JEE Main 2021 — Electromagnetic Induction Question with Solution
From: JEE Main 2021 (Online) 26th August Evening Shift
Question
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s1 in a uniform horizontal magnetic field of 3.0 102 T. The maximum emf induced the coil will be ................. 102 volt (rounded off to the nearest integer)
Enter your answer
Show full solutionCorrect answer: 60
Correct answer
60
Step-by-step explanation
Maximum emf
N = 20, = 50, B = 3 102 T
= 20 50 (0.08)2 3 102 = 60.28 102
Rounded off to nearest integer = 60
N = 20, = 50, B = 3 102 T
= 20 50 (0.08)2 3 102 = 60.28 102
Rounded off to nearest integer = 60
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 2021, 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.