A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other . The particle will remove in a

Options

A. straight line
B. circle
C. helix
D. cycloid

Answer

To solve the problem of a charged particle released from rest in a region of steady and uniform electric and magnetic fields that are parallel to each other, we can follow these steps:

### Step 1: Understand the Forces Acting on the Charged Particle
When a charged particle is placed in an electric field (\(E\)) and a magnetic field (\(B\)), it experiences forces due to both fields. The force due to the electric field (\(F_E\)) is given by:
\[ F_E = qE \]
where \(q\) is the charge of the particle.

The force due to the magnetic field (\(F_B\)) is given by:
\[ F_B = q(\mathbf{v} \times \mathbf{B}) \]
where \(\mathbf{v}\) is the velocity of the particle and \(\mathbf{B}\) is the magnetic field.

### Step 2: Analyze the Directions of the Fields
Since the electric field and magnetic field are parallel to each other, the angle (\(\theta\)) between the velocity vector (\(\mathbf{v}\)) and the magnetic field vector (\(\mathbf{B}\)) is \(0^\circ\). Therefore, the cross product \(\mathbf{v} \times \mathbf{B}\) becomes zero:
\[ F_B = qvB \sin(0^\circ) = 0 \]

### Step 3: Determine the Net Force on the Particle
Since the magnetic force is zero, the only force acting on the charged particle is the electric force:
\[ F_{\text{net}} = F_E = qE \]

### Step 4: Calculate the Acceleration of the Particle
Using Newton's second law, the acceleration (\(a\)) of the particle can be calculated as:
\[ a = \frac{F_{\text{net}}}{m} = \frac{qE}{m} \]
where \(m\) is the mass of the particle.

### Step 5: Determine the Motion of the Particle
Since the particle starts from rest and experiences a constant acceleration in the direction of the electric field, it will move in a straight line along the direction of the electric field.

### Conclusion
The charged particle will move in a straight line.

### Final Answer
The particle will move in a straight line.

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