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 radfius of curvature of the path of the point charge is `R_(1)`. In contrast, if the currentsd `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 need to find the ratio \( \frac{R_1}{R_2} \) given the conditions of the currents in the two parallel wires and the distances involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have two parallel wires separated by a distance \( X_0 \). A point charge is moving with speed \( u \) at a distance \( X_1 \) from one wire. The total distance between the wires is \( X_0 \), so the distance from the other wire is \( X_0 - X_1 \).

### Step 2: Magnetic Field Due to Wires
When the currents in both wires are in the same direction, the magnetic field \( B_1 \) at the location of the charge can be calculated using the formula for the magnetic field due to a long straight current-carrying wire:
\[ B = \frac{\mu_0 I}{2\pi r} \]
where \( r \) is the distance from the wire to the point where we are calculating the magnetic field.

For the first wire (at distance \( X_1 \)):
\[ B_{1,\text{left}} = \frac{\mu_0 I}{2\pi X_1} \]
For the second wire (at distance \( X_0 - X_1 \)):
\[ B_{1,\text{right}} = \frac{\mu_0 I}{2\pi (X_0 - X_1)} \]
Since both currents are in the same direction, the total magnetic field \( B_1 \) will be the difference:
\[ B_1 = B_{1,\text{left}} - B_{1,\text{right}} = \frac{\mu_0 I}{2\pi X_1} - \frac{\mu_0 I}{2\pi (X_0 - X_1)} \]

### Step 3: Magnetic Field When Currents Are Opposite
When the currents are in opposite directions, the magnetic fields add up:
\[ B_{2,\text{left}} = \frac{\mu_0 I}{2\pi X_1} \]
\[ B_{2,\text{right}} = \frac{\mu_0 I}{2\pi (X_0 - X_1)} \]
Thus, the total magnetic field \( B_2 \) is:
\[ B_2 = B_{2,\text{left}} + B_{2,\text{right}} = \frac{\mu_0 I}{2\pi X_1} + \frac{\mu_0 I}{2\pi (X_0 - X_1)} \]

### Step 4: Radius of Curvature
The radius of curvature \( R \) of the path of the charge moving in a magnetic field is given by:
\[ R = \frac{mv}{qB} \]
where \( m \) is the mass of the charge, \( v \) is its velocity, \( q \) is the charge, and \( B \) is the magnetic field.

Thus, we can express the radii \( R_1 \) and \( R_2 \) as:
\[ R_1 = \frac{mu}{qB_1} \]
\[ R_2 = \frac{mu}{qB_2} \]

### Step 5: Ratio of Radii
Now, we can find the ratio \( \frac{R_1}{R_2} \):
\[ \frac{R_1}{R_2} = \frac{B_2}{B_1} \]

### Step 6: Substitute and Simplify
From our earlier expressions for \( B_1 \) and \( B_2 \):
\[ \frac{R_1}{R_2} = \frac{\frac{\mu_0 I}{2\pi X_1} + \frac{\mu_0 I}{2\pi (X_0 - X_1)}}{\frac{\mu_0 I}{2\pi X_1} - \frac{\mu_0 I}{2\pi (X_0 - X_1)}} \]
The \( \frac{\mu_0 I}{2\pi} \) cancels out:
\[ \frac{R_1}{R_2} = \frac{\frac{1}{X_1} + \frac{1}{X_0 - X_1}}{\frac{1}{X_1} - \frac{1}{X_0 - X_1}} \]

### Step 7: Common Denominator
Let’s find a common denominator for the numerator and denominator:
- Numerator: \( \frac{(X_0 - X_1) + X_1}{X_1(X_0 - X_1)} = \frac{X_0}{X_1(X_0 - X_1)} \)
- Denominator: \( \frac{(X_0 - X_1) - X_1}{X_1(X_0 - X_1)} = \frac{X_0 - 2X_1}{X_1(X_0 - X_1)} \)

Thus:
\[ \frac{R_1}{R_2} = \frac{X_0}{X_0 - 2X_1} \]

### Step 8: Substitute Given Ratio
Given \( \frac{X_0}{X_1} = 3 \), we can express \( X_0 \) as \( 3X_1 \):
\[ \frac{R_1}{R_2} = \frac{3X_1}{3X_1 - 2X_1} = \frac{3X_1}{X_1} = 3 \]

### Final Answer
Thus, the value of \( \frac{R_1}{R_2} \) is:
\[ \frac{R_1}{R_2} = 3 \]