There exist uniform magnetic and electric fields of manitudes 1T and `1Vm^-1` respectively, along positive y-axis. A charged particle of mass 1kg and charge 1C is having velocity `1ms^-1` along x-axis and is at origin at t=0. Then, the coordinates of the particle at time `pis` will be

Options

A. `(0,1,2)`
B. `(0,-pi^(2)//2,-2)`
C. `(2,pi^(2)//2,2)`
D. `(0,pi^(2)//2,2)`

Explanation

To solve the problem, we need to analyze the motion of a charged particle in the presence of uniform electric and magnetic fields. Here are the steps to find the coordinates of the particle at time \( t = \pi \) seconds.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mass of the particle, \( m = 1 \, \text{kg} \)
- Charge of the particle, \( q = 1 \, \text{C} \)
- Velocity of the particle, \( v = 1 \, \text{m/s} \) along the x-axis
- Electric field, \( E = 1 \, \text{V/m} \) along the y-axis
- Magnetic field, \( B = 1 \, \text{T} \) along the y-axis

2. **Determine the Forces Acting on the Particle**:
- The force due to the electric field is given by:
\[ F_E = qE = 1 \, \text{C} \times 1 \, \text{V/m} = 1 \, \text{N} \]
- The magnetic force is given by:
\[ F_B = q(\vec{v} \times \vec{B}) \]
Since the velocity is along the x-axis and the magnetic field is along the y-axis, the magnetic force will act along the z-axis:
\[ F_B = 1 \, \text{C} \times (1 \, \text{m/s} \times 1 \, \text{T}) = 1 \, \text{N} \text{ (in the z-direction)} \]

3. **Calculate the Acceleration**:
- The total force acting on the particle is:
\[ F_{\text{total}} = F_E + F_B = 1 \, \text{N} \, \text{(y-direction)} + 1 \, \text{N} \, \text{(z-direction)} \]
- The acceleration due to the electric field is:
\[ a = \frac{F_E}{m} = \frac{1 \, \text{N}}{1 \, \text{kg}} = 1 \, \text{m/s}^2 \text{ (in the y-direction)} \]
- The acceleration due to the magnetic field does not contribute to the motion in the y-direction, but it causes circular motion in the x-z plane.

4. **Determine the Radius of Circular Motion**:
- The radius \( R \) of the circular motion can be calculated using:
\[ R = \frac{mv}{qB} = \frac{1 \, \text{kg} \times 1 \, \text{m/s}}{1 \, \text{C} \times 1 \, \text{T}} = 1 \, \text{m} \]

5. **Calculate the Time Period**:
- The time period \( T \) of the circular motion is given by:
\[ T = \frac{2\pi m}{qB} = \frac{2\pi \times 1 \, \text{kg}}{1 \, \text{C} \times 1 \, \text{T}} = 2\pi \, \text{s} \]

6. **Determine the Position at \( t = \pi \) seconds**:
- At \( t = \pi \) seconds, the particle completes half of its circular path (since \( \pi \) seconds is half of the time period \( 2\pi \) seconds).
- The coordinates in the x-z plane after half a rotation will be:
- \( x = 0 \) (since it moves from (1,0) to (0,1))
- \( z = 1 \) (it moves up in the z-direction)
- The y-coordinate due to the electric field after \( \pi \) seconds is calculated using:
\[ y = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \cdot 1 \, \text{m/s}^2 \cdot (\pi)^2 = \frac{\pi^2}{2} \]

7. **Final Coordinates**:
- Therefore, the coordinates of the particle at \( t = \pi \) seconds are:
\[ (x, y, z) = (0, \frac{\pi^2}{2}, 1) \]

### Final Answer:
The coordinates of the particle at time \( t = \pi \) seconds are:
\[ (0, \frac{\pi^2}{2}, 1) \]