In the previous problem, if time period, `T =2(pi) m/(BQ)` where Q is the charge of the particle and m is its mass, the ratio of time spent by the particle in field when `(theta)` is positive to when `(theta)` is negative is given by

Options

A. `((pi//2+theta)/(pi//2-theta))`
B. `((pi+theta)/(pi-theta))`
C. `((pi-theta)/(pi+theta))`
D. `((pi//2-theta)/(pi//2+theta))`

Explanation

To solve the problem, we need to find the ratio of the time spent by a charged particle in a magnetic field when the angle \( \theta \) is positive to when \( \theta \) is negative. The time period \( T \) is given by the formula:

\[ T = \frac{2\pi m}{qB} \]

where \( q \) is the charge of the particle, \( m \) is its mass, and \( B \) is the magnetic field strength.

### Step-by-Step Solution:

1. **Understand the Geometry of Motion**:
- When the angle \( \theta \) is positive, the particle traces a certain path in the magnetic field. The angle subtended by the path in the magnetic field can be denoted as \( \alpha_1 \).
- When the angle \( \theta \) is negative, the particle traces a different path, and the angle subtended in this case can be denoted as \( \alpha_2 \).

2. **Determine the Angles**:
- For the case where \( \theta \) is positive, the total angle traced in the magnetic field can be calculated as:
\[ \alpha_1 = 2\pi - (90^\circ + 90^\circ) + 2\theta = \pi + 2\theta \]
- For the case where \( \theta \) is negative, the angle traced can be calculated as:
\[ \alpha_2 = (90^\circ - \theta) + (90^\circ - \theta) = 180^\circ - 2\theta = \pi - 2\theta \]

3. **Calculate the Time Periods**:
- The time spent in the magnetic field when \( \theta \) is positive is proportional to \( \alpha_1 \):
\[ T_1 \propto \alpha_1 = \pi + 2\theta \]
- The time spent in the magnetic field when \( \theta \) is negative is proportional to \( \alpha_2 \):
\[ T_2 \propto \alpha_2 = \pi - 2\theta \]

4. **Form the Ratio**:
- The ratio of the time periods \( T_1 \) and \( T_2 \) can be expressed as:
\[ \frac{T_1}{T_2} = \frac{\pi + 2\theta}{\pi - 2\theta} \]

5. **Simplify the Ratio**:
- To express the ratio in a simpler form, we can divide both the numerator and denominator by 2:
\[ \frac{T_1}{T_2} = \frac{\frac{\pi}{2} + \theta}{\frac{\pi}{2} - \theta} \]

### Final Answer:
The ratio of the time spent by the particle in the field when \( \theta \) is positive to when \( \theta \) is negative is:

\[ \frac{T_1}{T_2} = \frac{\frac{\pi}{2} + \theta}{\frac{\pi}{2} - \theta} \]