Self-Inductance - Physics 2

1. An RL circuit consists of a resistor $R$ and an inductor $L$ connected in series with a battery of emf $\mathcal{E}_0$. Long after the switch is closed, the current has reached its steady-state value. The switch is then opened at $t = 0$, and the inductor drives current through a "snubber" resistor $R_s$ connected in parallel with the inductor (the battery branch is now open). Which of the following correctly describes the initial self-induced emf of the inductor at $t = 0^+$ and the time constant of the subsequent decay?
   Answer: The initial self-induced emf magnitude is $\mathcal{E}_0 R_s / R$, and the time constant is $L/R_s$, because only $R_s$ carries the decaying current after the battery branch opens. When analyzing RL circuits after a switch opens, your first job is to identify what's actually in the loop carrying current — the battery branch is now open, so only the inductor and snubber resistor $R_s$ form a closed circuit. The physics: Long after the switch was originally closed, the steady-state current is $I_0 = \mathcal{E}_0 / R$. Inductors resist changes in current, so at $t = 0^+$, this same current $I_0$ still flows — now entirely through $R_s$. The inductor's self-induced emf must equal the voltage across $R_s$, giving $\mathcal{E}_L = I_0 R_s = \frac{\mathcal{E}_0 R_s}{R}$. After the switch opens, the only resistance in the loop is $R_s$, so the decay time constant is $\tau = L/R_s$. This confirms answer D. Why the others fail: Choice A incorrectly claims the initial emf involves $(R + R_s)$, but $R$ is in the now-open battery branch — it carries zero current and contributes nothing to the voltage. Choice B gets the initial emf right but uses the wrong time constant $L/(R+R_s)$, again mistakenly including $R$ in the active loop. Choice C assumes the inductor behaves as if the battery is still present, producing emf $\mathcal{E}_0$ and time constant $L/R$ — this ignores that the circuit topology has fundamentally changed when the switch opened. Strategy tip: Whenever a switch opens in an RL circuit, redraw the circuit immediately to identify the new loop. The time constant and voltage are determined solely by components in that active loop — disconnected branches are irrelevant.
2. A long straight wire carrying current $I$ passes through the center of a toroidal coil of $N$ turns, mean radius $R$, and cross-sectional area $a$. A student claims that this arrangement has a nonzero self-inductance $L_{\text{self}}$ because the straight wire's magnetic field threads the toroid's core. Which of the following best evaluates this claim?
   Answer: The claim is incorrect. The self-inductance of the toroid is determined solely by the flux through the toroid due to its own current; the straight wire's field contributes only to mutual inductance between the wire and the toroid, not to the toroid's self-inductance. When you encounter a question mixing multiple circuit elements, always ask: whose current creates whose flux? Self-inductance and mutual inductance are defined by completely different flux linkages, and confusing them is exactly the trap this question sets. Self-inductance is defined as $L_{\text{self}} = N\Phi_{\text{self}}/I_{\text{toroid}}$, where $\Phi_{\text{self}}$ is the flux through the toroid produced by the toroid's own current. For a toroid, this gives the familiar $L = \mu_0 N^2 a / (2\pi R)$. This value is a geometric property of the coil alone — it doesn't change just because another current-carrying conductor is nearby. The straight wire does produce a magnetic field that threads the toroidal cross-sections, but that flux linkage defines the mutual inductance $M$ between the wire and toroid, not the toroid's self-inductance. Answer A correctly identifies this distinction. Answer B confuses total flux linkage with self-inductance. Yes, the wire's field threads the toroid, but that contribution belongs to $M$, not $L_{\text{self}}$. These are separate terms in the full flux-linkage equation $\Phi_{\text{total}} = LI_{\text{toroid}} + MI_{\text{wire}}$. Answer C introduces a fictional condition — self-inductance never depends on the ratio of currents between coupled elements. It's a fixed geometric quantity. Answer D is wrong in the opposite direction. A toroid absolutely has nonzero self-inductance; its confined field links with its own windings by definition. Study tip: Always separate $L$ (flux from your own current) from $M$ (flux from someone else's current). If an external source creates the flux, it's mutual inductance — full stop.
3. Which of the following correctly expresses the magnitude of the self-induced emf in the solenoid at time $t$?
   Answer: $\mathcal{E} = \frac{\mu_r \mu_0 N^2 A}{L} \cdot \frac{I_0}{\tau} e^{-t/\tau}$, because the self-inductance is $L_{\text{ind}} = \mu_r \mu_0 N^2 A / L$ and the emf magnitude equals $L_{\text{ind}} |dI/dt|$. When a solenoid carries a time-varying current, it generates a changing magnetic flux through its own coils, inducing a back-emf. The two key relationships you need are: the self-inductance of a solenoid filled with a magnetic material, $L_{\text{ind}} = \mu_r \mu_0 N^2 A / L$, and Faraday's law applied to inductors, $\mathcal{E} = L_{\text{ind}} |dI/dt|$. Keeping these two formulas straight — and knowing which one to apply — is the entire challenge of this question. Starting from the given current $I(t) = I_0 e^{-t/\tau}$, its time derivative is $dI/dt = -(I_0/\tau)e^{-t/\tau}$, so the magnitude is $(I_0/\tau)e^{-t/\tau}$. Multiplying by $L_{\text{ind}} = \mu_r \mu_0 N^2 A/L$ gives exactly what A states — confirming it as correct. B is wrong because it multiplies $L_{\text{ind}}$ by the current $I(t)$ itself rather than by $|dI/dt|$. This confuses the formula for magnetic flux linkage ($\Lambda = L_{\text{ind}} I$) with the formula for emf. C drops the $\mu_r$ factor, treating the core as free space. Relative permeability absolutely appears in the inductance of a solenoid with a magnetic core; ignoring it is only valid when $\mu_r = 1$ (vacuum or air). D introduces $d^2I/dt^2$, which would give a $1/\tau^2$ factor. The emf depends on the first derivative of current, not the second. Your study tip: always write down both formulas — $L_{\text{ind}}$ for the geometry and $\mathcal{E} = L_{\text{ind}}|dI/dt|$ for the physics — before plugging in. Mixing up $I$ and $dI/dt$ is the most common trap in inductor problems.
4. Compared to a single-wire solenoid of the same number of turns $N$, same length, and same cross-sectional area, the self-inductance of the bifilar coil is best described as:
   Answer: Approximately zero, because the magnetic fields from the two oppositely-wound wires nearly cancel throughout the interior, causing the net flux linkage — and hence the self-inductance — to be negligibly small. When analyzing inductive devices, always start by asking: what is the net magnetic flux through the coil? Self-inductance is defined as $L = N\Phi/I$, so if the net flux is zero, the inductance is zero — regardless of how many turns are wound. In a bifilar coil, two wires carry equal currents in opposite directions through adjacent, overlapping paths. By the right-hand rule, each wire produces a magnetic field that points in the opposite direction to its neighbor's field. Because the wires are wound side-by-side on the same form with the same geometry, these fields cancel almost perfectly throughout the interior. With essentially no net flux linkage, $L \approx 0$. Answer B is correct. Answer A is wrong because it conflates the physical geometry (two wires on the same form) with an equivalent 2N-turn coil — but that logic only applies when currents reinforce, not cancel. The "factor of four compensation" reasoning is invented and physically meaningless. Answer C treats the two windings as independent inductors in series with negligible mutual inductance, but this ignores the fact that mutual inductance is actually large and negative here — the two coils are tightly coupled with opposing polarity, so $L_{total} = L_1 + L_2 + 2M \approx L_0 + L_0 - 2L_0 = 0$. Answer D correctly recalls that inductance scales as $N^2$, but wrongly assumes the currents reinforce; in a bifilar winding they oppose, so you never get an effective 2N-turn coil. The key study tip: flux cancellation kills inductance. Bifilar windings are used precisely because they have nearly zero inductance — they appear as pure resistors at high frequencies, which is useful in precision resistors and certain sensor designs.
5. How does the self-inductance $L_2$ of solenoid 2 compare to $L_1$ of solenoid 1, assuming both solenoids are long enough that end effects are negligible?
   Answer: $L_2 = \tfrac{1}{2} L_1$, because halving the radius reduces the cross-sectional area by a factor of four, and doubling the length only partially compensates, so the product $N^2 A / \ell$ decreases by a factor of two. Whenever you see a solenoid inductance problem with geometric constraints, your first move should be to write out $L = \mu_0 n^2 A \ell$ and then carefully track how every variable changes — not just the ones the problem highlights. Here, the constraint is that both solenoids use the same total wire length. Solenoid 1 has $N$ turns of circumference $2\pi r$, so total wire length $\mathcal{L}_w = 2\pi r N$. Solenoid 2 has $2N$ turns of circumference $2\pi (r/2) = \pi r$, giving $\mathcal{L}_w = \pi r \cdot 2N = 2\pi r N$ — consistent. Since $n$ is unchanged and the number of turns doubles, the length of solenoid 2 must double: $\ell_2 = 2\ell$. Now apply the inductance formula. The cross-sectional area shrinks: $A_2 = \pi(r/2)^2 = A_1/4$. So: $L_2 = \mu_0 n^2 A_2 \ell_2 = \mu_0 n^2 \left(\frac{A_1}{4}\right)(2\ell) = \frac{1}{2}\mu_0 n^2 A_1 \ell = \frac{1}{2}L_1$ That confirms C is correct. A is wrong because it ignores the area change — even though $n$ and wire length are fixed, $A$ is not fixed, and it enters the formula directly. B misapplies the $N^2$ scaling: that form ($L = \mu_0 N^2 A/\ell$) requires you to also account for how $A$ and $\ell$ change, which this choice ignores entirely. D makes the same $N^2$ error and incorrectly claims wire-length conservation neutralizes the radius effect. Study tip: On inductance problems, never treat one variable as the "dominant" factor. Always substitute all changed quantities into a single formula and let the algebra decide.
6. At the instant the battery is short-circuited ($t = 0^+$), what is the magnitude of the self-induced emf across the inductor's inductive element (not including the drop across its internal resistance $r$)?
   Answer: $10 \text{ V}$, because at $t = 0^+$ the current is still $I_0 = 1 \text{ A}$ and the inductor must drive this current through the full series resistance $(R + r) = 10\ \Omega$, giving $\mathcal{E}_L = I_0(R+r) = (1)(10) = 10 \text{ V}$. When a circuit with an inductor reaches steady state, the inductor acts like a plain wire — it carries a constant current with no changing flux, so it produces no back-emf. Here, the steady-state current is $I_0 = \frac{10}{R + r} = \frac{10}{8 + 2} = 1\ \text{A}$. The instant the battery is replaced by a short circuit, the inductor's fundamental property kicks in: it cannot allow its current to change instantaneously. So at $t = 0^+$, the current is still exactly $1\ \text{A}$. Now the inductor becomes the sole "source" driving that 1 A through the only resistance left in the loop — the series combination $(R + r) = 10\ \Omega$. The self-induced emf must supply the voltage needed to sustain this current against the full loop resistance: $\mathcal{E}_L = I_0(R + r) = (1)(10) = 10\ \text{V}$. That makes C correct. Choice A is tempting but wrong — it only accounts for the drop across $R$, ignoring that the current also passes through the internal resistance $r$. The inductor must drive current through the entire closed loop. Choice B makes the opposite partial error, counting only the drop across $r$ and forgetting $R$. Choice D invents a halving rule that has no physical basis; the self-induced emf is determined by Kirchhoff's voltage law around the post-short-circuit loop, not by some fraction of the original battery voltage. Study tip: When the battery is removed and the inductor takes over, redraw the circuit and apply KVL to the new loop — the self-induced emf must balance all resistive drops that remain in that loop, not just some of them.
7. An ideal inductor of self-inductance $L$ carries a steady current $I_0$. At $t = 0$ the current begins increasing linearly: $I(t) = I_0 + \alpha t$ for $t > 0$, where $\alpha > 0$. Which of the following statements about the self-induced emf $\mathcal{E}$ and the energy stored in the inductor $U$ is correct for $t > 0$?
   Answer: $\mathcal{E}$ is constant and opposes the increase in current, while $U$ increases at a rate that grows linearly with time because $dU/dt = L I \alpha$ and $I$ itself increases linearly. When a question involves an inductor with changing current, your two key formulas are the self-induced emf $\mathcal{E} = -L\frac{dI}{dt}$ and the stored energy $U = \frac{1}{2}LI^2$. The critical insight is that these two quantities behave very differently even when $dI/dt$ is constant. Since $I(t) = I_0 + \alpha t$, the rate of change is $\frac{dI}{dt} = \alpha$, a constant. Therefore $\mathcal{E} = -L\alpha$ is constant in magnitude and opposes the increasing current by Lenz's law — that part is straightforward. For the energy, differentiate $U = \frac{1}{2}LI^2$ with respect to time: $\frac{dU}{dt} = LI\frac{dI}{dt} = LI\alpha$. Because $I = I_0 + \alpha t$ grows linearly, $\frac{dU}{dt}$ also grows linearly with time. This confirms B — the emf is constant, but the rate of energy storage increases over time because the same $\alpha$ is now pushing current through a growing $I$. A is wrong because it claims $dU/dt = L\alpha^2$, a constant — this ignores the $I$ factor in the power formula and is only valid instantaneously at $t = 0$. C makes the same error, asserting $dU/dt$ is constant, which contradicts the correct power expression. D is wrong in its very first claim: since $\frac{dI}{dt} = \alpha$ is constant, $\mathcal{E}$ is constant, not linearly increasing. Your study tip: always distinguish between $\mathcal{E}$ (which depends on $dI/dt$) and $dU/dt$ (which depends on both $I$ and $dI/dt$). A constant rate of current change does not mean a constant rate of energy change.
8. A coaxial cable of inner conductor radius $a$, outer conductor radius $b$, and length $\ell$ carries current $I$ along the inner conductor and return current $-I$ along the outer conductor. The self-inductance per unit length of this cable is $\Lambda = (\mu_0/2\pi)\ln(b/a)$. If both radii are scaled by a common factor $k > 1$ (so the new radii are $ka$ and $kb$) while the length $\ell$ is kept fixed, how does the total self-inductance change?
   Answer: The total self-inductance is unchanged, because $\ln(kb/ka) = \ln(b/a)$ and the length is fixed, so neither the inductance per unit length nor the total inductance changes. When a question asks how inductance changes under geometric scaling, your first instinct should be to write out the formula explicitly and track what actually changes — not what intuitively seems like it should change. The total self-inductance of the coaxial cable is simply the inductance per unit length multiplied by the length: $L = \Lambda \cdot \ell = \frac{\mu_0}{2\pi}\ln\!\left(\frac{b}{a}\right)\cdot\ell$. When both radii are scaled by $k$, the new ratio becomes $\frac{kb}{ka}$. Crucially, $k$ cancels: $\ln\!\left(\frac{kb}{ka}\right) = \ln\!\left(\frac{b}{a}\right)$. Since the length $\ell$ is also fixed, every factor in the formula is unchanged, and so is $L$. Answer D is correct. Answer A is a tempting but misapplied intuition — cross-sectional area scaling as $k^2$ is relevant for solenoids or flat coils where flux area appears directly in the formula, but coaxial inductance depends on a logarithmic ratio, not on area itself. Answer B reflects a true general dimensional argument (inductance does scale as length when all lengths scale), but here only the radii scale while $\ell$ is held fixed — and even if $\ell$ also scaled, the $\ln(b/a)$ term would remain the same, neutralizing the effect anyway. Answer C confuses $\ln(kb/ka)$ with $\ln(kb) - \ln(a)$, incorrectly splitting the logarithm to add $\ln(k)$ — this is an algebra error; the ratio $kb/ka$ simplifies before taking the log. Study tip: Whenever you see a logarithmic ratio in a physics formula, check whether the scaling factor cancels inside the ratio before assuming anything changes — logarithms of ratios are surprisingly robust to uniform scaling.

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