
As field is uniform we can replace the bent wire with straight wire from A to B.
So EMF :
ε=BvℓAB
=21×510 cm×2(10sin45∘)cmε=10mV
JEE Main 2025 — Physics Electromagnetism
Conductor wire ABCDE with each arm10 cm in length is placed in magnetic field of 21 Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is _____ mV.

Held on 4 Apr 2025 · Verified 6 Jul 2026.
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