A wire shaped y = cx² lies in a constant magnetic field B normal to the plane. A horizontal rod starts from rest at vertex and moves with upward acceleration a. Induced emf between the contacts is ε = 2By√(ka/c). Find k. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A wire shaped y = cx² lies in a constant magnetic field B normal to the plane. A horizontal rod starts from rest at vertex and moves with upward acceleration a. Induced emf between the contacts is ε = 2By√(ka/c). Find k. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — At height y, half-width of the parabola is x = √(y/c); instantaneous rod length between the rails is 2√(y/c). Step 2 — Rod starts at rest with acceleration a, so v² = 2ay and v = √(2ay). Step 3 — Motional emf is ε = Bℓv = B × 2√(y/c) × √(2ay) = 2By√(2a/c). Step 4 — Compare with 2By√(ka/c): k = 2.