Energy Stored in Inductors - Physics 2

1. Which of the following scenarios most directly refutes the student's claim by demonstrating a case where the inductor with smaller inductance stores more energy?
   Answer: Two inductors $L_1 = 1\text{ H}$ and $L_2 = 4\text{ H}$ are connected in parallel across a constant-current source of $I_{\text{total}} = 5\text{ A}$. The current divides inversely with inductance, giving $I_1 = 4\text{ A}$ and $I_2 = 1\text{ A}$, so $U_1 = 8\text{ J}$ and $U_2 = 2\text{ J}$: the smaller inductor stores more. When evaluating a claim about inductors, always ask: what constraint does the circuit impose? The student assumes larger $L$ always wins, but that ignores how current distributes depending on circuit configuration. The key insight in D is how inductors share current in a parallel configuration driven by a constant-current source. Unlike resistors, inductors in parallel divide current inversely proportional to their inductance (because equal voltage across both means $V = L\frac{dI}{dt}$, so the smaller inductor ramps up current faster). With $I_{\text{total}} = 5\text{ A}$, the split gives $I_1 = 4\text{ A}$ (smaller $L_1 = 1\text{ H}$) and $I_2 = 1\text{ A}$ (larger $L_2 = 4\text{ H}$). Plugging into $U = \frac{1}{2}LI^2$: $U_1 = \frac{1}{2}(1)(16) = 8\text{ J}$ and $U_2 = \frac{1}{2}(4)(1) = 2\text{ J}$. The smaller inductor stores more energy — directly refuting the student's claim within the same circuit. A is a trap: it correctly identifies that ideal inductors in parallel across a voltage source produce unbounded current, but this means no valid comparison exists — it doesn't refute anything. B actually confirms the student's claim because series inductors share the same current, so larger $L$ always wins. C presents a valid counterexample mathematically but fails the "same circuit" requirement the student's claim specifies, making it logically insufficient as a refutation. Your takeaway: whenever a claim says "always," look for the configuration that breaks the assumption. Parallel inductors with a current source are a classic case where current distribution flips the expected energy ranking.
2. A student claims: *'The inductor with the larger inductance stores more energy, so the ratio of energy stored in $L_2$ to energy stored in $L_1$ equals the ratio of their inductances, which is 4:1.'* A second student counters: 'Because they are in series, the voltage across each inductor is proportional to its inductance, so the energy ratio must equal the square of the inductance ratio, giving 16:1.' Which student, if either, is correct, and what is the actual energy ratio $U_2 : U_1$?
   Answer: The first student is correct: $U_2 : U_1 = 4:1$, because $U = \frac{1}{2}LI^2$ and the current $I$ is the same for both, so the energy ratio equals the inductance ratio directly. When inductors are connected in series with no mutual inductance, every element shares the same current at every instant. This is the critical insight that unlocks the entire question — start there whenever you see energy storage in series inductors. The energy stored in any inductor is $U = \frac{1}{2}LI^2$. Since both inductors carry the same current $I = 3\text{ A}$, you can write the ratio directly: $\frac{U_2}{U_1} = \frac{\frac{1}{2}L_2 I^2}{\frac{1}{2}L_1 I^2} = \frac{L_2}{L_1} = \frac{8}{2} = 4$ So $U_2 : U_1 = 4:1$, confirming that answer A is correct. The first student's reasoning is sound. The second student (answer B) confuses two separate ideas. Yes, the voltage across each inductor is proportional to its inductance ($V = L\frac{dI}{dt}$), but energy stored in an inductor is not $\frac{1}{2}\frac{V^2}{L}$ in the way resistive power is. That formula applies to resistors with steady-state voltages — not to inductors storing magnetic energy. Squaring the inductance ratio to get 16:1 has no physical basis here. Answer C is a fabricated rule — energy in series inductors does not scale with the square root of inductance. No such relationship exists. Answer D reverses the logic entirely; a larger inductance doesn't reduce stored energy when current is identical — it increases it. Study tip: In series circuits, current is always the equalizer. When a question asks you to compare energy in series inductors, immediately recognize that $I$ cancels in the ratio, leaving only $L_2/L_1$.
3. What fraction of the energy that was stored in the inductor at $t = 0$ has been dissipated in the resistor by time $t = 2\tau$, where $\tau = L/R$?
   Answer: $1 - e^{-4}$, because the current decays as $e^{-t/\tau}$, the energy decays as $e^{-2t/\tau}$, and the fraction dissipated is $1$ minus the remaining energy fraction $e^{-2(2\tau)/\tau} = e^{-4}$. When an inductor discharges through a resistor, you need to track energy, not just current — and these decay at different rates. That distinction is what this question is testing. At steady state, the initial current is $I_0 = \mathcal{E}/R$, and the energy stored is $U_0 = \frac{1}{2}LI_0^2$. After the battery is removed, the current decays as $I(t) = I_0\, e^{-t/\tau}$. The energy remaining in the inductor at any time is $U(t) = \frac{1}{2}L I(t)^2 = \frac{1}{2}L I_0^2\, e^{-2t/\tau} = U_0\, e^{-2t/\tau}$. Notice the exponent is $-2t/\tau$, not $-t/\tau$, because energy goes as current squared. At $t = 2\tau$, the remaining energy fraction is $e^{-2(2\tau)/\tau} = e^{-4}$. Since all energy lost by the inductor is dissipated in the resistor, the fraction dissipated is $1 - e^{-4}$, confirming D. Choice A confuses the current decay factor $e^{-2}$ with an energy fraction — current and energy decay on different timescales. Choice B correctly identifies that energy (not current) governs the answer, but then applies a time constant of $\tau$ instead of $\tau/2$, arriving at the remaining fraction rather than accounting for the squared relationship properly. Choice C invents a false physical rule — all stored energy is eventually dissipated as heat; the inductor has no mechanism to "keep" half of it. Study tip: Whenever you see energy and exponential decay together, ask yourself: is the exponent coming from a quantity, or from that quantity squared? Energy always decays twice as fast as the current in an RL circuit.
4. At the instant when the energy stored in the inductor equals three times the energy stored in the capacitor, what is the magnitude of the current through the inductor?
   Answer: $I = V_0\sqrt{\frac{3C}{4L}} = 50\sqrt{\frac{3(40\times10^{-6})}{4(0.01)}} \approx 2.74\text{ A}$, found by setting the inductor energy equal to three-fourths of the total initial energy. When you see an LC circuit energy problem, your instinct should be to write two equations: the energy condition given, and total energy conservation. Together, they fully determine the unknowns. Here's the logic for choice B. The total initial energy is stored entirely in the capacitor: $U_{total} = \frac{1}{2}CV_0^2$. At any later moment, this energy splits between the inductor and capacitor. If the inductor holds three times the capacitor's energy, and those two portions must sum to the total, then $U_L + U_C = U_{total}$ becomes $3U_C + U_C = \frac{1}{2}CV_0^2$, giving $U_C = \frac{1}{8}CV_0^2$ and $U_L = \frac{3}{4} \cdot \frac{1}{2}CV_0^2$. Setting $\frac{1}{2}LI^2 = \frac{3}{4} \cdot \frac{1}{2}CV_0^2$ and solving: $I = V_0\sqrt{\frac{3C}{4L}} \approx 2.74\text{ A}$. That's B, and it's correct. Choice A makes a critical algebra error: it sets $U_L = 3 \cdot \frac{1}{2}CV_0^2$, treating the initial capacitor energy as the benchmark instead of the current capacitor energy. This ignores the constraint that both energies must sum to $U_{total}$, badly overcounting. Choice C computes the maximum possible current (when all energy is in the inductor), which is a useful reference but doesn't satisfy the stated condition — it's an upper bound, not the answer. Choice D assumes the capacitor's energy dropped to one-third of its initial value, which misreads the condition; the problem says the inductor energy is three times the current capacitor energy, not one-third of the original. The key strategy: whenever a problem says one energy is a multiple of the other, immediately write that relationship and the conservation equation simultaneously. Two unknowns, two equations — don't skip either step.
5. An engineer stores energy in two inductors by connecting them in series to a current source delivering $I = 5\text{ A}$. Inductor 1 has $L_1 = 3\text{ H}$ and Inductor 2 has $L_2 = 7\text{ H}$ with a mutual inductance $M = 2\text{ H}$ between them (the coils are wound so their fields add). What is the total magnetic energy stored in this coupled system?
   Answer: $U = \frac{1}{2}(L_1 + L_2 + 2M)I^2 = \frac{1}{2}(14)(25) = 175\text{ J}$, because for magnetically coupled series inductors with aiding fields, the effective inductance is $L_1 + L_2 + 2M$. When two inductors are magnetically coupled and carry the same current (series connection), you cannot treat them as isolated — their magnetic fields interact, and that interaction contributes real, physical energy to the system. The total energy stored is: $U = \frac{1}{2}L_1 I^2 + \frac{1}{2}L_2 I^2 \pm MI^2$ where the $\pm$ depends on whether the fields aid (+) or oppose (−) each other. Notice the mutual energy term carries no $\frac{1}{2}$ — this is because the coupling energy between coil 1 and coil 2 is counted once from each coil's perspective, and when you derive it from first principles (integrating power), those two contributions combine to give a single $MI^2$. Factoring out, the effective inductance for aiding fields is $L_{eff} = L_1 + L_2 + 2M$, making the total energy $\frac{1}{2}(3 + 7 + 4)(25) = \frac{1}{2}(14)(25) = \mathbf{175\text{ J}}$, confirming C is correct. A is wrong because it ignores mutual inductance entirely — a critical error whenever the problem states the coils are magnetically coupled. B adds $M$ only once, as if the coupling appears in just one coil's equation, but physically both coils experience the mutual flux. D makes the subtle mistake of writing the mutual energy as $\frac{1}{2}MI^2$, applying the $\frac{1}{2}$ factor that belongs only to self-inductance terms — the mutual energy term is $MI^2$, not $\frac{1}{2}MI^2$. A reliable memory rule: self-energy terms get $\frac{1}{2}$, mutual energy terms do not. For aiding series inductors, always add $2M$ to the inductances before computing energy.
6. A long solenoid of length $\ell$, cross-sectional area $A$, and $n$ turns per unit length carries current $I$. The energy stored per unit volume in the magnetic field inside the solenoid is $u_B$. If both the number of turns per unit length and the current are simultaneously doubled while $\ell$ and $A$ remain fixed, which of the following correctly describes the changes in total stored energy $U$ and energy density $u_B$?
   Answer: Both $U$ and $u_B$ increase by a factor of 16, because the magnetic field doubles due to doubling $n$ and doubles again due to doubling $I$, so $B$ increases by 4, and both energy and energy density scale as $B^2$. Whenever you see a question about energy stored in a magnetic field, your two essential formulas are the magnetic field inside a solenoid, $B = \mu_0 n I$, and the energy density, $u_B = \frac{B^2}{2\mu_0}$. Everything else follows from these. When you double both $n$ and $I$, the field becomes $B' = \mu_0 (2n)(2I) = 4\mu_0 nI = 4B$. Since energy density scales as $B^2$, you get $u_B' = \frac{(4B)^2}{2\mu_0} = 16\,u_B$ — a factor of 16. For total stored energy, since the geometry ($\ell$ and $A$) is unchanged, the volume $V = A\ell$ is constant, so $U = u_B \cdot V$ also scales by exactly the same factor of 16. Answer A is correct. Answer B claims $u_B$ only increases by 4, confusing energy density (which depends purely on $B$) with inductance arguments — but $u_B = B^2/2\mu_0$ has nothing to do with $L$ directly. Answer C incorrectly states inductance merely doubles when $n$ doubles; in fact $L = \mu_0 n^2 A \ell$, so doubling $n$ quadruples $L$, and the full energy $U = \frac{1}{2}LI^2$ gives $4 \times 4 = 16$, not 4. Answer D states the same final answer as A but invents a flawed "doubles twice" mechanism — $B$ does not "double twice"; it quadruples in a single step because both $n$ and $I$ double simultaneously. A reliable strategy: always go back to $B = \mu_0 nI$ first, compute the new $B$, then apply $u_B \propto B^2$. This single chain of reasoning handles both $u_B$ and $U$ without needing inductance formulas.
7. How does the energy stored in the inductor at time $t = \tau$ compare to the initial stored energy $U_0$?
   Answer: $U(\tau) = U_0 / e^2$, because the current has decayed to $I_0/e$ and energy depends on the square of the current. When a question links exponential decay to stored energy, your first instinct should be to track the current carefully, then apply the energy formula — don't jump straight to the energy decay without thinking about the exponent. Here's the chain of reasoning: when the resistive segment is introduced, the current decays as $I(t) = I_0 e^{-t/\tau}$. At $t = \tau$, the current is $I(\tau) = I_0 e^{-1} = I_0/e$. Now apply the energy formula for an inductor: $U = \frac{1}{2}LI^2$. Substituting the decayed current gives $U(\tau) = \frac{1}{2}L\left(\frac{I_0}{e}\right)^2 = \frac{1}{2}LI_0^2 \cdot \frac{1}{e^2} = \frac{U_0}{e^2}$. The squaring of the exponential factor is the key step, confirming answer A. Answer B claims energy is proportional to current (not current squared), so it only divides by $e$ once. This confuses the linear relationship between current and, say, flux, with the quadratic relationship between current and energy — a classic trap. Answer C describes the energy dissipated, not the energy remaining. It's true that roughly $(1 - 1/e)$ of the energy has been lost by $t = \tau$, but the question asks what's still stored, which is $U_0/e^2$, not $U_0(1-1/e)$. Answer D introduces a false concept: one time constant is not equivalent to a half-life. The half-life would require $e^{-t/\tau} = 1/2$, which doesn't occur at $t = \tau$. Study tip: Whenever energy depends on a decaying quantity, always square the decay factor. Energy goes as $e^{-2t/\tau}$, not $e^{-t/\tau}$.

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