In the given figure, a bar magnet is quickly moved towards a conducting loop having a capacitor. Predict the polarity of the plates A and B of the capacitor

When a bar magnet is quickly moved towards a conducting loop containing a capacitor, the magnetic flux through the loop changes. According to Faraday's law of electromagnetic induction, this change in magnetic flux induces an electromotive force (emf) and consequently an induced current in the loop.

The induced current in the loop will flow in such a direction as to oppose the change in magnetic flux that produced it, following Lenz's law.

As the bar magnet is moved towards the conducting loop, the magnetic flux through the loop increases. To oppose this increase in magnetic flux, the induced current will flow in such a way that it creates a magnetic field opposing the motion of the magnet towards the loop.

By the right-hand rule for the direction of induced current, if we imagine the magnetic field lines pointing towards the loop, the induced current will flow in such a direction as to generate a magnetic field opposing the approaching magnet.

Now, let's consider the capacitor in the loop. When a current flows through a capacitor, it charges the plates of the capacitor. The direction of the induced current in the loop will cause positive charge to accumulate on one plate of the capacitor and negative charge to accumulate on the other plate.

Since the induced current will create a magnetic field opposing the motion of the approaching magnet, the induced current will flow in such a way that it creates a magnetic field that repels the approaching magnet. This means that the induced current will generate a magnetic field that flows away from the approaching magnet.

Using the right-hand rule for the direction of the magnetic field around a current-carrying conductor, we find that the induced current will circulate counterclockwise in the loop. This counterclockwise current flow will cause positive charge to accumulate on plate A and negative charge to accumulate on plate B of the capacitor.

Therefore, the plate A of the capacitor will be positively charged, and plate B will be negatively charged. This results in the polarity of the capacitor being such that plate A is positive, and plate B is negative.