Figure shows a rectangular conductor PQRS in which the conductor PQ i…

The correct answer is Option (3) → Magnetic flux = \( Blx \); Motional emf = \( BlV \)

The field is perpendicular to the loop, so the magnetic flux is the field times the area of the loop. The loop has width \( l \) and length \( x \).

    \( \displaystyle \Phi = B \times (l\,x) = Blx \)

By Faraday's law the motional emf is the rate of change of this flux. The rod moves so that \( x \) changes at the rate \( V \), so \( |dx/dt| = V \).

    \( \displaystyle \varepsilon = \left| \frac{d\Phi}{dt} \right| = Bl\,\left| \frac{dx}{dt} \right| = BlV \)

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27):
Motional Electromotive Force – “… Figure 6.10 shows a rectangular conductor PQRS in which the conductor PQ is free to move. The rod PQ is moved towards the left with a constant velocity v as shown in the figure.

… If the length RQ = x and RS = l, the magnetic flux Φ_B enclosed by the loop PQRS will be

    \( \displaystyle \Phi_B = Blx \)

Since x is changing with time, the rate of change of flux Φ_B will induce an emf given by:

    \( \displaystyle \varepsilon = \frac{-\,d\Phi_B}{dt} = -\frac{d}{dt}(Blx) = -Bl\frac{dx}{dt} = Blv \qquad (6.5) \)”