Q 12-06-102JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
In the given figure the magnetic flux through the loop increases according to the relation $\phi_B(t) = 10t^2 + 20t$, where $\phi_B$ is in milliwebers and $t$ is in seconds. The magnitude of current through $R = 2\ \Omega$ resistor at $t = 5$ s is ______ mA.
Numerical value type. Enter your answer.
Answer: 60
$|\varepsilon| = \dfrac{d\phi_B}{dt} = 20t + 20$ mV. At $t = 5$ s: $\varepsilon = 120$ mV.
$$I = \frac{120\ \text{mV}}{2\ \Omega} = 60\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics