Two parallel wires in the plane of the paper are distance `X_0` apart. A point charge is moving with speed u between the wires in the same plane at a distance `X_1` from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is `R_1` In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is `R_2`. If `(X_0)/(X_1)=3`, the value of `(R_1)/(R_2)` is

Explanation

To solve the problem, we will analyze the two cases of the magnetic field produced by the parallel wires and how they affect the radius of curvature of a moving point charge.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two parallel wires separated by a distance \(X_0\).
- A point charge moves with speed \(u\) at a distance \(X_1\) from one wire.
- The ratio \(\frac{X_0}{X_1} = 3\), which implies that \(X_1 = \frac{X_0}{3}\) and the distance from the other wire, \(X_2 = X_0 - X_1 = \frac{2X_0}{3}\).

2. **Magnetic Field Calculation for Case 1 (Same Direction of Currents)**:
- The magnetic field \(B_1\) at the location of the charge due to wire 1 is:
\[ B_1 = \frac{\mu_0 I}{2\pi X_1} \]
- The magnetic field \(B_2\) at the location of the charge due to wire 2 is:
\[ B_2 = -\frac{\mu_0 I}{2\pi X_2} \]
- Since the currents are in the same direction, the total magnetic field \(B\) at the position of the charge is:
\[ B = B_1 + B_2 = \frac{\mu_0 I}{2\pi X_1} - \frac{\mu_0 I}{2\pi X_2} \]
- Substituting \(X_1\) and \(X_2\):
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{1}{\frac{X_0}{3}} - \frac{1}{\frac{2X_0}{3}} \right) = \frac{\mu_0 I}{2\pi} \left( \frac{3}{X_0} - \frac{3}{2X_0} \right) = \frac{\mu_0 I}{2\pi} \cdot \frac{3}{X_0} \left(1 - \frac{1}{2}\right) = \frac{\mu_0 I}{2\pi} \cdot \frac{3}{2X_0} \]
- Thus, we have:
\[ B_1 = \frac{3\mu_0 I}{4\pi X_0} \]

3. **Magnetic Field Calculation for Case 2 (Opposite Direction of Currents)**:
- In this case, both magnetic fields point in the same direction:
\[ B = B_1 + B_2 = \frac{\mu_0 I}{2\pi X_1} + \frac{\mu_0 I}{2\pi X_2} \]
- Substituting \(X_1\) and \(X_2\):
\[ B = \frac{\mu_0 I}{2\pi} \left( \frac{3}{X_0} + \frac{3}{2X_0} \right) = \frac{\mu_0 I}{2\pi} \cdot \frac{3}{X_0} \left(1 + \frac{1}{2}\right) = \frac{\mu_0 I}{2\pi} \cdot \frac{3}{X_0} \cdot \frac{3}{2} = \frac{9\mu_0 I}{4\pi X_0} \]
- Thus, we have:
\[ B_2 = \frac{9\mu_0 I}{4\pi X_0} \]

4. **Finding the Ratio of Radii of Curvature**:
- The radius of curvature \(R\) of the path of the charge is given by the formula:
\[ R = \frac{mv}{qB} \]
- For the two cases:
\[ R_1 = \frac{mu}{qB_1} \quad \text{and} \quad R_2 = \frac{mu}{qB_2} \]
- Therefore, the ratio of the radii of curvature is:
\[ \frac{R_1}{R_2} = \frac{B_2}{B_1} \]
- Substituting the values of \(B_1\) and \(B_2\):
\[ \frac{R_1}{R_2} = \frac{\frac{9\mu_0 I}{4\pi X_0}}{\frac{3\mu_0 I}{4\pi X_0}} = \frac{9}{3} = 3 \]

### Final Answer:
\[ \frac{R_1}{R_2} = 3 \]