A coil of wire having finite inductance and resistance has a conducting ring placed co-axially withi

Q: Why does the current $I_1(t)$ grow exponentially instead of instantly?

The coil has finite inductance, which opposes any change in current according to Lenz's law. This creates a back EMF that slows down the current growth, resulting in an exponential rise.

Q: Why is the magnetic field $B(t)$ directly proportional to $I_1(t)$?

According to the Biot-Savart law and Ampere's law, the magnetic field produced by a current-carrying coil at any point in space is directly proportional to the current flowing through it.

Q: How do we know the induced current $I_2(t)$ decays exponentially?

The induced EMF is proportional to the derivative of the magnetic field. Since $B(t) \propto (1 - e^{-kt})$, its derivative is proportional to $e^{-kt}$, which represents an exponential decay.

Q: Why does the product $I_2(t)B(t)$ start at zero?

At $t=0$, the current $I_1(t)$ is zero, so the magnetic field $B(t)$ is also zero. Even though the induced current $I_2(t)$ is at its maximum, multiplying it by zero gives a product of zero.

Q: How can we mathematically prove it passes through a maximum?

By taking the derivative of the product $P(t) = C e^{-kt}(1 - e^{-kt})$ with respect to time and setting it to zero, we find a critical point at $e^{-kt} = 1/2$. Since the second derivative is negative, this point is a maximum.

Question

A coil of wire having finite inductance and resistance has a conducting ring placed co-axially within it. The coil is connected to a battery at time $t = 0$ so that a time dependent current $I_1(t)$ starts flowing through the coil. If $I_2(t)$ is the current induced in the ring and $B(t)$ is the magnetic field at the axis of the coil due to $I_1(t)$, then as a function of time ($t > 0$), the product $I_2(t) B(t)$

Options:

(A) increases with time

(B) decreases with time

(C) does not vary with time

(D) passes through a maximum

Sigma Percentile · Accepted Answer

The behavior of coupled circuits is one of the most fascinating topics in electromagnetism. In this problem, we explore the dynamic relationship between a primary coil and a secondary conducting ring. Let's break down the physics step-by-step.

### Analyzing the Setup

We have a primary coil with finite inductance $L$ and resistance $R$. Inside it, a conducting ring is placed coaxially. At $t = 0$, the coil is connected to a battery.

Because of the coil's self-inductance, the current $I_1(t)$ does not reach its steady-state value instantly. Instead, it grows exponentially according to the classic L-R circuit equation:
$$I_1(t) = I_0 (1 - e^{-kt})$$
where $k = \frac{R}{L}$ is the inverse of the time constant.

### The Magnetic Field and Induced Current

The magnetic field $B(t)$ at the axis of the coil is directly proportional to the current $I_1(t)$ producing it. Therefore, the magnetic field also exhibits exponential growth:
$$B(t) = B_0 (1 - e^{-kt})$$

Now, what happens to the conducting ring? According to **Faraday's Law of Induction**, the changing magnetic flux through the ring induces an electromotive force (EMF). This induced EMF, $e_2$, is proportional to the rate of change of the magnetic field:
$$e_2 \propto -\frac{dB}{dt}$$

Taking the derivative of $B(t)$, we find:
$$\frac{dB}{dt} = B_0 k e^{-kt}$$

Since the ring has some resistance, the induced current $I_2(t)$ is simply the induced EMF divided by the resistance. Thus, $I_2(t)$ decays exponentially:
$$I_2(t) = I_{20} e^{-kt}$$

### The Master Equation

The question asks us to analyze the product of the induced current and the magnetic field, $P(t) = I_2(t) B(t)$. Let's multiply our two expressions:
$$P(t) = (I_{20} e^{-kt}) \cdot (B_0 (1 - e^{-kt}))$$
$$P(t) = C e^{-kt} (1 - e^{-kt})$$
where $C = I_{20} B_0$ is a positive constant.

### Final Calculation

To understand how this product behaves over time, let's look at its limits:
1. **At $t = 0$:** The term $(1 - e^{-kt})$ becomes $(1 - 1) = 0$. Therefore, $P(0) = 0$.
2. **As $t \rightarrow \infty$:** The term $e^{-kt}$ approaches $0$. Therefore, $P(\infty) = 0$.

Since the product is zero at both extremes and is strictly positive for all $t > 0$, it must rise to a peak before falling back down. Mathematically, it **passes through a maximum**.

If you want to find the exact moment this happens, you can set the derivative $\frac{dP}{dt}$ to zero, which yields $e^{-kt} = \frac{1}{2}$, or $t = \frac{\ln 2}{k}$. This beautiful interplay between exponential growth and decay creates a transient pulse of energy transfer between the coil and the ring!

Select Answer:

Visualized Solution

$t = 0$

$I_{1} (t)$

$B (t)$

$I_{2} (t)$

$P (t) = I_{2} (t) B (t)$

$P (t) Limits$

$P_{ma x}$

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup

The Magnetic Field and Induced Current

The Master Equation

Final Calculation

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