Uniform electric field E and magnetic field B, respectively, are along y-axis as shown in the figure. A particle with specific charge q/m leaves the origin O in the direction of x-axis with an initial non-relativistic speed `v_(0). The coordinate `Y_(n)` of the particle when it crosses the y-axis for the nth time is:

Options

A. `(2pi^(2)mn^(2)E)/(qB^(2))`
B. `(pi^(2)mn^(2)E)/(qB^(2))`
C. `(2pi^(2)mn^(2)E)/(3qB^(2))`
D. `(sqrt(3)pi^(2)mn^(2) E)/(qB^(2))`

Explanation

To solve the problem, we need to determine the coordinate \( Y_n \) of a particle when it crosses the y-axis for the nth time, given that it is subjected to uniform electric and magnetic fields.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle**:
The particle experiences two forces:
- Due to the electric field \( E \), the force is \( F_E = qE \).
- Due to the magnetic field \( B \), the force is \( F_B = q(v \times B) \). However, since the particle moves along the x-axis initially, we will focus on the electric force for vertical motion.

2. **Calculate Acceleration**:
The acceleration \( a \) of the particle in the y-direction due to the electric field is given by:
\[ a = \frac{F_E}{m} = \frac{qE}{m} \]

3. **Determine the Time Period for Circular Motion**:
The particle will undergo circular motion in the x-z plane due to the magnetic field. The time period \( T \) for one complete revolution in the x-z plane is:
\[ T = \frac{2\pi m}{qB} \]

4. **Calculate Total Time for n Revolutions**:
For \( n \) revolutions, the total time \( t \) is:
\[ t = nT = n \cdot \frac{2\pi m}{qB} \]

5. **Calculate the Displacement in y-direction**:
The displacement \( Y_n \) in the y-direction when the particle crosses the y-axis for the nth time can be calculated using the equation of motion:
\[ Y_n = \frac{1}{2} a t^2 \]
Substituting the values of \( a \) and \( t \):
\[ Y_n = \frac{1}{2} \left(\frac{qE}{m}\right) \left(n \cdot \frac{2\pi m}{qB}\right)^2 \]

6. **Simplify the Expression**:
Now, simplifying the expression:
\[ Y_n = \frac{1}{2} \cdot \frac{qE}{m} \cdot \left(n^2 \cdot \frac{4\pi^2 m^2}{q^2 B^2}\right) \]
\[ Y_n = \frac{2\pi^2 m n^2 E}{q B^2} \]

### Final Result:
Thus, the coordinate \( Y_n \) of the particle when it crosses the y-axis for the nth time is:
\[ Y_n = \frac{2\pi^2 m n^2 E}{q B^2} \]