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

Explanation

To solve the problem step by step, we will analyze the motion of the charged particle in the presence of uniform electric and magnetic fields.

### Step 1: Identify the given parameters
- **Charge (q)** = 1 C
- **Mass (m)** = 1 kg
- **Velocity (v)** = 1 m/s (along the x-axis)
- **Magnetic field (B)** = 1 T (along the y-axis)
- **Electric field (E)** = 1 V/m (along the y-axis)
- **Initial position** = (0, 0, 0) at time \( t = 0 \)

### Step 2: Determine the force acting on the particle
The total force \( F \) acting on the charged particle can be calculated using the Lorentz force equation:
\[ F = q(E + v \times B) \]
Here, since the electric field \( E \) and magnetic field \( B \) are both in the y-direction, we need to calculate \( v \times B \):
- The velocity vector \( \vec{v} = (1, 0, 0) \) m/s
- The magnetic field vector \( \vec{B} = (0, 1, 0) \) T

Calculating the cross product:
\[ v \times B = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} = (0, 0, 1) \]

Thus, the magnetic force is:
\[ F_B = q(v \times B) = 1 \cdot (0, 0, 1) = (0, 0, 1) \text{ N} \]

The electric force is:
\[ F_E = qE = 1 \cdot (0, 1, 0) = (0, 1, 0) \text{ N} \]

Adding these forces gives:
\[ F = F_E + F_B = (0, 1, 0) + (0, 0, 1) = (0, 1, 1) \text{ N} \]

### Step 3: Calculate acceleration
Using Newton's second law \( F = ma \):
\[ a = \frac{F}{m} = \frac{(0, 1, 1)}{1} = (0, 1, 1) \text{ m/s}^2 \]

### Step 4: Determine the motion in the y and z directions
The particle moves linearly in the y and z directions due to the electric and magnetic forces. The equations of motion for the y and z coordinates can be derived from the kinematic equations.

For the y-coordinate:
\[ y(t) = u_y t + \frac{1}{2} a_y t^2 \]
Here, \( u_y = 0 \) (initial velocity in the y-direction) and \( a_y = 1 \):
\[ y(t) = 0 \cdot t + \frac{1}{2} \cdot 1 \cdot t^2 = \frac{1}{2} t^2 \]

For the z-coordinate:
\[ z(t) = u_z t + \frac{1}{2} a_z t^2 \]
Here, \( u_z = 0 \) (initial velocity in the z-direction) and \( a_z = 1 \):
\[ z(t) = 0 \cdot t + \frac{1}{2} \cdot 1 \cdot t^2 = \frac{1}{2} t^2 \]

### Step 5: Calculate the coordinates at \( t = \pi \)
Substituting \( t = \pi \):
- For y-coordinate:
\[ y(\pi) = \frac{1}{2} \pi^2 \]
- For z-coordinate:
\[ z(\pi) = \frac{1}{2} \pi^2 \]

### Step 6: Determine the x-coordinate
Since the particle is moving in a circular path in the x-y plane due to the magnetic field, we need to find the x-coordinate. The time period \( T \) for the circular motion is given by:
\[ T = \frac{2\pi m}{qB} = \frac{2\pi \cdot 1}{1 \cdot 1} = 2\pi \text{ seconds} \]
At \( t = \pi \), the particle completes half of the circular path, which means the x-coordinate will be 0 (it will be at the y-axis).

### Final coordinates
Thus, the coordinates of the particle at \( t = \pi \) seconds are:
\[ (x, y, z) = (0, \frac{1}{2} \pi^2, \frac{1}{2} \pi^2) \]

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