An electron is travelling with a constant velocity `vecv=v_(0)hatj` in the presence of a uniform electric field `vecE=E_(0)hatk`, where `E_(0)=(mg)/e` (Here, m denotes the mass of the electron and e denotes its charge g denotes the acceleration due to gravity). Gravity acts in the -Z direction. At `t=0`, when the electron is at the origin, a uniform magnetic fiedl `vecB=B_(x)hati+B_(y)hatj` is switched on. Here `B_(x)` and `B_(y)` are fixed positive constants and `B_(0)=sqrt(B_(x)^(2)+B_(y)^(2))`. The coordinates of the point whre the electron intersects the first time after `t=0` are given by:

Options

A. `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))`
B. `((2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))`
C. `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(y)^(2))/(B_(0)^(3))))`
D. `((2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))),(2pimv_(0))/e((B_(y)^(2))/(B_(0)^(3))))`

Explanation

To solve the problem step-by-step, we need to analyze the motion of the electron in the presence of both electric and magnetic fields.

### Step 1: Understand the Initial Conditions
The electron is moving with a constant velocity in the positive y-direction, given by:
\[ \vec{v} = v_0 \hat{j} \]
It is also subjected to a uniform electric field in the positive z-direction:
\[ \vec{E} = E_0 \hat{k} \]
where \( E_0 = \frac{mg}{e} \). Here, \( m \) is the mass of the electron, \( g \) is the acceleration due to gravity, and \( e \) is the charge of the electron.

### Step 2: Analyze the Forces Acting on the Electron
When the magnetic field is turned on at \( t = 0 \), the force acting on the electron due to the electric field is:
\[ \vec{F_E} = e \vec{E} = e E_0 \hat{k} = \frac{mg}{e} e \hat{k} = mg \hat{k} \]
This force acts in the positive z-direction.

The magnetic field is given by:
\[ \vec{B} = B_x \hat{i} + B_y \hat{j} \]
The magnetic force acting on the electron can be calculated using the Lorentz force equation:
\[ \vec{F_B} = -e (\vec{v} \times \vec{B}) \]

### Step 3: Calculate the Magnetic Force
Substituting the values of \( \vec{v} \) and \( \vec{B} \):
\[ \vec{F_B} = -e (v_0 \hat{j} \times (B_x \hat{i} + B_y \hat{j})) \]
Calculating the cross product:
\[ \vec{v} \times \vec{B} = v_0 \hat{j} \times B_x \hat{i} + v_0 \hat{j} \times B_y \hat{j} = v_0 B_x (\hat{j} \times \hat{i}) + 0 = -v_0 B_x \hat{k} \]
Thus, the magnetic force becomes:
\[ \vec{F_B} = -e (-v_0 B_x \hat{k}) = e v_0 B_x \hat{k} \]

### Step 4: Determine the Net Force
The net force acting on the electron is:
\[ \vec{F_{net}} = \vec{F_E} + \vec{F_B} = mg \hat{k} + e v_0 B_x \hat{k} = (mg + e v_0 B_x) \hat{k} \]

### Step 5: Motion Analysis
The electron will undergo circular motion in the x-y plane due to the magnetic field while moving linearly in the z-direction due to the electric field. The radius \( r \) of the circular motion can be given by:
\[ r = \frac{mv_{\perp}}{eB} \]
where \( v_{\perp} = v_0 \sin(\theta) \) and \( B = B_0 = \sqrt{B_x^2 + B_y^2} \).

### Step 6: Time Period of Circular Motion
The time period \( T \) of the circular motion is given by:
\[ T = \frac{2\pi m}{eB} \]

### Step 7: Calculate the Displacement
The displacement in the y-direction after one complete revolution is:
\[ \Delta y = v_0 T = v_0 \cdot \frac{2\pi m}{eB} \]

### Step 8: Find the Coordinates
The coordinates of the point where the electron intersects for the first time after \( t=0 \) will be:
- In the x-direction: \( x = r \cos(\theta) \)
- In the y-direction: \( y = r \sin(\theta) \)

Substituting \( r \) and the values of \( \sin(\theta) \) and \( \cos(\theta) \):
\[ x = \frac{mv_0 \sin(\theta)}{eB} \cos(\theta) \]
\[ y = \frac{mv_0 \sin(\theta)}{eB} \sin(\theta) \]

### Final Result
The coordinates of the point where the electron intersects for the first time after \( t=0 \) are given by:
\[ \left( \frac{2\pi m v_0 B_x}{e B_0^3}, \frac{2\pi m v_0 B_y}{e B_0^3} \right) \]