Consider a circuit consisting of a battery, a resistor and a solenoid inductor. Then, the emf $\mathcal{E}$, is defined as:

$$\mathcal{E} = L\frac{di}{dt} + iR$$

Multiplying both sides by $i$ gives:

$$\mathcal{E}i = Li\frac{di}{dt} + i^2R$$

The term on the left side gives the rate at which the battery does work. Since the second term on the right side gives the rate at which energy appears as thermal energy in the resistor, the second term gives the rate at which magnetic potential energy is stored in the magnetic field.

Therefore $$\frac{dU_B}{dt} = Li\frac{di}{dt}$$ $$\int^{U_B}_{0} dU_B = \int^i_0 Li\text{ }di$$ $$U_B = \frac{1}{2}Li^2$$

Q1) I'm assuming there finding the energy in the steady state. I thought the current was constant in the steady state so shouldn't $\frac{di}{dt}$ be zero?

Q2) Why isn't the emf:

$$\mathcal{E} = -L\frac{di}{dt} + iR$$

Since the self-induced emf generated by an inductor tries to oppose the flow of current, shouldn't the emf be the opposite way?

Q3)The bounds of the integral: $U_B$ and $i$. How are they related? Are they the energy and current at the same point in time $t$? Or is $U_B$ the energy at any point in time and $i$ the current at some other point in time (not necessarily the same times)?


1 Answer 1


1) With a constant and DC power source eventually the solenoid will become fully 'charged'. At that point its 'resistance' term vanishes because it no longer produces an emf against the battery. At this point, the $\frac{di}{dt}$ term will be zero, because the current isn't changing.

2) When you cut power, the magnetic flux is no longer maintained by the current. However, the flux will try to stay constant, so that means the current will continue as it did before, powered by the magnetic field of the solenoid.

3) $U_B$ is simply all the $dU_B$ added up over time. You'd have to integrate until $t =\infty$, aka until the current behaves as if there's no solenoid at all (the integral at the righthand side).

  • $\begingroup$ Could you please break your answer up into pieces that answer each individual question? I can't connect the pieces of your answer to my questions. $\endgroup$
    – dfg
    Commented Mar 5, 2014 at 20:28
  • $\begingroup$ I wanted to explain the broad concept rather than address them directly. Anyway, the emf already is the opposite way as it's on the other side of the equation. if the battery pushes to the right then the solenoid and the resistance push to the left. $\endgroup$
    – Kvothe
    Commented Mar 5, 2014 at 20:40
  • $\begingroup$ If the current isn't constant, then the final current is not equal to the current current (the current right now). So $i_f$ doesn't equal $i$. And since your calculating the energy at the final moment in time, the upper bound of the integral is $i_f$. However since $i_f$ doesn't equal the current current $i$(the current right now), why does $(i_f)(i) = i^2$? $\endgroup$
    – dfg
    Commented Mar 6, 2014 at 16:44
  • $\begingroup$ It's confusing that they used $I$ for the upper limit of the integral, perhaps using $I_{final}$ would be clearer. Also, they're not multiplying $i_f$ and $i$ anywhere; $i$ is the integration variable. $\endgroup$
    – Kvothe
    Commented Mar 6, 2014 at 17:50

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