Consider an LC circuit, with inductance L = 0.1 H and capacitance C = 10^-3 F, kept on a plane. The

Q: Why is the induced EMF constant?

The magnetic field is given by $B = B_0 + \beta t$. The rate of change of the magnetic field, $\frac{dB}{dt}$, is simply the constant $\beta$. Since the area is also constant, the induced EMF, which is Area $\times \frac{dB}{dt}$, remains constant over time.

Q: Why can we treat the induced EMF like a battery?

Faraday's Law tells us that a changing magnetic flux creates an electromotive force around the loop. In a lumped-element circuit model, this distributed EMF can be perfectly represented as an ideal battery in series with the other components.

Q: At maximum current, why isn't the charge on the capacitor zero?

In a standard LC circuit with *no* external voltage, maximum current occurs when the capacitor is fully discharged ($q=0$). However, here we have a constant external EMF $\varepsilon$. The equilibrium position shifts! Maximum current now occurs when the voltage across the capacitor equals the external EMF ($\frac{q}{C} = \varepsilon$).

Q: How did we get the energy conservation equation $\varepsilon q = \frac{q^2}{2C} + \frac{1}{2} L i^2$?

The total work done by the constant EMF source as a charge $q$ flows through it is $W = \varepsilon q$. This work must equal the total energy stored in the circuit, which is the sum of the electrostatic energy in the capacitor ($\frac{q^2}{2C}$) and the magnetic energy in the inductor ($\frac{1}{2} L i^2$).

Q: What happens to the current after it reaches its maximum?

It will start to decrease, eventually reaching zero, and then reverse direction! The circuit will oscillate indefinitely (since there's no resistance) around the new equilibrium charge $q = C\varepsilon$.

Question

Consider an LC circuit, with inductance $L = 0.1	ext{ H}$ and capacitance $C = 10^{-3}	ext{ F}$, kept on a plane. The area of the circuit is $1	ext{ m}^2$. 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 $B = B_0 + \beta t$ with $\beta = 0.04	ext{ Ts}^{-1}$. The maximum magnitude of the current in the circuit is____$	ext{mA}$.

Sigma Percentile · Accepted Answer

The problem presents us with a fascinating intersection of two major concepts in electromagnetism: **Faraday's Law of Induction** and **LC Circuit Dynamics**.

### The Invisible Battery

Imagine the setup: we have a closed LC circuit resting in a plane, and a magnetic field is piercing through it. But this isn't just any magnetic field; it's growing stronger linearly with time. According to Faraday's Law, a changing magnetic flux induces an electromotive force (EMF) in any closed loop.

The magnetic flux $\Phi$ is the product of the magnetic field $B$ and the area $A$. Since the area is a constant $1 	ext{ m}^2$, the rate of change of flux is entirely driven by the changing magnetic field.

$$ \varepsilon = \left| -\frac{d\Phi}{dt} ight| = A \frac{dB}{dt} $$

We are given that $B = B_0 + \beta t$, which means the rate of change $\frac{dB}{dt}$ is simply the constant $\beta = 0.04 	ext{ Ts}^{-1}$.

$$ \varepsilon = 1 	imes 0.04 = 0.04 	ext{ V} $$

This is a profound realization! The changing magnetic field acts exactly like a **constant DC battery** of $0.04 	ext{ V}$ suddenly connected to our LC circuit at $t=0$.

### The LC Circuit Awakens

Now, let's analyze the circuit itself. We have an inductor $L$ and a capacitor $C$ connected in series with our newly found "invisible battery" $\varepsilon$. Applying Kirchhoff's Voltage Law, we get the differential equation governing the circuit:

$$ L\frac{di}{dt} + \frac{q}{C} = \varepsilon $$

Because the circuit starts with no charge and no current, the solution to this differential equation reveals that the charge $q(t)$ and current $i(t)$ will oscillate. The energy from the constant EMF source is continuously traded between the electric field of the capacitor and the magnetic field of the inductor.

To find the maximum current, we can use the principle of energy conservation. The total work done by the EMF source ($\varepsilon q$) is stored in the capacitor ($\frac{q^2}{2C}$) and the inductor ($\frac{1}{2}Li^2$).

$$ \varepsilon q = \frac{q^2}{2C} + \frac{1}{2}Li^2 $$

The current reaches its maximum when its derivative is zero ($\frac{di}{dt} = 0$). Looking back at our voltage equation, this happens when $\frac{q}{C} = \varepsilon$, or $q = C\varepsilon$. Substituting this specific charge back into our energy equation yields the elegant formula for maximum current:

$$ I_{\max} = \varepsilon \sqrt{\frac{C}{L}} $$

### The Final Calculation

We have all the pieces of the puzzle. We just need to carefully substitute the given values into our master equation. We know $\varepsilon = 0.04 	ext{ V}$, $C = 10^{-3} 	ext{ F}$, and $L = 0.1 	ext{ H}$.

$$ I_{\max} = 0.04 	imes \sqrt{\frac{10^{-3}}{0.1}} $$

$$ I_{\max} = 0.04 	imes \sqrt{10^{-2}} $$

$$ I_{\max} = 0.04 	imes 0.1 = 0.004 	ext{ A} $$

Converting this to milliamperes, we get our final, satisfying answer:

$$ I_{\max} = 4 	ext{ mA} $$

The beauty of this problem lies in its seamless transition from a changing magnetic field to a driven harmonic oscillator. It's a perfect example of how different physical phenomena are deeply interconnected!

Enter Numerical Value:

Visualized Solution

$The Setup$

$Faraday’s Law$

$Induced EMF$

$Calculating EMF$

$LC Circuit Dynamics$

$Maximum Current Formula$

$Substitution$

$Final Answer$

$The Way Forward$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

The Invisible Battery

The LC Circuit Awakens

The Final Calculation

Similar Questions

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

At time $t = 0$ magnetic field of 1000 gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 gauss, in the next 5 s, then induced emf in the loop is

A planar loop of wire rotates in a uniform magnetic field. Initially at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane, then the magnitude of induced emf will be maximum and minimum respectively at

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Faraday’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Induced EMF

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating EMF

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LC Circuit Dynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Current Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward