Physics 2 Quiz: Self Inductance

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Self InductanceQuestion 1 of 8

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 $t > 0$ ?

\mathcal{E}

is constant and opposes the increase in current, while

U

increases at a constant rate equal to

L \alpha^2

because the power delivered depends only on the constant rate of current change.

\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.

Both

\mathcal{E}

and

dU/dt

are constant, with

\mathcal{E} = L\alpha

and

dU/dt = L\alpha^2

, so the energy increases quadratically but at a rate that does not depend on the instantaneous current.

\mathcal{E}

increases linearly with time because the current increases linearly, while

U

increases quadratically with time because energy is proportional to

I^2

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Physics 2 Quiz

Physics 2 Quiz: Self Inductance

Practice Self Inductance in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Self Inductance, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\mathcal{E}$ is constant and opposes the increase in current, while $U$ increases at a constant rate equal to $L \alpha^2$ because the power delivered depends only on the constant rate of current change.
$\mathcal{E}$ 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. (correct answer)
Both $\mathcal{E}$ and $dU/dt$ are constant, with $\mathcal{E} = L\alpha$ and $dU/dt = L\alpha^2$ , so the energy increases quadratically but at a rate that does not depend on the instantaneous current.
$\mathcal{E}$ increases linearly with time because the current increases linearly, while $U$ increases quadratically with time because energy is proportional to $I^2$ .

Explanation: 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.

Question 2

A solenoid of length $L$ , cross-sectional area $A$ , and $N$ total turns is filled with a magnetic material of relative permeability $\mu_r$ . The solenoid carries a current $I(t) = I_0 e^{-t/\tau}$ , where $I_0$ and $\tau$ are positive constants.

Which of the following correctly expresses the magnitude of the self-induced emf in the solenoid at time $t$ ?

$\mathcal{E} = \frac{\mu_r \mu_0 N^2 A}{L} \cdot \frac{I_0}{\tau} e^{-t/\tau}$ , because the self-inductance is $L_{\text{ind}} = \mu_r \mu_0 N^2 A / L$ and the emf magnitude equals $L_{\text{ind}} |dI/dt|$ . (correct answer)
$\mathcal{E} = \frac{\mu_r \mu_0 N^2 A}{L} \cdot I_0 e^{-t/\tau}$ , because the emf equals the self-inductance multiplied by the instantaneous current rather than its time derivative.
$\mathcal{E} = \frac{\mu_0 N^2 A}{L} \cdot \frac{I_0}{\tau} e^{-t/\tau}$ , because the relative permeability of the core material does not appear in the expression for self-inductance when the material is linear and isotropic.
$\mathcal{E} = \frac{\mu_r \mu_0 N^2 A}{L} \cdot \frac{I_0}{\tau^2} e^{-t/\tau}$ , because the emf involves the second time derivative of current when the current decays exponentially.

Explanation: 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.

Question 3

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?

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. (correct answer)
The claim is correct. Because the straight wire's flux passes through the toroidal cross-sections, it adds to the total flux linkage of the toroid and therefore increases the toroid's self-inductance beyond what it would have in the absence of the wire.
The claim is correct only if the current in the straight wire equals the current in the toroid; otherwise the two fields partially cancel and the effective self-inductance is reduced.
The claim is incorrect because a toroidal coil has zero self-inductance by symmetry; its magnetic field is entirely confined within the core and therefore cannot link with any winding of the coil itself.

Explanation: 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.

Question 4

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?

The initial self-induced emf magnitude is $\mathcal{E}_0(R + R_s)/R$ , and the time constant is $L/R_s$ , because the initial current through $R_s$ is $\mathcal{E}_0/R$ and the voltage across $R_s$ at $t = 0^+$ involves both resistors.
The initial self-induced emf magnitude is $\mathcal{E}_0 R_s / R$ , and the time constant is $L/(R + R_s)$ , because the two resistors appear in series once the battery branch opens.
The initial self-induced emf magnitude is $\mathcal{E}_0$ , and the time constant is $L/R$ , because the inductor simply continues to behave as though the battery were still present.
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. (correct answer)

Explanation: 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.

Question 5

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?

The total self-inductance increases by a factor of $k^2$ , because the cross-sectional area enclosed by the cable scales as $k^2$ and inductance is proportional to the area through which flux passes.
The total self-inductance increases by a factor of $k$ , because scaling all lengths by $k$ scales every geometric parameter of the inductor by $k$ , and inductance has dimensions of length.
The total self-inductance increases by a factor of $\ln(k) + \ln(b/a)$ , because the logarithm of the new ratio adds $\ln(k)$ to the original logarithm term.
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. (correct answer)

Explanation: 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.

Question 6

Two solenoids, labeled 1 and 2, are made from the same total length of wire. Solenoid 1 has $N$ turns, length $\ell$ , and radius $r$ . Solenoid 2 is wound with wire of the same total length but with twice as many turns ( $2N$ ) by halving the radius and adjusting the length so that the number of turns per unit length $n$ remains the same as solenoid 1.

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?

$L_2 = L_1$ , because both solenoids use the same total wire length and the same turns per unit length $n$ , and since $L = \mu_0 n^2 A \ell$ , the inductance depends only on these fixed quantities.
$L_2 = 2 L_1$ , because the number of turns is doubled and inductance scales as $N^2$ , so the fourfold increase in $N^2$ dominates all other geometric changes.
$L_2 = \tfrac{1}{2} L_1$ , 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. (correct answer)
$L_2 = 4 L_1$ , because inductance scales as $N^2$ and the number of turns is doubled, producing a fourfold increase that is not affected by the change in radius when wire length is conserved.

Explanation: 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.

Question 7

A physicist winds a "bifilar" coil by winding two wires side-by-side on the same form, then connecting them so that the current in one wire flows in the opposite direction to the current in the adjacent wire. The resulting double-wound coil has $N$ effective turns on each wire.

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:

Approximately equal to the single-wire solenoid's inductance, because the bifilar winding doubles the total number of turns on the form, increasing $L$ by a factor of four relative to one wire alone, which compensates for the opposing currents.
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. (correct answer)
Exactly twice the single-wire solenoid's inductance, because each wire independently contributes an inductance $L_0$ and the total self-inductance of two series-connected coils is additive when their mutual inductance is negligible.
Exactly four times the single-wire solenoid's inductance, because connecting two $N$ -turn coils in series on the same form gives a $2N$ -turn coil, and inductance scales as $N^2$ , yielding $L_{2N} = 4L_N$ .

Explanation: 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.

Question 8

An inductor with self-inductance $L = 50 \text{ mH}$ and internal resistance $r = 2\ \Omega$ is connected in series with an external resistor $R = 8\ \Omega$ and a 10 V ideal battery. After the circuit reaches steady state, the battery is replaced instantaneously by a short circuit (a wire of zero resistance).

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 $r$ )?

$8 \text{ V}$ , because at $t = 0^+$ the steady-state current of 1 A flows entirely through the external resistor $R = 8\ \Omega$ , and the self-induced emf equals the voltage drop across $R$ alone.
$2 \text{ V}$ , because at $t = 0^+$ the current is 1 A and the self-induced emf equals only the voltage drop across the internal resistance: $\mathcal{E}_L = I_0 r = (1)(2) = 2 \text{ V}$ .
$10 \text{ V}$ , 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 $\mathcal{E}_L = I_0(R+r) = (1)(10) = 10 \text{ V}$ . (correct answer)
$5 \text{ V}$ , because the self-induced emf at $t = 0^+$ equals half the original battery emf, since the inductor's internal resistance $r$ accounts for half the total voltage drop in the original steady-state circuit.

Explanation: 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.