- \( \mathbf{v} \) is the velocity vector of the particle,
- \( \mathbf{B} \) is the magnetic field vector.

### Step-by-Step Solution:

1. **Identify the Charge, Velocity, and Magnetic Field:**
- The charge of the electron is \( Q = -e \) (negative because it is an electron).
- The velocity \( \mathbf{v} \) is in the positive x-direction, which can be represented as \( \mathbf{v} = v \hat{i} \).
- The magnetic field \( \mathbf{B} \) is in the positive y-direction, represented as \( \mathbf{B} = B \hat{j} \).

2. **Apply the Lorentz Force Formula:**
- Substitute the values into the Lorentz force equation:
\[ \mathbf{F} = Q (\mathbf{v} \times \mathbf{B}) = -e (\mathbf{v} \times \mathbf{B}) \]
- This becomes:
\[ \mathbf{F} = -e (v \hat{i} \times B \hat{j}) \]

3. **Calculate the Cross Product:**
- We need to calculate \( \hat{i} \times \hat{j} \):
\[ \hat{i} \times \hat{j} = \hat{k} \]
- Therefore, we have:
\[ \mathbf{F} = -e (vB \hat{k}) \]

4. **Determine the Direction of the Force:**
- Since \( \mathbf{F} = -e vB \hat{k} \), the negative sign indicates that the force is in the negative z-direction.
- Thus, the force on the electron acts in the negative z-direction.

5. **Conclusion:**
- The force on the electron acts in the negative z-direction.

### Final Answer:
The force on the electron acts in the negative z-direction.