1. Magnetic field due to a long straight wire

At a perpendicular distance $r$ from a long straight wire carrying current $I$, the magnetic field magnitude is

$B = \frac{\mu_0 I}{2\pi r}$

The two wires are parallel and lie in the plane of the paper. The charge moves in the same plane between them, so the magnetic fields at the charge due to both wires are perpendicular to the plane of the paper.

2. Geometry of the setup

Let the distance between the wires be $X_0$. The charge is between the wires at distance $X_1$ from one wire. So its distance from the other wire is

$X_0 - X_1$

Given:

$\frac{X_0}{X_1} = 3 \implies X_0 = 3X_1$

Hence,

$X_0 - X_1 = 3X_1 - X_1 = 2X_1$

So the point is at distances:

• from wire 1: $X_1$
• from wire 2: $2X_1$

3. Relation between radius and magnetic field

For a charged particle moving perpendicular to magnetic field,

$R = \frac{mu}{qB}$

Thus,

$R \propto \frac{1}{B}$

So,

$\frac{R_1}{R_2} = \frac{B_2}{B_1}$

where:

• $B_1$ = net field when currents are in the same direction
• $B_2$ = net field when currents are in opposite directions

4. Case 1: currents in the same direction

At a point between the wires, the magnetic fields due to the two wires are in opposite directions (by right-hand rule), so they subtract.

Magnitudes:

$B_a = \frac{\mu_0 I}{2\pi X_1}, \qquad B_b = \frac{\mu_0 I}{2\pi (2X_1)} = \frac{\mu_0 I}{4\pi X_1}$

Therefore,