JEE Main 2021 — Electromagnetic Induction Question with Solution
From: JEE Main 2021 (Online) 27th July Evening Shift
Question
In the given figure the magnetic flux through the loop increases according to the relation B(t) = 10t2 + 20t, where B is in milliwebers and t is in seconds.
The magnitude of current through R = 2 resistor at t = 5 s is ___________ mA.
The magnitude of current through R = 2 resistor at t = 5 s is ___________ mA.
This question includes a diagram. The text above accompanies the figure.
Enter your answer
Show full solutionCorrect answer: 60
Correct answer
60
Step-by-step explanation
= 20t + 20 mV
= 10t + 10 mA
at t = 5
= 60 mA
= 10t + 10 mA
at t = 5
= 60 mA
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