In a region of space a uniform magnetic field exist in positive z direction and there also exists a uniform electric field along positive y direction. A particle having charge + q and mass m is released from rest at the origin. The particle moves on a curve known as cycloid. If a wheel of radius R were to roll on the X axis, a fixed point on the circumference of the wheel would generate this cycloid. Find the radius R. It is given that strength of magnetic and electric fields are B0 and E0 respectively.

Answer

To solve the problem, we need to analyze the motion of a charged particle in the presence of both electric and magnetic fields. The particle is subject to forces due to these fields, and we will derive the radius \( R \) of the cycloid generated by the motion of the particle.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The particle with charge \( +q \) and mass \( m \) is released from rest at the origin.
- The electric field \( \vec{E} \) is directed along the positive \( y \)-axis, and the magnetic field \( \vec{B} \) is directed along the positive \( z \)-axis.
- The force due to the electric field is given by \( \vec{F}_E = q \vec{E} \).
- The force due to the magnetic field is given by \( \vec{F}_B = q \vec{v} \times \vec{B} \), where \( \vec{v} \) is the velocity of the particle.

2. **Setting Up the Equations of Motion**:
- The total force acting on the particle is the sum of the electric and magnetic forces:
\[ \vec{F} = \vec{F}_E + \vec{F}_B = q \vec{E} + q \vec{v} \times \vec{B} \]
- In component form, this gives:
\[ m \frac{d\vec{v}}{dt} = q \vec{E} + q \vec{v} \times \vec{B} \]

3. **Defining the Motion**:
- Let the velocity of the particle be \( \vec{v} = v_x \hat{i} + v_y \hat{j} \).
- The electric field \( \vec{E} = E_0 \hat{j} \) and the magnetic field \( \vec{B} = B_0 \hat{k} \).

4. **Writing the Force Components**:
- The force in the \( x \)-direction is:
\[ m \frac{dv_x}{dt} = q v_y B_0 \]
- The force in the \( y \)-direction is:
\[ m \frac{dv_y}{dt} = q E_0 - q v_x B_0 \]

5. **Differentiating the Equations**:
- Differentiate the equation for \( v_y \):
\[ m \frac{d^2v_y}{dt^2} = -q B_0 \frac{dv_x}{dt} \]
- Substitute \( \frac{dv_x}{dt} \) from the first equation into this equation.

6. **Solving the Differential Equations**:
- By solving the differential equations, we find:
\[ v_y = \frac{E_0}{B_0} \sin\left(\frac{q B_0 t}{m}\right) \]
- Integrating \( v_y \) gives the position \( y \):
\[ y = \frac{m E_0}{q B_0^2} \left( 1 - \cos\left(\frac{q B_0 t}{m}\right) \right) \]

7. **Finding the Maximum Value of \( y \)**:
- The maximum value of \( y \) occurs when \( \cos\left(\frac{q B_0 t}{m}\right) = -1 \):
\[ y_{\text{max}} = \frac{2 m E_0}{q B_0^2} \]
- This maximum height corresponds to \( 2R \) (the height of the cycloid).

8. **Calculating the Radius \( R \)**:
- Therefore, we can express the radius \( R \) as:
\[ R = \frac{m E_0}{q B_0^2} \]

### Final Answer:
The radius \( R \) of the cycloid is given by:
\[ R = \frac{m E_0}{q B_0^2} \]