Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move. The conductor PQ is moved towards the left with a constant velocity V as shown in the figure. Assume that there is no loss of energy due to friction. What will be the magnetic flux linked with the loop PQRS and the motional emf?

The correct option is: Magnetic flux = Blx: Motional emf = BlV

Step-by-Step Explanation:

1. Magnetic Flux Linked with the Loop:
From the provided diagram, we can observe the rectangular conductor loop labeled as PQRS. The uniform magnetic field, denoted by B, is directed perpendicularly into the page (represented by the array of blue 'X' marks).
The movable conductor bar PQ has a length of $l$, and its position at any instant is at a distance $x$ from the side RS.
Consequently, the area A of the rectangular loop enclosed by PQRS at that instant is given by:
$$A=l·x$$
The magnetic flux (${\Phi }_B$) linked with the loop is the product of the magnetic field strength and the area perpendicular to it:
$${\Phi }_B=B·A=Blx$$

2. Motional Electromotive Force (emf):
According to Faraday's law of electromagnetic induction, the induced electromotive force ($\epsilon$) is equal to the negative rate of change of magnetic flux through the loop:
$$\epsilon =-\frac{d{\Phi }_B}{dt}$$
Substituting the expression for magnetic flux into Faraday's equation:
$$\epsilon =-\frac{d}{dt}\left(Blx\right)$$
Since the magnetic field $B$ and the length $l$ remain constant with time:
$$\epsilon =-Bl\frac{dx}{dt}$$
The conductor PQ is moving towards the left with a constant velocity $V$ (as indicated by the arrow for $v$ in the image). This motion reduces the distance $x$ over time. Thus, the rate of change of distance is:
$$\frac{dx}{dt}=-V$$
Substituting this back into the emf equation yields:
$$\epsilon =-Bl(-V)=BlV$$

Therefore, the magnetic flux linked with the loop is Blx and the motional emf is BlV.