A particle of mass `1xx 10^(-26) kg` and charge `+1.6xx 10^(-19) C` travelling with a velocity `1.28xx 10^6 ms^-1` in the `+x` direction enters a region in which uniform electric field E and a uniform magnetic field of induction B are present such that `E_x = E_y = 0, E_z= -102.4 kV m^-1, and B_x = B_z =0, B_y = 8xx 10^-2.` The particle enters this region at time `t=0.` Determine the location (x,y,z coordinates) of the particle at `t= 5xx10^-6 s.` If the electric field is switched off at this instant (with the magnetic field present), what will be the position of the particle at `t= 7.45xx10^-6 s` ?

Options

A. net force acts o the particle long the +ve z-direction.
B. net force acts on the particle alog -ve z-direction
C. net force acting on particle is zero
D. net force acts in xy-plane

Explanation

To solve the problem, we need to determine the position of the particle at two different times: first when both electric and magnetic fields are present, and then when only the magnetic field is present.

### Step 1: Determine the forces acting on the particle

1. **Electric Force (F_E)**: The electric force acting on the particle is given by:
\[ F_E = qE \]
where \( q = 1.6 \times 10^{-19} \, C \) and \( E_z = -102.4 \times 10^3 \, V/m \).
\[ F_E = (1.6 \times 10^{-19}) \times (-102.4 \times 10^3) = -1.6384 \times 10^{-14} \, N \, \text{(in the -z direction)} \]

2. **Magnetic Force (F_B)**: The magnetic force acting on the particle is given by:
\[ F_B = q(\mathbf{v} \times \mathbf{B}) \]
where \( \mathbf{v} = (1.28 \times 10^6, 0, 0) \, m/s \) and \( \mathbf{B} = (0, 8 \times 10^{-2}, 0) \, T \).
\[ F_B = q(v_x B_y) = (1.6 \times 10^{-19}) \times (1.28 \times 10^6) \times (8 \times 10^{-2}) \]
\[ F_B = (1.6 \times 10^{-19}) \times (1.28 \times 10^6) \times (0.08) = 1.6384 \times 10^{-14} \, N \, \text{(in the +z direction)} \]

### Step 2: Analyze the forces

Since \( F_E \) and \( F_B \) are equal in magnitude but opposite in direction, the net force on the particle is zero. Therefore, the particle will continue to move in the +x direction without any deflection.

### Step 3: Calculate the position at \( t = 5 \times 10^{-6} \, s \)

The position of the particle in the x-direction can be calculated using:
\[ x = v_x \cdot t \]
Substituting the values:
\[ x = (1.28 \times 10^6) \cdot (5 \times 10^{-6}) = 6.4 \, m \]
Since there is no motion in the y and z directions (as the forces cancel each other), we have:
\[ y = 0, \quad z = 0 \]

Thus, the position of the particle at \( t = 5 \times 10^{-6} \, s \) is:
\[ (x, y, z) = (6.4, 0, 0) \, m \]

### Step 4: Determine the position when the electric field is switched off

When the electric field is switched off, only the magnetic field is present. The particle will continue to move in the +x direction with the same velocity \( v_x = 1.28 \times 10^6 \, m/s \).

Now we need to calculate the position at \( t = 7.45 \times 10^{-6} \, s \):
\[ x = 1.28 \times 10^6 \cdot (7.45 \times 10^{-6}) = 9.536 \, m \]
Since there is still no motion in the y and z directions:
\[ y = 0, \quad z = 0 \]

Thus, the position of the particle at \( t = 7.45 \times 10^{-6} \, s \) is:
\[ (x, y, z) = (9.536, 0, 0) \, m \]

### Final Answer:
1. At \( t = 5 \times 10^{-6} \, s \): \( (x, y, z) = (6.4, 0, 0) \, m \)
2. At \( t = 7.45 \times 10^{-6} \, s \): \( (x, y, z) = (9.536, 0, 0) \, m \)