- #1

zenterix

- 555

- 76

- Homework Statement
- Consider the circuit shown below consisting of a battery with emf ##\mathcal{E}##, a resistor with resistance ##R##, a switch ##S##, and a long solenoid of radius ##a##, height ##H## that has ##N## turns.

Coaxial with the solenoid at the center of the solenoid is a circular copper ring of wire of radius ##b## with ##b>a## and resistance ##R_1##.

At ##t=0##, the switch is closed.

- Relevant Equations
- What is the induced current in the copper ring at the instant the switch is closed?

Assume the solenoid is very long such that the magnetic field can be approximated as constant inside.

If there weren't any ring surrounding the solenoid, then when the switch were closed we would have the following equation from Faraday's law

$$\oint \vec{E}\cdot d\vec{l}=-\mathcal{E}+IR=-L\dot{I}$$

$$\dot{I}+\frac{R}{L}I=\frac{\mathcal{E}}{L}$$

$$I(t)=\frac{\mathcal{E}}{R}\left (1-e^{-\frac{Rt}{L}}\right )$$

$$\dot{I}(t)=\frac{\mathcal{E}}{L}e^{-\frac{Rt}{L}}$$

We see that the current would initially be zero and would increase to a maximum value of ##\frac{\mathcal{E}}{R}## as ##t\to\infty##.

But we do have the surrounding copper wire.

My question is about the calculations when we take this wire into account.

Qualitatively, it seems that when the switch is closed there is a large ##\dot{I}## due to the battery.

The magnetic flux through the solenoid is

$$\Phi_{sol}=N\Phi_{turn}=N\frac{\mu_0 NI\pi a^2}{H}=\frac{N^2\mu_0\pi a^2}{H}I$$

and the rate of change of this flux is

$$\dot{\Phi}_{sol}=\frac{N^2\mu_0\pi a^2}{H}\dot{I}$$

Thus, the self-inductance of the solenoid is

$$L=\frac{N^2\mu_0\pi a^2}{H}$$

The back emf due to the solenoid is at its strongest when the switch is just closed because ##\mathcal{E}_L=-\dot{\Phi}=-L\dot{I}##.

Now consider the emf induced in the outer ring.

The flux through the ring due to the solenoid is simply the flux through one turn of the solenoid.

$$\Phi_{sol,ring}=\frac{\mu_0 N\pi a^2}{H}I$$

Thus

$$\dot{\Phi}_{sol,ring}=\frac{\mu_0 N\pi a^2}{H}\dot{I}=M\dot{I}$$

where ##M## is the mutual inductance for the solenoid and the outer ring.

At this point, what I initially thought to do was compute the current in the ring as follows

$$\mathcal{E}_{ring}=RI_{ring}=-\dot{\Phi}_{sol,ring}=-M\dot{I}=-\frac{\mu_0 N\pi a^2}{H}\dot{I}$$

$$I_{ring}(t)=-\frac{\mu_0 N\pi a^2}{RH}\frac{\mathcal{E}}{L}e^{-Rt/L}$$

$$=-\frac{\mathcal{E}}{NR}e^{-\frac{Rt}{n^2 H\mu_0\pi a^2}}$$

Note that in these calculations I assuming the following for the circulation direction and am assuming that the solenoid magnetic field is pointing in the same direction as ##\hat{n}_{rhr}##.

Therefore, we have a counterclockwise current (due to the negative sign and our clockwise positive circulation direction).

This makes sense qualitatively, but doesn't seem to be correct quantitatively.

With this reasoning, the answer to the problem (induced ring current at ##t=0## is ##-\frac{\mathcal{E}}{NR}##.

I think there is something missing: the self-inductance of the ring.

**Is this so?**

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