JEE Main 2017 — Electromagnetic Induction Question with Solution
From: JEE Main 2017 (Online) 8th April Morning Slot
Question
A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos (t). The emf induced in the smaller inner loop is nearly :
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
Mutual inductance,
M =
here = N2 = 1
M =
Current I = I0 cos (t)
According to Faraday's law,
e = M
= (I0 cos t)
= + I0 sin t
= . sin t
M =
here = N2 = 1
M =
Current I = I0 cos (t)
According to Faraday's law,
e = M
= (I0 cos t)
= + I0 sin t
= . sin t
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This is a previous-year question from JEE Main 2017, 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.