An electron of charge `-e`, mass m, enters a uniform magnetic field `vec(B)= B hat(i)` with an initial velocity `vec(v) = v_(x) hat(i) + v_(y) hat(j)`. What is the velocity of the electron after a time interval of t seconds?

Options

A. `v_(x) hat(i) + v_(y) hat(j) + (e )/(m) v_(y) Bt hat(k)`
B. `v_(x) hat(i) + v_(y) hat(j) - (e )/(m) v_(y) B t hat(k)`
C. `v_(x) hat(i) + (v_(y) + (e )/(m) v_(y) B t) hat(j)`
D. `(v_(x) + (e )/(m) v_(y) B t) hat(i) + v_(y) hat(j)`

Explanation

To find the velocity of the electron after a time interval of \( t \) seconds when it enters a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Charge of the electron: \( q = -e \)
- Mass of the electron: \( m \)
- Initial velocity: \( \vec{v} = v_x \hat{i} + v_y \hat{j} \)
- Magnetic field: \( \vec{B} = B \hat{i} \)

2. **Determine the Magnetic Force:**
The force acting on a charged particle moving in a magnetic field is given by the Lorentz force equation:
\[ \vec{F} = q \vec{v} \times \vec{B} \]
Substituting the values:
\[ \vec{F} = -e (v_x \hat{i} + v_y \hat{j}) \times (B \hat{i}) \]

3. **Calculate the Cross Product:**
Using the properties of the cross product:
\[ \hat{i} \times \hat{i} = 0 \quad \text{and} \quad \hat{j} \times \hat{i} = -\hat{k} \]
Therefore:
\[ \vec{F} = -e \left( v_y (-B \hat{k}) \right) = e v_y B \hat{k} \]

4. **Use Newton's Second Law:**
According to Newton's second law, \( \vec{F} = m \vec{a} \):
\[ m \vec{a} = e v_y B \hat{k} \]
Thus, the acceleration \( \vec{a} \) is:
\[ \vec{a} = \frac{e v_y B}{m} \hat{k} \]

5. **Find the Velocity After Time \( t \):**
To find the velocity after time \( t \), we use the equation:
\[ \vec{v}(t) = \vec{v}_0 + \vec{a} t \]
where \( \vec{v}_0 = v_x \hat{i} + v_y \hat{j} \) is the initial velocity. Substituting the acceleration:
\[ \vec{v}(t) = (v_x \hat{i} + v_y \hat{j}) + \left(\frac{e v_y B}{m} \hat{k}\right) t \]

6. **Final Expression for the Velocity:**
Thus, the velocity of the electron after time \( t \) is:
\[ \vec{v}(t) = v_x \hat{i} + v_y \hat{j} + \frac{e v_y B t}{m} \hat{k} \]

### Final Answer:
\[ \vec{v}(t) = v_x \hat{i} + v_y \hat{j} + \frac{e v_y B t}{m} \hat{k} \]