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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1
A magnetic field is directed into the page and changes with time according to the expression B=(0.03t2+1.4)TB=\left(0.03 t^2+1.4\right) T, where tt is in seconds. The field has a circular cross section of radius RR when t=3.0 s\mathrm{t}=3.0 \mathrm{~s} and r=0.02 m\mathrm{r}=0.02 \mathrm{~m}.
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Correct option: A
18×10−4N/C18 \times 10^{-4} N / C
Option A (18×10−4N/C18 \times 10^{-4} N / C) is correct because it's the induced electric field at r = 0.02 m when t = 3.0 s, calculated using Faraday's law of induction.

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