A very long solenoid of radius R is carrying current I(t) = kte^(-alpha t) (k 0), as a function of

Q: Why did we use $\pi R^2$ for the area when calculating flux? The outer coil has a radius of $2R$!

Magnetic flux is the product of the magnetic field and the area *through which the field passes*. Since the magnetic field outside an ideal solenoid is zero, there are no magnetic field lines passing through the region between $R$ and $2R$. Therefore, the only area contributing to the flux is the cross-sectional area of the solenoid itself, which is $\pi R^2$.

Q: Why is there a negative sign in Faraday's Law?

The negative sign represents Lenz's Law. It indicates that the induced EMF (and consequently the induced current) will always flow in a direction that opposes the change in magnetic flux that produced it.

Q: I'm confused about the differentiation step. How did we get $(1 - \alpha t)$?

We used the product rule: $\frac{d}{dt}(u \cdot v) = u'v + uv'$. Here, $u = t$ and $v = e^{-\alpha t}$. The derivative of $t$ is $1$, and the derivative of $e^{-\alpha t}$ is $-\alpha e^{-\alpha t}$ (using the chain rule). So, $\frac{d}{dt}(t e^{-\alpha t}) = (1)(e^{-\alpha t}) + (t)(-\alpha e^{-\alpha t}) = e^{-\alpha t} - \alpha t e^{-\alpha t} = e^{-\alpha t}(1 - \alpha t)$.

Q: What does a negative current mean physically in this context?

The problem states that counter-clockwise current is taken to be positive. Therefore, a negative value for the induced current simply means that the current is flowing in the clockwise direction.

Q: Why does the induced current eventually go back to zero?

The induced current depends on the *rate of change* of the magnetic flux. As $t \to \infty$, the primary current $I(t) = kte^{-\alpha t}$ decays to zero. Because the primary current stops changing (it just stays at zero), the magnetic flux stops changing, and thus the induced EMF and current also become zero.

Question

A very long solenoid of radius $R$ is carrying current $I(t) = kte^{-\alpha t} (k > 0)$, as a function of time $(t \ge 0)$. Counter clockwise current is taken to be positive. A circular conducting coil of radius $2R$ is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a very long solenoid of radius $R$. Inside this solenoid, a time-varying current $I(t) = kte^{-\alpha t}$ is flowing. This current creates a magnetic field inside the solenoid. Now, placed concentrically in the equatorial plane of this solenoid is a larger circular conducting coil of radius $2R$. Our mission is to determine how the induced current in this outer coil behaves over time.

### The Magnetic Flux Trap
The first step is to calculate the magnetic flux $\phi$ passing through the outer coil. The magnetic field inside an ideal long solenoid is uniform and given by $B = \mu_0 n I(t)$. Outside the solenoid, the magnetic field is practically zero.

Here lies a classic trap! The outer coil has an area of $\pi (2R)^2$, but the magnetic field does not fill this entire area. The field only exists up to the radius $R$ of the solenoid. Therefore, the effective area through which the magnetic flux passes is just $\pi R^2$.

The magnetic flux is:
$$\phi = B 	imes 	ext{Area} = (\mu_0 n I(t)) 	imes (\pi R^2)$$

Substituting the given current $I(t)$:
$$\phi(t) = \mu_0 n \pi R^2 (kte^{-\alpha t})$$

To make our lives easier, let's bundle all the constant terms ($\mu_0, n, \pi, R^2, k$) into a single positive constant $C$. So, our flux equation simplifies beautifully to:
$$\phi(t) = C t e^{-\alpha t}$$

### The Master Equation: Faraday's Law
According to Faraday's Law of Electromagnetic Induction, the induced EMF $e$ in the outer coil is the negative rate of change of magnetic flux:
$$e = -\frac{d\phi}{dt}$$

Let's differentiate our flux expression using the product rule:
$$e = -\frac{d}{dt}(C t e^{-\alpha t})$$
$$e = -C \left[ 1 \cdot e^{-\alpha t} + t \cdot (-\alpha e^{-\alpha t}) ight]$$
$$e = -C e^{-\alpha t}(1 - \alpha t)$$

The induced current $i(t)$ is simply the induced EMF divided by the resistance $R_{coil}$ of the outer coil. Let $C' = \frac{C}{R_{coil}}$:
$$i(t) = -C' e^{-\alpha t}(1 - \alpha t)$$
$$i(t) = C' e^{-\alpha t}(\alpha t - 1)$$

### Decoding the Graph
Now, we don't need to plot this entire function point by point. We can act like physicists and check the boundary conditions to eliminate the wrong options.

Let's check the current at the very beginning, $t = 0$:
$$i(0) = C' e^0 (\alpha \cdot 0 - 1)$$
$$i(0) = C'(1)(-1) = -C'$$

This is a massive revelation! The induced current must start at a **negative** value.

If we look at the given options:
- Graph (a) starts at $0$.
- Graph (b) starts at a positive value.
- Graph (c) starts at $0$.
- Graph (d) starts at a negative value.

Only graph (d) satisfies our initial condition. Furthermore, if we look at the expression $i(t) = C' e^{-\alpha t}(\alpha t - 1)$, we can see that at $t = \frac{1}{\alpha}$, the current becomes zero, and for $t > \frac{1}{\alpha}$, the current becomes positive, eventually decaying to zero as $t 	o \infty$. This perfectly matches the entire trajectory shown in option (d).

Select Answer:

Visualized Solution

Visualizing the Setup

Magnetic Field of the Solenoid

Magnetic Flux through the Outer Coil

Substituting the Current Function

Applying Faraday's Law

Expression for Induced Current

Analyzing the Graph at $t = 0$

Final Conclusion

The Sigma Insight: Faraday's Laws of Electromagnetic Induction

Analyzing the Setup Imagine you are looking at a very long solenoid of radius $R$

The Magnetic Flux Trap The first step is to calculate the magnetic flux $ϕ$ passing through the outer coil

The Master Equation

Decoding the Graph Now, we don't need to plot this entire function point by point

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