JEE Advanced
Physics
Alternating Current
2022
JEE Advanced 2022 (Paper 1)
JEE Advanced Physics Question (2022) — Solution
Question
Consider an LC circuit, with inductance L = 0 . 1   H and capacitance C = 10 - 3   F , kept on a plane. The area of the circuit is 1   m 2 . It is placed in a constant magnetic field of strength B 0 which is perpendicular to the plane of the circuit. At time t = 0 , the magnetic field strength starts increasing linearly as B = B 0 + β t with β = 0 . 04   T   s - 1 . The maximum magnitude of the current in the circuit is____ m   A .
Step-by-step solution
Emf induced in the circuit is E = d ϕ d t = d d t B 0 + β t A = β × A = 0 . 04   V So the circuit can be rearranged as Using Kirchhoff's law we can write E = L d i d t + q C L d i d t = E - q C For maximum current, d i d t = 0 , Or q = C E Using work energy theorem, q E = 1 2 L i 2 + q 2 2 C At i max , C E E = 1 2 L i m a x 2 + E 2 C 2 2 C Or, i max = C L E So, i max = 10 - 3 × 0 . 04 0 . 1 × 10 - 3 = 4   mA
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Related: Physics — Alternating Current · All PYQ Banks