In the given figure the magnetic flux through the loop increases according to the relation φ B(t) = 10t2 + 20t, where φ B is in milliwebers and t is in seconds. The magnitude of current through R = 2 Ω resistor at t = 5 s is ___________ mA. Numerical answer type. Work it out, then open the answer.

Answer: 60. e = (dφB)/(dt) = 20t + 20 mV At t = 5 s: e = 100 + 20 = 120 mV I = e/R = 120/2 = 60 mA So the current is 60 mA.

In the given figure the magnetic flux through the loop increases… | JEE Main 2021 (27 Jul, Shift 2)

Q65JEE Main 2021 · 27 JUL · SHIFT 2Medium

Magnetic Flux & Faraday's Law

In the given figure the magnetic flux through the loop increases according to the relation $\phi$ B(t) = 10t2 + 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 answer type. Work it out, then open the answer.

Show answer and solution

Answer: 60

$e = \frac{d\phi_B}{dt} = 20t + 20$ mV
At $t = 5$ s: $e = 100 + 20 = 120$ mV
$I = \frac{e}{R} = \frac{120}{2} = 60$ mA
So the current is $60$ mA.

Report a mistake

Magnetic Flux & Faraday's Law question from JEE Main 2021 (27 Jul, Shift 2)

More questions on Magnetic Flux & Faraday's Law