A

Step 1: Analyze the forces acting on the connector. The magnetic force acting on the connector when it moves in the magnetic field is given by the Lorentz force $$ F = B imes I $$, where I is the current flowing through the connector.
Step 2: Since the connector is released from a compressed position, it will experience a spring force directed back towards its equilibrium position, which is given by Hooke's Law as $$ F = -kx $$, where k is the spring constant and x is the displacement from the equilibrium position.
Step 3: Write the equation of motion for the connector. The net force will be the sum of the magnetic force and the spring force:
$$ m \frac{d^2x}{dt^2} = B \times I - kx \text{ (Hooke's Law)} $$
Substitute the expression for current I as $$ I = \frac{dQ}{dt} = \frac{Bl}{dt} $$, leading to a differential equation that resembles simple harmonic motion (SHM).
Step 4: The time period for SHM is given by $$ T = 2\pi \sqrt{\frac{m}{k}} $$. To find the time to return to the original position, note that after half a period, it will be at the equilibrium point. This means the time taken to return to the original position after being released is given by:
$$ t = \frac{T}{2} = \frac{\pi}{2} \sqrt{\frac{m}{k}} $$. Therefore, it takes a time of $$ \frac{\pi}{2} \sqrt{\frac{m}{k}} $$ to return to its original position.