The problem occurs in two distinct phases:

• Phase 1 (Impulse): The magnetic field builds up from $0$ to $B$. This changing flux induces an EMF, driving a current. The interaction of this induced current with the magnetic field creates an impulsive force that gives both jumpers an initial inward velocity, $v_0$.
• Phase 2 (Retardation): The magnetic field is now steady at $B$. The jumpers move inward with velocity $v$. This motion cuts field lines, generating a motional EMF that drives a current in the opposite direction. This creates a retarding magnetic force that eventually stops the jumpers.

Let the distance between the rails be $l$. The total resistance of the loop formed by the two jumpers is $R + R = 2R$.
During the time interval $dt$ when the magnetic field increases by $dB$, the change in flux is $d\Phi = l x_0 dB$.

The induced current $i$ is given by Faraday’s Law: $$ i = \frac{\varepsilon}{R_{total}} = \frac{1}{2R} \frac{d\Phi}{dt} = \frac{l x_0}{2R} \frac{dB}{dt} $$

The force $F$ on one jumper due to this current is: $$ F = i l B = \left( \frac{l x_0}{2R} \frac{dB}{dt} \right) l B $$

Using the Impulse-Momentum Theorem ($\int F dt = \Delta p$): $$ \int F dt = m v_0 – 0 $$ $$ \int_{0}^{t} \frac{l^2 x_0}{2R} B \frac{dB}{dt} dt = m v_0 $$ $$ \frac{l^2 x_0}{2R} \int_{0}^{B} B dB = m v_0 $$ $$ \frac{l^2 x_0}{2R} \left[ \frac{B^2}{2} \right] = m v_0 $$ $$ m v_0 = \frac{B^2 l^2 x_0}{4R} \quad \text{— (Equation 1)} $$

Now the field is constant at $B$, and the jumpers are moving inward with speed $v$. The rate of decrease of the area is $l(2v)$ (since both move inward).

The motional EMF is: $$ \varepsilon_{motional} = B l (2v) $$ The current in the loop is: $$ i’ = \frac{\varepsilon_{motional}}{2R} = \frac{2 B l v}{2R} = \frac{B l v}{R} $$ The retarding force on each jumper is: $$ F_{ret} = i’ l B = \frac{B l v}{R} l B = \frac{B^2 l^2 v}{R} $$

This force causes the jumper to decelerate. We apply the Impulse-Momentum theorem again for the entire stopping process (from $v_0$ to $0$): $$ \int F_{ret} dt = \Delta p $$ $$ \int \frac{B^2 l^2}{R} v dt = m(v_0 – 0) $$ $$ \frac{B^2 l^2}{R} \int v dt = m v_0 $$

Here, $\int v dt$ represents the distance traveled by one jumper, let’s call it $y$. $$ \frac{B^2 l^2}{R} y = m v_0 $$

Substitute the value of $m v_0$ from Equation 1 into the stopping equation: $$ \frac{B^2 l^2}{R} y = \frac{B^2 l^2 x_0}{4R} $$ $$ y = \frac{x_0}{4} $$

Each jumper moves inward by a distance $y = x_0/4$. Since there are two jumpers moving towards each other, the total reduction in separation is: $$ \Delta x = 2y = 2 \left( \frac{x_0}{4} \right) = \frac{x_0}{2} $$

The final separation $x_f$ is: $$ x_f = x_0 – \Delta x = x_0 – \frac{x_0}{2} = \frac{x_0}{2} $$