In a region of space uniform electric field is present as `vec(E)=E_(0)hat(i)` and uniform magnetci field is present as `vec(B)=-B_(0)hat(j)` . An electron is released from rest at origin. What is the path followed by electron after released. `(E_(0) & B_(0)`are positive constants )

Options

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Answer

To solve the problem, we need to analyze the forces acting on the electron in the presence of both electric and magnetic fields. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Electron
The electron is subjected to two forces due to the electric and magnetic fields:
1. **Electric Force (F_E)**: The force on the electron due to the electric field is given by:
\[ \vec{F_E} = -e \vec{E} \]
where \( e \) is the charge of the electron (which is negative). Given that the electric field \( \vec{E} = E_0 \hat{i} \), the force will be:
\[ \vec{F_E} = -e E_0 \hat{i} \]
This force acts in the negative x-direction.

2. **Magnetic Force (F_B)**: The force on the electron due to the magnetic field is given by:
\[ \vec{F_B} = -e (\vec{V} \times \vec{B}) \]
where \( \vec{B} = -B_0 \hat{j} \). The direction of the magnetic force depends on the velocity of the electron, which will change as it moves.

### Step 2: Analyze the Motion of the Electron
Initially, the electron is at rest, so its initial velocity \( \vec{V} = 0 \). Therefore, the magnetic force \( \vec{F_B} \) will also be zero at the moment of release. The only force acting on the electron at the start is the electric force \( \vec{F_E} \).

### Step 3: Determine the Acceleration
Using Newton's second law:
\[ \vec{F} = m \vec{a} \]
where \( m \) is the mass of the electron and \( \vec{F} = \vec{F_E} \). Thus, the acceleration \( \vec{a} \) can be calculated as:
\[ \vec{a} = \frac{\vec{F_E}}{m} = \frac{-e E_0 \hat{i}}{m} \]
This indicates that the electron will start accelerating in the negative x-direction.

### Step 4: Consider the Effect of the Magnetic Field
As the electron accelerates in the negative x-direction, it will gain a velocity \( \vec{V} \) in the negative x-direction. Once it has a velocity, the magnetic force will come into play. The magnetic force will be calculated as:
\[ \vec{F_B} = -e (\vec{V} \times \vec{B}) \]
Since \( \vec{B} = -B_0 \hat{j} \) and \( \vec{V} \) is in the negative x-direction, the cross product \( \vec{V} \times \vec{B} \) will yield a force in the negative z-direction (using the right-hand rule).

### Step 5: Combine the Forces
Now, the electron experiences two forces:
1. A constant force in the negative x-direction due to the electric field.
2. A force in the negative z-direction due to the magnetic field.

### Step 6: Determine the Path of the Electron
The combination of these two forces results in a parabolic trajectory. The electron will continue to accelerate in the negative x-direction while also moving in the negative z-direction due to the magnetic force. Thus, the path followed by the electron will be a parabola in the negative x and negative z plane.

### Final Answer
The path followed by the electron after being released is a parabola in the negative x and negative z plane.

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