Irodov 3.334: A Ramp Induces a Steady Current – Amit & Deepali Sinha Foundation

Problem Irodov 3.334

Two loops that do not move have mutual inductance $L_{12}$ . From the moment $t = 0$ the current in the first loop is made to grow as $I_1 = \alpha t$ , with $\alpha$ a constant. The second loop has inductance $L_2$ and resistance $R$ . Find the current $I_2(t)$ in the second loop as a function of time.

The two loops as coupled coils: loop 1 carries the prescribed current I_1 = α t (sense chosen), loop 2 is the coil L_2 closed through R; the dots mark the ends for which L_12 > 0. — The two loops as coupled coils: loop 1 carries the prescribed current $I_1 = \alpha t$ (sense chosen), loop 2 is the coil $L_2$ closed through $R$ ; the dots mark the ends for which $L_{12} > 0$ .

Given

$L_{12}$
$I_1 = \alpha t$ from $t = 0$
loop 2: $L_2$ , $R$
no current in loop 2 at $t = 0$ (assumed)

Find

the current

I_2(t)

Solution

Step 1The emf in the second loop

The reference sense of I_2, out of the dotted end of its coil while I_1 enters its own: the two currents send flux through loop 2 opposite ways. — The reference sense of $I_2$ , out of the dotted end of its coil while $I_1$ enters its own: the two currents send flux through loop 2 opposite ways.

Take $I_2$ positive in the sense of the figure, out of the dotted end of its coil. By the dot convention $I_1$ , entering its dot, then sends through loop 2, oriented along $I_2$ , the flux $-L_{12}I_1$ , while $I_2$ sends $L_2I_2$ . Faraday’s law gives the emf round loop 2 as minus the rate of change of the total, and Ohm’s law sets it equal to $RI_2$ :

-\frac{d}{dt}\left(L_2I_2 - L_{12}I_1\right) = RI_2

The loops do not move, so $L_{12}$ and $L_2$ are constants, and $dI_1/dt = \alpha$ :

L_2\,\frac{dI_2}{dt} + RI_2 = \alpha L_{12}

(1)

Step 2Integrating from the start of the ramp

The right-hand side of (1) is constant, so its steady solution is $I_2 = \alpha L_{12}/R$ , and the departure from it decays. Separating the variables and integrating from $I_2 = 0$ at $t = 0$ (assumed):

\begin{aligned} \int_0^{I_2}\frac{di}{\alpha L_{12}/R - i} &= \frac{R}{L_2}\int_0^t dt'\\[2pt] -\ln\frac{\alpha L_{12}/R - I_2}{\alpha L_{12}/R} &= \frac{tR}{L_2} \end{aligned}

Taking the exponential and solving for $I_2$ :

I_2(t) = \frac{\alpha L_{12}}{R}\left(1 - e^{-tR/L_2}\right)

(2)

The current flows in the drawn sense, round loop 2 against $I_1$ (each leaving or entering its coil at the dot oppositely), so it opposes the growth of the flux from $I_1$ , as Lenz’s rule requires.

The induced current: a rise from zero toward α L_12/R with the time constant L_2/R. — The induced current: a rise from zero toward $\alpha L_{12}/R$ with the time constant $L_2/R$ .

Check

Substituting (2) back: $L_2\,dI_2/dt = \alpha L_{12}e^{-tR/L_2}$ and $RI_2 = \alpha L_{12} - \alpha L_{12}e^{-tR/L_2}$ , which add to $\alpha L_{12}$ , as (1) requires, and $I_2(0) = 0$ . With $R \to 0$ the exponential expands to $I_2 \approx \alpha L_{12}t/L_2$ , so $L_2I_2 - L_{12}I_1 = 0$ : a loop with no resistance keeps its flux. Units: $\alpha L_{12}$ is $(\text{A/s})\cdot\text{H} = \text{V}$ , divided by $R$ a current.

Remark

A steadily growing current induces a constant emf, $\alpha L_{12}$ in magnitude, so loop 2 ends up with a steady current $\alpha L_{12}/R$ , as if it held a battery. What the inductance $L_2$ adds is the delay: for $t \ll L_2/R$ the loop behaves as a superconducting one and its current cancels the incoming flux, and only after a few $L_2/R$ does the resistance take over and set the current.