In the given figure the magnetic flux through the loop increases according to the relation $\phi_{\beta}(t) 10 t^{2}+20 t$, where $\phi_{\beta}$ is in milliwebers and t is in seconds. The magnitude of current through $\mathrm{R}=2 \Omega$ resistor at $\mathrm{t}=5 \mathrm{~s}$ is \_\_\_\_ mA.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Electromagnetic Induction

In the given figure the magnetic flux through the loop increases according to the relation

ϕ_{β} (t) 10 t^{2} + 20 t

, where

ϕ_{β}

is in milliwebers and t is in seconds. The magnitude of current through

R = 2Ω

resistor at

t = 5 s

is
____ mA.

Answer: 60

Solution

Faraday's law gives the induced emf as the rate of change of flux:

∣ ε ∣ = \frac{d ϕ _{B}}{d t} = \frac{d}{d t} (10 t^{2} + 20 t) = (20 t + 20) mV

t = 5

∣ ε ∣ = 120

mV. The loop carries only the

2 Ω

resistor, so

I = \frac{120 mV}{2 Ω} = 60 mA

The magnitude of the current through the resistor is

60

mA.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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