In the previous problem, what will be acceleration of the system (in` m//sec^(2))`?

Options

A. `5sqrt(2)`
B. `0.5sqrt(2)`
C. zero
D. 2.5

Explanation

To find the acceleration of the system, we need to analyze the forces acting on the charged particle moving in a magnetic field.

### Step-by-Step Solution:

1. **Identify the Forces**:
The charged particle is moving in a magnetic field. The magnetic force acting on the particle can be represented by the equation \( F = q(v \times B) \), where \( q \) is the charge of the particle, \( v \) is the velocity, and \( B \) is the magnetic field.

2. **Direction of Velocity and Magnetic Field**:
In this problem, the velocity \( v \) of the particle is in the y-direction, and the magnetic field \( B \) is out of the plane (which we can assume to be in the positive z-direction).

3. **Determine the Magnetic Force**:
Using the right-hand rule for the cross product \( v \times B \):
- Point your fingers in the direction of \( v \) (upwards along the y-axis).
- Curl your fingers in the direction of \( B \) (out of the plane, towards you).
- Your thumb will point in the direction of the magnetic force \( F \), which will be in the x-direction.

4. **Centripetal Force and Circular Motion**:
Since the magnetic force is always perpendicular to the velocity of the particle, it does not do work on the particle. Instead, it changes the direction of the particle's velocity, causing it to move in a circular path. The magnetic force acts as the centripetal force, which is given by:
\[ F = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is the speed, and \( r \) is the radius of the circular path.

5. **Equating Forces**:
The magnetic force can be equated to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

6. **Acceleration Calculation**:
In circular motion, the acceleration \( a \) is given by the centripetal acceleration formula:
\[ a = \frac{v^2}{r} \]
However, since the magnetic force does not change the speed of the particle (only its direction), the net acceleration in the direction of the velocity is zero. Thus, we conclude that:
\[ a = 0 \]

### Final Answer:
The acceleration of the system is \( 0 \, \text{m/s}^2 \).

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