College Physics Quiz: Inductance
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InductanceQuestion 1 of 16

A toroidal inductor has a rectangular cross-section with inner radius a=2.0 cma = 2.0 \text{ cm}, outer radius b=3.0 cmb = 3.0 \text{ cm}, and height h=1.5 cmh = 1.5 \text{ cm}. If the toroid has 200 turns, what is its self-inductance? (Use μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \text{ H/m})

5.1×10−5 H5.1 \times 10^{-5} \text{ H}
1.0×10−4 H1.0 \times 10^{-4} \text{ H}
2.1×10−4 H2.1 \times 10^{-4} \text{ H}
3.8×10−4 H3.8 \times 10^{-4} \text{ H}
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College Physics Quiz: Inductance

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

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

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Question 1

A toroidal inductor has a rectangular cross-section with inner radius a=2.0 cma = 2.0 \text{ cm}, outer radius b=3.0 cmb = 3.0 \text{ cm}, and height h=1.5 cmh = 1.5 \text{ cm}. If the toroid has 200 turns, what is its self-inductance? (Use μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7} \text{ H/m})

  1. 5.1×10−5 H5.1 \times 10^{-5} \text{ H} (correct answer)
  2. 1.0×10−4 H1.0 \times 10^{-4} \text{ H}
  3. 2.1×10−4 H2.1 \times 10^{-4} \text{ H}
  4. 3.8×10−4 H3.8 \times 10^{-4} \text{ H}
Explanation: For a toroidal inductor, L=μ0N2h2πln⁡(b/a)L = \frac{\mu_0 N^2 h}{2\pi} \ln(b/a). Substituting values: L=(4π×10−7)(200)2(0.015)2πln⁡(3.0/2.0)=(4π×10−7)(40000)(0.015)2πln⁡(1.5)=(2×10−7)(600)ln⁡(1.5)=1.2×10−4ln⁡(1.5)≈1.2×10−4(0.405)≈5.1×10−5 HL = \frac{(4\pi \times 10^{-7})(200)^2(0.015)}{2\pi} \ln(3.0/2.0) = \frac{(4\pi \times 10^{-7})(40000)(0.015)}{2\pi} \ln(1.5) = (2 \times 10^{-7})(600) \ln(1.5) = 1.2 \times 10^{-4} \ln(1.5) \approx 1.2 \times 10^{-4}(0.405) \approx 5.1 \times 10^{-5} \text{ H}. Choice B omits the logarithmic factor. Choice C uses an incorrect formula with linear dimensions. Choice D incorrectly uses (b−a)(b-a) instead of ln⁡(b/a)\ln(b/a).

Question 2

A solenoid with 500 turns and a cross-sectional area of 2.0×10−3 m22.0 \times 10^{-3} \, \text{m}^2 has a length of 0.20 m. When the current through the solenoid changes at a rate of 3.0 A/s, what is the magnitude of the self-induced EMF? (μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A})

  1. 9.4×10−3 V9.4 \times 10^{-3} \, \text{V} (correct answer)
  2. 1.9×10−2 V1.9 \times 10^{-2} \, \text{V}
  3. 3.8×10−2 V3.8 \times 10^{-2} \, \text{V}
  4. 7.5×10−2 V7.5 \times 10^{-2} \, \text{V}
  5. 1.5×10−1 V1.5 \times 10^{-1} \, \text{V}
Explanation: When you encounter electromagnetic induction problems involving solenoids, you're dealing with Faraday's law and self-inductance. The key relationship is that a changing current in a coil induces an EMF that opposes the change. The self-induced EMF is given by ∣E∣=L∣dIdt∣|\mathcal{E}| = L \left|\frac{dI}{dt}\right|, where LL is the self-inductance. For a solenoid, the inductance is L=μ0N2AlL = \frac{\mu_0 N^2 A}{l}, where NN is the number of turns, AA is the cross-sectional area, and ll is the length. First, calculate the inductance: L=(4π×10−7)(500)2(2.0×10−3)0.20=(4π×10−7)(250,000)(2.0×10−3)0.20=3.14×10−3 HL = \frac{(4\pi \times 10^{-7})(500)^2(2.0 \times 10^{-3})}{0.20} = \frac{(4\pi \times 10^{-7})(250,000)(2.0 \times 10^{-3})}{0.20} = 3.14 \times 10^{-3} \text{ H} Then find the EMF: ∣E∣=(3.14×10−3)(3.0)=9.4×10−3 V|\mathcal{E}| = (3.14 \times 10^{-3})(3.0) = 9.4 \times 10^{-3} \text{ V} This confirms answer A is correct. Answer B (1.9×10−2 V1.9 \times 10^{-2} \, \text{V}) likely results from doubling the correct answer, perhaps from incorrectly including a factor of 2 in the inductance formula. Answer C (3.8×10−2 V3.8 \times 10^{-2} \, \text{V}) represents roughly four times the correct value, possibly from squaring an incorrect factor. Answer D (7.5×10−2 V7.5 \times 10^{-2} \, \text{V}) is about eight times too large, suggesting multiple computational errors. Remember: self-inductance problems require careful unit tracking and precise application of the solenoid inductance formula. Always double-check that your calculated inductance has units of henries before proceeding to find the EMF.

Question 3

A circular coil of 200 turns has a radius of 5.0 cm. If the coil is placed in a uniform magnetic field and the current through it changes at a rate of 1.5 A/s, what is the approximate self-inductance of the coil? (Assume μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} and neglect end effects)

  1. 7.9×10−5 H7.9 \times 10^{-5} \, \text{H} (correct answer)
  2. 1.6×10−4 H1.6 \times 10^{-4} \, \text{H}
  3. 3.1×10−4 H3.1 \times 10^{-4} \, \text{H}
  4. 6.3×10−4 H6.3 \times 10^{-4} \, \text{H}
  5. 1.3×10−3 H1.3 \times 10^{-3} \, \text{H}
Explanation: Self-inductance questions test your understanding of how coils create magnetic flux that opposes current changes. When you see a circular coil problem, you need the formula for self-inductance: L=μ0n2AℓL = \mu_0 n^2 A \ell, where nn is turns per unit length, AA is cross-sectional area, and ℓ\ell is coil length. However, for a single-layer circular coil (like this problem), we use the approximation: L=μ0N2A2RL = \frac{\mu_0 N^2 A}{2R}, where NN is total turns, A=πr2A = \pi r^2 is the coil's area, and RR is the radius. Let's calculate: N=200N = 200, r=0.05 mr = 0.05 \text{ m}, so A=π(0.05)2=7.85×10−3 m2A = \pi (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2. L=(4π×10−7)(200)2(7.85×10−3)2(0.05)L = \frac{(4\pi \times 10^{-7})(200)^2(7.85 \times 10^{-3})}{2(0.05)} L=(4π×10−7)(40000)(7.85×10−3)0.1=7.9×10−5 HL = \frac{(4\pi \times 10^{-7})(40000)(7.85 \times 10^{-3})}{0.1} = 7.9 \times 10^{-5} \text{ H} This confirms answer A is correct. Answer B (1.6×10−4 H1.6 \times 10^{-4} \text{ H}) results from using the wrong geometric factor—likely forgetting the factor of 2 in the denominator. Answer C (3.1×10−4 H3.1 \times 10^{-4} \text{ H}) comes from incorrectly using the coil circumference instead of area. Answer D (6.3×10−4 H6.3 \times 10^{-4} \text{ H}) suggests using radius instead of area in the numerator. Remember: the rate of current change given (1.5 A/s) is a red herring—self-inductance depends only on coil geometry, not current behavior.

Question 4

Two coils have a mutual inductance of 0.030 H. If the current in the first coil decreases uniformly from 4.0 A to zero in 0.20 s, what is the average EMF induced in the second coil?

  1. 0.30 V0.30 \, \text{V}
  2. 0.60 V0.60 \, \text{V} (correct answer)
  3. 1.2 V1.2 \, \text{V}
  4. 2.4 V2.4 \, \text{V}
  5. 4.8 V4.8 \, \text{V}
Explanation: When you encounter mutual inductance problems, you're dealing with how changing current in one coil induces EMF in a nearby coil. The key relationship is Faraday's law applied to mutual inductance: ε=−MdIdt\varepsilon = -M \frac{dI}{dt}, where MM is the mutual inductance and dIdt\frac{dI}{dt} is the rate of current change. First, find the rate of current change. The current decreases from 4.0 A to 0 A over 0.20 s, so dIdt=0−4.00.20=−20 A/s\frac{dI}{dt} = \frac{0 - 4.0}{0.20} = -20 \, \text{A/s}. The induced EMF magnitude is ∣ε∣=M∣dIdt∣=0.030 H×20 A/s=0.60 V|\varepsilon| = M \left|\frac{dI}{dt}\right| = 0.030 \, \text{H} \times 20 \, \text{A/s} = 0.60 \, \text{V}. Answer B (0.60 V) is correct. Answer A (0.30 V) likely results from using only half the current change (perhaps 2.0 A instead of 4.0 A) or making an arithmetic error in the calculation. Answer C (1.2 V) suggests doubling the correct answer, possibly from incorrectly using the total current (4.0 A) instead of the rate of change, or miscalculating 0.030×400.030 \times 40. Answer D (2.4 V) appears to come from using 4.00.20=20\frac{4.0}{0.20} = 20 but then multiplying by 0.120.12 instead of 0.0300.030, or other computational mistakes. Remember: mutual inductance problems always require finding dIdt\frac{dI}{dt} first. Calculate the current change divided by time, then multiply by the mutual inductance. Watch your arithmetic carefully—these problems often include tempting wrong answers that result from common calculation errors.

Question 5

The self-inductance of a coil depends on which of the following factors?

  1. The number of turns only
  2. The cross-sectional area and length only
  3. The number of turns, cross-sectional area, and length only
  4. The number of turns, cross-sectional area, length, and core material only (correct answer)
  5. The number of turns, cross-sectional area, length, core material, and current magnitude
Explanation: When you encounter questions about self-inductance, remember that inductance measures a coil's ability to oppose changes in current by generating a back EMF. This property depends on the coil's geometry and the magnetic properties of its environment. The self-inductance of a coil is given by L=μN2AlL = \mu \frac{N^2 A}{l}, where N is the number of turns, A is the cross-sectional area, l is the length, and μ is the permeability of the core material. This formula reveals that all four factors directly influence inductance. More turns (N²) create stronger magnetic coupling between loops, increasing inductance. A larger cross-sectional area allows more magnetic flux, while greater length spreads the same number of turns over more distance, reducing flux density and inductance. The core material's permeability dramatically affects how easily magnetic field lines form—an iron core can increase inductance by hundreds of times compared to air. Option A is incomplete because it ignores geometry and core material effects. Option B overlooks both the number of turns (which appears squared in the formula) and the core material—two of the most significant factors. Option C includes the three geometric factors but misses the core material, which can have the largest impact on inductance values in practical applications. For physics exams, remember that inductance questions often test whether you recognize all the variables in the inductance formula. Don't fall for answers that only include some factors—inductance depends on both the coil's physical geometry and the magnetic properties of its core material.

Question 6

A coil with self-inductance LL carries current II. If both the number of turns and the radius are doubled while keeping the length constant, what happens to the stored magnetic energy when the same current II flows through the modified coil?

  1. It increases by a factor of 4
  2. It increases by a factor of 8
  3. It increases by a factor of 16 (correct answer)
  4. It increases by a factor of 32
  5. It increases by a factor of 64
Explanation: When you encounter problems about inductance modifications, focus on how self-inductance depends on the coil's geometry and how that affects stored energy. Self-inductance scales as L∝N2AlL \propto \frac{N^2 A}{l}, where NN is the number of turns, AA is the cross-sectional area, and ll is the length. Since area A=πr2A = \pi r^2, we have L∝N2r2lL \propto \frac{N^2 r^2}{l}. When both the number of turns and radius are doubled while length stays constant, the new inductance becomes L′∝(2N)2(2r)2l=4N2⋅4r2l=16⋅N2r2lL' \propto \frac{(2N)^2(2r)^2}{l} = \frac{4N^2 \cdot 4r^2}{l} = 16 \cdot \frac{N^2 r^2}{l}. Therefore, L′=16LL' = 16L. The magnetic energy stored in an inductor is U=12LI2U = \frac{1}{2}LI^2. With the same current II flowing through the modified coil, the new energy is U′=12L′I2=12(16L)I2=16UU' = \frac{1}{2}L'I^2 = \frac{1}{2}(16L)I^2 = 16U. The energy increases by a factor of 16, making C correct. Option A (factor of 4) only accounts for doubling either the turns or radius, not both. Option B (factor of 8) might result from incorrectly treating the area as linear in radius rather than quadratic. Option D (factor of 32) could come from miscounting the geometric factors or confusing this with a different scaling relationship. Remember: inductance problems often involve squared relationships (N2N^2 for turns, r2r^2 for area), so geometric changes can produce surprisingly large effects on stored energy.

Question 7

A solenoid has an inductance of 8.0 mH. If the current through the solenoid increases linearly from 0 to 3.0 A in 0.15 s, what is the magnitude of the back EMF during this time interval?

  1. 0.080 V0.080 \, \text{V}
  2. 0.16 V0.16 \, \text{V} (correct answer)
  3. 0.24 V0.24 \, \text{V}
  4. 0.32 V0.32 \, \text{V}
  5. 0.48 V0.48 \, \text{V}
Explanation: When you encounter inductance problems involving changing current, you're dealing with Faraday's law of electromagnetic induction. The key relationship here is that a changing current through an inductor creates a back EMF that opposes the change. The back EMF in an inductor is given by ε=−LdIdt\varepsilon = -L \frac{dI}{dt}, where LL is inductance and dIdt\frac{dI}{dt} is the rate of current change. Since the current increases linearly from 0 to 3.0 A in 0.15 s, the rate of change is constant: dIdt=3.0 A−0 A0.15 s=20 A/s\frac{dI}{dt} = \frac{3.0 \, \text{A} - 0 \, \text{A}}{0.15 \, \text{s}} = 20 \, \text{A/s} The magnitude of back EMF is: ∣ε∣=L×dIdt=8.0×10−3 H×20 A/s=0.16 V|\varepsilon| = L \times \frac{dI}{dt} = 8.0 \times 10^{-3} \, \text{H} \times 20 \, \text{A/s} = 0.16 \, \text{V} Looking at the wrong answers: Choice A (0.080 V) uses half the correct rate of change, possibly from incorrectly calculating dIdt=3.00.3\frac{dI}{dt} = \frac{3.0}{0.3} instead of 3.00.15\frac{3.0}{0.15}. Choice C (0.24 V) might result from using 30 A/s as the rate instead of 20 A/s. Choice D (0.32 V) doubles the correct answer, possibly from a calculation error or unit confusion. The correct answer is B: 0.16 V. Study tip: For linear current changes in inductors, always calculate dIdt\frac{dI}{dt} carefully as ΔIΔt\frac{\Delta I}{\Delta t}, then multiply by inductance. The negative sign in Lenz's law affects direction, not magnitude calculations.

Question 8

An inductor with self-inductance 0.50 H carries a current of 2.0 A. How much magnetic energy is stored in the inductor?

  1. 0.50 J0.50 \, \text{J}
  2. 1.0 J1.0 \, \text{J} (correct answer)
  3. 1.5 J1.5 \, \text{J}
  4. 2.0 J2.0 \, \text{J}
  5. 4.0 J4.0 \, \text{J}
Explanation: When you encounter questions about magnetic energy storage in inductors, you're dealing with the electromagnetic equivalent of kinetic energy storage. Just as a moving mass stores kinetic energy, a current-carrying inductor stores magnetic energy in its magnetic field. The magnetic energy stored in an inductor is given by the formula U=12LI2U = \frac{1}{2}LI^2, where LL is the self-inductance and II is the current. This formula is analogous to the kinetic energy formula KE=12mv2KE = \frac{1}{2}mv^2, with inductance playing the role of "magnetic mass." Substituting the given values: U=12(0.50 H)(2.0 A)2=12(0.50)(4.0)=1.0 JU = \frac{1}{2}(0.50 \, \text{H})(2.0 \, \text{A})^2 = \frac{1}{2}(0.50)(4.0) = 1.0 \, \text{J}. This confirms answer B. Looking at the wrong answers: A) 0.50 J0.50 \, \text{J} results from forgetting to square the current, calculating 12LI\frac{1}{2}LI instead of 12LI2\frac{1}{2}LI^2. C) 1.5 J1.5 \, \text{J} comes from incorrectly using 34LI2\frac{3}{4}LI^2 instead of the correct 12\frac{1}{2} factor. D) 2.0 J2.0 \, \text{J} results from omitting the 12\frac{1}{2} factor entirely, calculating just LI2LI^2. Study tip: Remember that energy formulas in physics often contain a 12\frac{1}{2} factor (kinetic energy, spring potential energy, capacitor energy, and inductor energy). Always double-check that you've included this factor and properly squared the relevant variable—current for inductors, velocity for kinetic energy.

Question 9

Two identical coils are placed near each other. When the current in coil A changes from 2.0 A to 6.0 A in 0.10 s, an EMF of 0.20 V is induced in coil B. What is the mutual inductance between the coils?

  1. 5.0×10−3 H5.0 \times 10^{-3} \, \text{H}
  2. 2.0×10−2 H2.0 \times 10^{-2} \, \text{H}
  3. 4.0×10−2 H4.0 \times 10^{-2} \, \text{H}
  4. 5.0×10−2 H5.0 \times 10^{-2} \, \text{H} (correct answer)
  5. 8.0×10−2 H8.0 \times 10^{-2} \, \text{H}
Explanation: When you encounter a mutual inductance problem, you're dealing with how changing current in one coil induces an EMF in a nearby coil. The key relationship is Faraday's law for mutual inductance: E=−MdIdt\mathcal{E} = -M \frac{dI}{dt}, where MM is the mutual inductance, and dIdt\frac{dI}{dt} is the rate of current change. First, calculate the rate of current change in coil A: dIdt=6.0 A−2.0 A0.10 s=4.0 A0.10 s=40 A/s\frac{dI}{dt} = \frac{6.0 \, \text{A} - 2.0 \, \text{A}}{0.10 \, \text{s}} = \frac{4.0 \, \text{A}}{0.10 \, \text{s}} = 40 \, \text{A/s} Now solve for mutual inductance: M=∣E∣∣dIdt∣=0.20 V40 A/s=0.005 H=5.0×10−2 HM = \frac{|\mathcal{E}|}{|\frac{dI}{dt}|} = \frac{0.20 \, \text{V}}{40 \, \text{A/s}} = 0.005 \, \text{H} = 5.0 \times 10^{-2} \, \text{H} Looking at the wrong answers: Choice A (5.0×10−3 H5.0 \times 10^{-3} \, \text{H}) results from incorrectly using the total current change (4.0 A) instead of the rate of change (40 A/s). Choice B (2.0×10−2 H2.0 \times 10^{-2} \, \text{H}) comes from using only the initial current (2.0 A) in place of the current change rate. Choice C (4.0×10−2 H4.0 \times 10^{-2} \, \text{H}) results from using the current change (4.0 A) directly instead of dividing by time to get the rate. The answer is D: 5.0×10−2 H5.0 \times 10^{-2} \, \text{H}. Study tip: Always remember that Faraday's law involves the rate of change, not just the change itself. Calculate dIdt\frac{dI}{dt} first, then solve for inductance.

Question 10

A long solenoid has 1000 turns per meter and carries a current that increases at a rate of 50 A/s50 \text{ A/s}. A single-turn circular loop of radius 2.0 cm2.0 \text{ cm} is placed inside the solenoid with its axis parallel to the solenoid's axis. What is the magnitude of the induced EMF in the loop?

  1. 7.9×10−6 V7.9 \times 10^{-6} \text{ V} (correct answer)
  2. 1.6×10−5 V1.6 \times 10^{-5} \text{ V}
  3. 2.5×10−5 V2.5 \times 10^{-5} \text{ V}
  4. 5.0×10−5 V5.0 \times 10^{-5} \text{ V}
Explanation: The magnetic field inside a long solenoid is B=μ0nIB = \mu_0 nI where nn is turns per unit length. The flux through the loop is Φ=BA=μ0nIπr2\Phi = BA = \mu_0 nI \pi r^2. The induced EMF is ∣ε∣=∣dΦ/dt∣=μ0nπr2∣dI/dt∣|\varepsilon| = |d\Phi/dt| = \mu_0 n \pi r^2 |dI/dt|. Substituting: ∣ε∣=(4π×10−7)(1000)π(0.02)2(50)=(4π×10−7)(1000)π(4×10−4)(50)=4π2×10−7×2×10−2=8π2×10−9≈7.9×10−6 V|\varepsilon| = (4\pi \times 10^{-7})(1000)\pi(0.02)^2(50) = (4\pi \times 10^{-7})(1000)\pi(4 \times 10^{-4})(50) = 4\pi^2 \times 10^{-7} \times 2 \times 10^{-2} = 8\pi^2 \times 10^{-9} \approx 7.9 \times 10^{-6} \text{ V}. Choice B omits one factor of π\pi. Choice C uses the diameter instead of radius. Choice D ignores the π\pi factors entirely.

Question 11

An RL circuit consists of a 25 mH25 \text{ mH} inductor and a 50 Ω50 \text{ Ω} resistor connected in series with a battery. At t=0t = 0, the switch closes and current begins to flow. What fraction of the final steady-state current is reached after a time equal to twice the time constant?

  1. 1−e−2≈0.861 - e^{-2} \approx 0.86 (correct answer)
  2. 1−e−1≈0.631 - e^{-1} \approx 0.63
  3. e−2≈0.14e^{-2} \approx 0.14
  4. 2(1−e−1)≈1.262(1 - e^{-1}) \approx 1.26
Explanation: In an RL circuit, current grows as I(t)=If(1−e−t/τ)I(t) = I_f(1 - e^{-t/\tau}) where τ=L/R\tau = L/R is the time constant and IfI_f is the final current. The time constant is τ=(25×10−3)/(50)=5.0×10−4 s\tau = (25 \times 10^{-3})/(50) = 5.0 \times 10^{-4} \text{ s}. At t=2τt = 2\tau, the fraction is (1−e−2τ/τ)=1−e−2≈0.86(1 - e^{-2\tau/\tau}) = 1 - e^{-2} \approx 0.86. Choice B gives the fraction at t=τt = \tau, not 2τ2\tau. Choice C gives the fraction that has NOT been reached (the exponential decay term). Choice D incorrectly applies a factor of 2 to the single time constant result.

Question 12

A toroidal coil has 800 turns wound on a core with inner radius 4.0 cm, outer radius 6.0 cm, and height 2.0 cm. If the relative permeability of the core material is 2000, what is the approximate self-inductance of the toroidal coil? (μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A})

  1. 0.13 H0.13 \, \text{H} (correct answer)
  2. 0.26 H0.26 \, \text{H}
  3. 0.52 H0.52 \, \text{H}
  4. 1.04 H1.04 \, \text{H}
  5. 2.08 H2.08 \, \text{H}
Explanation: When you encounter a toroidal inductor problem, you need to use the specific formula for toroidal geometry, which accounts for the varying magnetic field strength at different radii within the core. For a toroidal coil, the self-inductance is given by: L=μ0μrN2h2πln⁡(r2r1)L = \frac{\mu_0 \mu_r N^2 h}{2\pi} \ln\left(\frac{r_2}{r_1}\right) where NN is the number of turns, hh is the height, r2r_2 is the outer radius, r1r_1 is the inner radius, and μr\mu_r is the relative permeability. Substituting the given values: L=(4π×10−7)(2000)(800)2(0.02)2πln⁡(0.060.04)L = \frac{(4\pi \times 10^{-7})(2000)(800)^2(0.02)}{2\pi} \ln\left(\frac{0.06}{0.04}\right) L=(4π×10−7)(2000)(640,000)(0.02)2πln⁡(1.5)L = \frac{(4\pi \times 10^{-7})(2000)(640,000)(0.02)}{2\pi} \ln(1.5) L=(4×10−7)(2000)(640,000)(0.02)×0.405L = (4 \times 10^{-7})(2000)(640,000)(0.02) \times 0.405 L=0.13 HL = 0.13 \, \text{H} This confirms answer A is correct. Answer B (0.26 H) would result from forgetting the natural logarithm term or miscalculating it. Answer C (0.52 H) likely comes from using an incorrect geometric factor, perhaps treating this as a solenoid instead of a toroid. Answer D (1.04 H) represents a calculation error, possibly doubling one of the terms or mishandling the units. Remember that toroidal inductors require the logarithmic term due to their geometry—the magnetic field isn't uniform across the core's cross-section. Always double-check that you're using the toroidal formula, not the simpler solenoid approximation.

Question 13

A student is investigating the behavior of an air-core solenoid inductor. The solenoid has a length of 20 cm, a diameter of 4.0 cm, and is wound with 800 turns of wire. The student wants to understand how changes to the solenoid's geometry affect its inductance.

If the student compresses the solenoid to half its original length while keeping the same number of turns and the same diameter, by what factor does the inductance change?

  1. The inductance decreases by a factor of 2
  2. The inductance remains the same
  3. The inductance increases by a factor of 2 (correct answer)
  4. The inductance increases by a factor of 4
Explanation: For a solenoid, L=μ0n2AℓL = \mu_0 n^2 A \ell where n=N/ℓn = N/\ell is the turn density. When length is halved: nn doubles (from N/ℓN/\ell to N/(ℓ/2)=2N/ℓN/(\ell/2) = 2N/\ell), so n2n^2 increases by factor of 4. However, ℓ\ell decreases by factor of 2. Net change: Lnew=μ0(2n)2A(ℓ/2)=μ0(4n2)A(ℓ/2)=2μ0n2Aℓ=2LoriginalL_{new} = \mu_0 (2n)^2 A (\ell/2) = \mu_0 (4n^2) A (\ell/2) = 2\mu_0 n^2 A \ell = 2L_{original}. Choice A incorrectly focuses only on length decrease. Choice B assumes inductance is independent of geometry. Choice D incorrectly considers only the n2n^2 increase while ignoring the length decrease.

Question 14

An inductor stores 0.25 J0.25 \text{ J} of energy when the current through it is 2.0 A2.0 \text{ A}. If this inductor is connected in series with a 100 Ω100 \text{ Ω} resistor and a 12 V12 \text{ V} battery, what is the time constant of the resulting RL circuit?

  1. 1.00×10−2 s1.00 \times 10^{-2} \text{ s}
  2. 2.50×10−3 s2.50 \times 10^{-3} \text{ s}
  3. 5.00×10−3 s5.00 \times 10^{-3} \text{ s}
  4. 1.25×10−3 s1.25 \times 10^{-3} \text{ s} (correct answer)
Explanation: RL circuits combine inductance and resistance, and understanding their time constants requires connecting energy storage in inductors to circuit dynamics. When you encounter problems involving both inductor energy and circuit time constants, you need to find the inductance first, then apply the RL time constant formula. Start by finding the inductance using the energy formula. The energy stored in an inductor is U=12LI2U = \frac{1}{2}LI^2. With U=0.25 JU = 0.25 \text{ J} and I=2.0 AI = 2.0 \text{ A}: 0.25=12L(2.0)20.25 = \frac{1}{2}L(2.0)^2 0.25=2L0.25 = 2L L=0.125 HL = 0.125 \text{ H} The time constant of an RL circuit is τ=LR\tau = \frac{L}{R}. With L=0.125 HL = 0.125 \text{ H} and R=100 ΩR = 100 \text{ Ω}: τ=0.125100=1.25×10−3 s\tau = \frac{0.125}{100} = 1.25 \times 10^{-3} \text{ s} This confirms answer D is correct. Answer A (1.00×10−2 s1.00 \times 10^{-2} \text{ s}) suggests using L=1.0 HL = 1.0 \text{ H}, which would result from incorrectly solving the energy equation. Answer B (2.50×10−3 s2.50 \times 10^{-3} \text{ s}) comes from using L=0.25 HL = 0.25 \text{ H}, likely from setting L=UL = U directly without proper calculation. Answer C (5.00×10−3 s5.00 \times 10^{-3} \text{ s}) results from using L=0.5 HL = 0.5 \text{ H}, possibly from omitting the factor of 12\frac{1}{2} in the energy formula. Remember: RL circuit problems often require two steps—first extract the inductance from given information (energy, flux, etc.), then apply the specific circuit formula you need. Always double-check your algebra in the energy equation since small errors multiply through the final calculation.

Question 15

In an LR circuit with L=40 mHL = 40 \text{ mH} and R=80 ΩR = 80 \text{ Ω}, the current is decreasing from an initial value. When the current has decreased to 37% of its initial value, how much energy remains stored in the magnetic field compared to the initial energy?

  1. 86%86\% of the initial energy remains
  2. 37%37\% of the initial energy remains
  3. 63%63\% of the initial energy remains
  4. 14%14\% of the initial energy remains (correct answer)
Explanation: When analyzing energy in LR circuits, remember that magnetic energy depends on the square of the current, not the current itself. This quadratic relationship is the key to solving problems involving energy changes as current decays. The energy stored in an inductor's magnetic field is given by U=12LI2U = \frac{1}{2}LI^2. Initially, the energy is U0=12LI02U_0 = \frac{1}{2}LI_0^2. When the current decreases to 37% of its initial value, the new current is I=0.37I0I = 0.37I_0. The remaining energy becomes: U=12L(0.37I0)2=12LI02(0.37)2=U0(0.37)2U = \frac{1}{2}L(0.37I_0)^2 = \frac{1}{2}LI_0^2(0.37)^2 = U_0(0.37)^2 Calculating: (0.37)2=0.1369=13.7%≈14%(0.37)^2 = 0.1369 = 13.7\% \approx 14\% Therefore, 14% of the initial energy remains, making D correct. A (86%) incorrectly assumes the energy decreases linearly with current, calculating 100% - 37% = 63% lost, leaving 37% remaining. This ignores the quadratic relationship. B (37%) represents the most common error: confusing current percentage with energy percentage. Students often mistakenly think that if current drops to 37%, energy also drops to 37%. C (63%) likely comes from calculating how much energy was lost (100% - 37% = 63%) rather than what remains. Study tip: Always remember that energy storage formulas containing squared terms (like 12LI2\frac{1}{2}LI^2 or 12CV2\frac{1}{2}CV^2) mean the energy changes as the square of the variable. If a quantity drops to 37%, the energy drops to (0.37)2=14%(0.37)^2 = 14\%.

Question 16

Two identical circular coils are positioned coaxially. When a current of 2.0 A2.0 \text{ A} flows through coil 1, the magnetic flux through coil 2 is 1.5×10−4 Wb1.5 \times 10^{-4} \text{ Wb}. If the coils are moved closer together so that the flux through coil 2 increases to 2.4×10−4 Wb2.4 \times 10^{-4} \text{ Wb} for the same current, what is the new mutual inductance between the coils?

  1. 7.5×10−5 H7.5 \times 10^{-5} \text{ H}
  2. 1.2×10−4 H1.2 \times 10^{-4} \text{ H} (correct answer)
  3. 1.95×10−4 H1.95 \times 10^{-4} \text{ H}
  4. 3.9×10−4 H3.9 \times 10^{-4} \text{ H}
Explanation: Mutual inductance is defined as M=Φ21/I1M = \Phi_{21}/I_1, where Φ21\Phi_{21} is the flux through coil 2 due to current I1I_1 in coil 1. In the new configuration, M=Φ21/I1=(2.4×10−4 Wb)/(2.0 A)=1.2×10−4 HM = \Phi_{21}/I_1 = (2.4 \times 10^{-4} \text{ Wb})/(2.0 \text{ A}) = 1.2 \times 10^{-4} \text{ H}. Choice A uses the original flux value instead of the new one. Choice C incorrectly averages the two flux values. Choice D uses the wrong relationship, multiplying flux and current instead of dividing.