JEE Main 2023 — Electromagnetic Induction Question with Solution
From: JEE Main 2023 (Online) 25th January Evening Shift
Question
A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.
Choose an option
Show full solutionCorrect option: C
Correct answer
C16 V
Step-by-step explanation
Induced emf across the ends = Bv
= 2 × 8 × 1 = 16 V
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Electromagnetic Induction chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2023, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.