Consider an LC circuit, with inductance L = 0.1 H and capacitance , kept on a plane. The area of the circuit is . It is placed in a constant magnetic field of strength , which is perpendicular to the plane of the circuit. At time , the magnetic field stre

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 $β = 0.04 {Ts}^{- 1} .$ The maximum magnitude of the current in the circuit is _______ $m A$ . [2022]

Smart Achievers · Accepted Answer

(4)

$Induced emf, E = | \frac{d ϕ}{d t} | = \frac{d}{d t} (B A) = \frac{A d}{d t} (B_{0} + β t) = A β = 1 \times 0.04 = 0.04 V$

So, the circuit can be drawn as

$By KVL, E = L \frac{d i}{d t} + \frac{q}{C}$

$\Rightarrow L \frac{d i}{d t} = E - \frac{q}{C}$

Since $i = \frac{d q}{d t}$ ,

$\Rightarrow \frac{d^{2} q}{d t^{2}} = - \frac{1}{L C} (q - C E)$

Comparing it with the equation of SHM, we get

$q = C E + A \sin (ω t + ϕ), where ω = \frac{1}{\sqrt{L C}}$

So, $i = A ω \cos (ω t + ϕ)$

At $t = 0$ , $q = 0$ and $i = 0$

So, $O = C E + A \sin ϕ \Rightarrow A \sin ϕ = - C E$ $\dots (i)$

$O = \cos ϕ \Rightarrow ϕ = \frac{π}{2}$ $\dots (i i)$

From (i) and (ii), we get

$A = - C E$

So, $i = - C E ω \cos (ω t + \frac{π}{2}) = C E ω \sin ω t$

Therefore, $i_{\max} = C E ω = 10^{- 3} \times 0.04 \times \frac{1}{\sqrt{0.1 \times 10^{- 3}}} = 4 mA$

Solution

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