A charge particle is moving in a plane perpendicular to uniform magnetic field, then

Options

A. Its momentum changes but kinetic energy remains the same
B. Both momentum and kinetic energy remain the same
C. Both will change
D. Kinetic energy change but momentum remains the same

Answer

To solve the problem of a charged particle moving in a plane perpendicular to a uniform magnetic field, we can follow these steps:

### Step 1: Understand the Setup
We have a charged particle (let's assume it has a positive charge \( Q \)) moving with a velocity \( \mathbf{V} \) in a plane that is perpendicular to a uniform magnetic field \( \mathbf{B} \). The magnetic field is directed along the z-axis.

**Hint:** Visualize the scenario by drawing a coordinate system with the x-y plane for the motion of the particle and the z-axis for the magnetic field.

### Step 2: Apply the Lorentz Force Law
The force acting on the charged particle due to the magnetic field is given by the Lorentz force equation:
\[ \mathbf{F} = Q (\mathbf{V} \times \mathbf{B}) \]
Here, \( \mathbf{V} \) is the velocity vector of the particle, and \( \mathbf{B} \) is the magnetic field vector.

**Hint:** Remember that the cross product \( \mathbf{V} \times \mathbf{B} \) will give a vector that is perpendicular to both \( \mathbf{V} \) and \( \mathbf{B} \).

### Step 3: Determine the Direction of the Force
Assuming the velocity \( \mathbf{V} \) is in the x-direction (let's say \( \mathbf{V} = V \hat{i} \)) and the magnetic field \( \mathbf{B} \) is in the negative z-direction (i.e., \( \mathbf{B} = -B \hat{k} \)), we can compute the cross product:
\[ \mathbf{F} = Q (V \hat{i} \times (-B \hat{k})) = QVB (\hat{i} \times -\hat{k}) = QVB \hat{j} \]
This indicates that the force \( \mathbf{F} \) acts in the positive y-direction.

**Hint:** Use the right-hand rule to determine the direction of the force from the cross product.

### Step 4: Analyze the Motion of the Particle
Since the magnetic force is always perpendicular to the velocity of the particle, it does not do work on the particle. Therefore, the speed of the particle remains constant, but its direction changes, causing the particle to move in a circular path.

**Hint:** Recognize that uniform circular motion occurs when a constant force acts perpendicular to the direction of motion.

### Step 5: Consider the Kinetic Energy
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} m V^2 \]
Since the speed \( V \) remains constant (as established), the kinetic energy of the particle also remains constant throughout its motion.

**Hint:** Kinetic energy is a scalar quantity and is not affected by changes in direction, only by changes in speed.

### Conclusion
The charged particle will undergo uniform circular motion due to the magnetic force acting perpendicular to its velocity. The speed remains constant, and thus the kinetic energy remains unchanged.

### Final Answer
The charged particle performs circular motion, the speed remains constant, and the kinetic energy remains the same.

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