Question:

The arm PQ of a rectangular conductor is moving from \(x = 0\) to \(x = 2b\) outwards and then inwards from \(x = 2b\) to \(x = 0\) as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from \(x = 0\) to \(x = b\). Identify the graph showing the variation of different quantities with distance.

Show Hint

Flux grows linearly while PQ is in the field, stays constant afterwards, and falls on return; the EMF is \(-d\Phi/dt\) (opposite signs going out and coming back), and power \(\propto \mathcal{E}^2\) is never negative.
Updated On: Nov 4, 2026
  • A-Flux, B-Power dissipated, C-EMF
  • A-Flux, B-EMF, C-Power dissipated
  • A-Power dissipated, B-Flux, C-EMF
  • A-EMF, B-Power dissipated, C-Flux
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Flux \(\Phi = Blx\) rises linearly from \(x = 0\) to \(x = b\), stays constant from \(b\) to \(2b\) (no field there), and on the return falls linearly from \(b\) to \(0\). This is the trapezoid shape, graph A.
EMF \(\mathcal{E} = -\frac{d\Phi}{dt}\) is constant in magnitude while PQ moves in the field. It is zero where the flux is constant, and it reverses sign on the return trip (flux decreasing). The graph with one level below the zero line, a zero stretch, and one level above it is graph B.
Power dissipated \(P = \frac{\mathcal{E}^2}{R}\) is non-zero only in the field region and is always positive in both directions: graph C.
So A = Flux, B = EMF, C = Power dissipated.
Answer: (2) A-Flux, B-EMF, C-Power dissipated
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Electromagnetic induction Questions

View More Questions