Skip to content
dronacharya.ai
← JEE Main 2026 · 5 Apr · Shift 2
PhysicsInteger+4 / -1Question 50 of 75

JEE Main 2026 · 5 Apr · Shift 2, Physics question 50

A circular loop of radius 20 cm and resistance 2 Ω is placed in a time varying magnetic field B⃗=(2t2+2t+3)\vec B=\left(2t^2+2t+3\right) T. At t=0t=0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t=3t=3 s is α50\dfrac\alpha{50} A. The value of α\alpha is ________.

(Take π=22/7\pi=22/7)

Answer and worked solution

Official answer: 44

EMF at t = 3 s. dBdt=4t+2=14\dfrac{dB}{dt}=4t+2=14 T/s, and the area is π(0.2)2=0.04×227\pi(0.2)^2=0.04\times\tfrac{22}7 m², so
ε=14×0.04×227=1.76 V.\varepsilon=14\times0.04\times\frac{22}7=1.76\text{ V}.

Current. I=1.762=0.88=4450I=\dfrac{1.76}2=0.88=\dfrac{44}{50} A.

Answer: 44