A particle of charge 'q' and mass 'm' is projected from the origin with velocity `(u_0hati+v_0 hatj) ` in a gravity free region where uniform electric field `-E_0hati` and uniform magnetic field `-B_0hati` exist. Find the condition so that the particle would return to origin at least for once .

Options

A. `(mu_0B_0)/(2piE_0)` is an integer
B. `sqrt(u_0^2+v_0^2)B_0/(piE_0)` is an integer
C. `(v_0B_0)/(piB_0)` is an integer
D. `(mu_0B_0)/(piE_0)` is an integer

Answer

To solve the problem, we need to analyze the motion of a charged particle in the presence of both an electric field and a magnetic field. The particle is projected from the origin with an initial velocity, and we want to find the condition under which it returns to the origin at least once.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
- The electric force acting on the particle due to the electric field \( \vec{E} = -E_0 \hat{i} \) is given by:
\[ \vec{F}_E = q \vec{E} = -q E_0 \hat{i} \]
- The magnetic force acting on the particle due to the magnetic field \( \vec{B} = -B_0 \hat{i} \) is given by:
\[ \vec{F}_B = q \vec{v} \times \vec{B} \]
Here, \( \vec{v} = u_0 \hat{i} + v_0 \hat{j} \). The magnetic force will only have a component in the \( \hat{k} \) direction since the magnetic field is in the \( \hat{i} \) direction.

2. **Calculate the Magnetic Force:**
- The magnetic force can be calculated as:
\[ \vec{F}_B = q (u_0 \hat{i} + v_0 \hat{j}) \times (-B_0 \hat{i}) = q v_0 (-B_0 \hat{k}) \]
- Thus, the total force acting on the particle is:
\[ \vec{F}_{\text{total}} = \vec{F}_E + \vec{F}_B = -q E_0 \hat{i} - q v_0 B_0 \hat{k} \]

3. **Determine the Acceleration:**
- Using Newton's second law, the acceleration \( \vec{a} \) of the particle is given by:
\[ \vec{a} = \frac{\vec{F}_{\text{total}}}{m} = -\frac{q E_0}{m} \hat{i} - \frac{q v_0 B_0}{m} \hat{k} \]
- The acceleration in the \( x \)-direction is:
\[ a_x = -\frac{q E_0}{m} \]
- The acceleration in the \( y \)-direction is zero since there is no force acting in that direction.

4. **Analyze the Motion in the \( x \)-Direction:**
- The motion in the \( x \)-direction is uniformly accelerated motion. The equation of motion can be used:
\[ v_x = u_x + a_x t \]
- Setting \( v_x = -u_0 \) (the particle must return to the origin), we have:
\[ -u_0 = u_0 - \frac{q E_0}{m} t \]
- Rearranging gives:
\[ \frac{q E_0}{m} t = 2u_0 \implies t = \frac{2u_0 m}{q E_0} \]

5. **Determine the Time Period of Circular Motion:**
- The magnetic force provides centripetal acceleration. The radius \( r \) of the circular motion can be derived from:
\[ r = \frac{mv}{qB} \]
- The time period \( T \) for one complete circular motion is given by:
\[ T = \frac{2 \pi r}{v} = \frac{2 \pi m}{q B} \]

6. **Condition for Returning to the Origin:**
- For the particle to return to the origin, the time taken to return must be an integer multiple of the time period:
\[ \frac{2u_0 m}{q E_0} = n \cdot \frac{2 \pi m}{q B} \]
- Simplifying, we get:
\[ n = \frac{u_0 B}{\pi E_0} \]
- For \( n \) to be an integer, the condition is:
\[ \frac{u_0 B}{\pi E_0} \text{ must be an integer.} \]

### Final Condition:
The condition for the particle to return to the origin at least once is:
\[ \frac{u_0 B}{\pi E_0} = n \quad (n \in \mathbb{Z}) \]