JEE Main 2019 — Electromagnetic Induction Question with Solution
From: JEE Main 2019 (Online) 9th January Morning Slot
Question
A conducting circular loop made of a thin wire, has area 3.5 103 m2 and resistance 10 . It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to :
Choose an option
Show full solutionCorrect option: A
Correct answer
A0.14 mC
Step-by-step explanation
At t = 0 s
B(0) = 0.4 sin (0) = 0
and at t = 10 ms
B(10) = 0.4 sin (501010-3)
= 0.4 sin
= 0.4
As q =
=
=
= 0.14 mC
B(0) = 0.4 sin (0) = 0
and at t = 10 ms
B(10) = 0.4 sin (501010-3)
= 0.4 sin
= 0.4
As q =
=
=
= 0.14 mC
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This is a previous-year question from JEE Main 2019, 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.