Inductance - College Physics

1. A long solenoid has 1000 turns per meter and carries a current that increases at a rate of $50 \text{ A/s}$. A single-turn circular loop of radius $2.0 \text{ cm}$ 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?
   Answer: $7.9 \times 10^{-6} \text{ V}$ The magnetic field inside a long solenoid is $B = \mu_0 nI$ where $n$ is turns per unit length. The flux through the loop is $\Phi = BA = \mu_0 nI \pi r^2$. The induced EMF is $|\varepsilon| = |d\Phi/dt| = \mu_0 n \pi r^2 |dI/dt|$. Substituting: $|\varepsilon| = (4\pi \times 10^{-7})(1000)\pi(0.02)^2(50) = (4\pi \times 10^{-7})(1000)\pi(4 \times 10^{-4})(50) = 4\pi^2 \times 10^{-7} \times 2 \times 10^{-2} = 8\pi^2 \times 10^{-9} \approx 7.9 \times 10^{-6} \text{ V}$. Choice B omits one factor of $\pi$. Choice C uses the diameter instead of radius. Choice D ignores the $\pi$ factors entirely.
2. In an LR circuit with $L = 40 \text{ mH}$ and $R = 80 \text{ Ω}$, 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?
   Answer: $14\%$ of the initial energy remains 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 $U = \frac{1}{2}LI^2$. Initially, the energy is $U_0 = \frac{1}{2}LI_0^2$. When the current decreases to 37% of its initial value, the new current is $I = 0.37I_0$. The remaining energy becomes: $U = \frac{1}{2}L(0.37I_0)^2 = \frac{1}{2}LI_0^2(0.37)^2 = U_0(0.37)^2$ Calculating: $(0.37)^2 = 0.1369 = 13.7\% \approx 14\%$ 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 $\frac{1}{2}LI^2$ or $\frac{1}{2}CV^2$) mean the energy changes as the square of the variable. If a quantity drops to 37%, the energy drops to $(0.37)^2 = 14\%$.
3. 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?
   Answer: $0.60 \, \text{V}$ 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: $\varepsilon = -M \frac{dI}{dt}$, where $M$ is the mutual inductance and $\frac{dI}{dt}$ 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 $\frac{dI}{dt} = \frac{0 - 4.0}{0.20} = -20 \, \text{A/s}$. The induced EMF magnitude is $|\varepsilon| = M \left|\frac{dI}{dt}\right| = 0.030 \, \text{H} \times 20 \, \text{A/s} = 0.60 \, \text{V}$. 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 $0.030 \times 40$. Answer D (2.4 V) appears to come from using $\frac{4.0}{0.20} = 20$ but then multiplying by $0.12$ instead of $0.030$, or other computational mistakes. Remember: mutual inductance problems always require finding $\frac{dI}{dt}$ 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.
4. 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?
   Answer: $0.16 \, \text{V}$ 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 $\varepsilon = -L \frac{dI}{dt}$, where $L$ is inductance and $\frac{dI}{dt}$ 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: $\frac{dI}{dt} = \frac{3.0 \, \text{A} - 0 \, \text{A}}{0.15 \, \text{s}} = 20 \, \text{A/s}$ The magnitude of back EMF is: $|\varepsilon| = L \times \frac{dI}{dt} = 8.0 \times 10^{-3} \, \text{H} \times 20 \, \text{A/s} = 0.16 \, \text{V}$ Looking at the wrong answers: Choice A (0.080 V) uses half the correct rate of change, possibly from incorrectly calculating $\frac{dI}{dt} = \frac{3.0}{0.3}$ instead of $\frac{3.0}{0.15}$. 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 $\frac{dI}{dt}$ carefully as $\frac{\Delta I}{\Delta t}$, then multiply by inductance. The negative sign in Lenz's law affects direction, not magnitude calculations.
5. Two identical circular coils are positioned coaxially. When a current of $2.0 \text{ A}$ flows through coil 1, the magnetic flux through coil 2 is $1.5 \times 10^{-4} \text{ Wb}$. If the coils are moved closer together so that the flux through coil 2 increases to $2.4 \times 10^{-4} \text{ Wb}$ for the same current, what is the new mutual inductance between the coils?
   Answer: $1.2 \times 10^{-4} \text{ H}$ Mutual inductance is defined as $M = \Phi_{21}/I_1$, where $\Phi_{21}$ is the flux through coil 2 due to current $I_1$ in coil 1. In the new configuration, $M = \Phi_{21}/I_1 = (2.4 \times 10^{-4} \text{ Wb})/(2.0 \text{ A}) = 1.2 \times 10^{-4} \text{ H}$. 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.
6. The self-inductance of a coil depends on which of the following factors?
   Answer: The number of turns, cross-sectional area, length, and core material only 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 $L = \mu \frac{N^2 A}{l}$, 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.
7. A coil with self-inductance $L$ carries current $I$. 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 $I$ flows through the modified coil?
   Answer: It increases by a factor of 16 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 $L \propto \frac{N^2 A}{l}$, where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length. Since area $A = \pi r^2$, we have $L \propto \frac{N^2 r^2}{l}$. When both the number of turns and radius are doubled while length stays constant, the new inductance becomes $L' \propto \frac{(2N)^2(2r)^2}{l} = \frac{4N^2 \cdot 4r^2}{l} = 16 \cdot \frac{N^2 r^2}{l}$. Therefore, $L' = 16L$. The magnetic energy stored in an inductor is $U = \frac{1}{2}LI^2$. With the same current $I$ flowing through the modified coil, the new energy is $U' = \frac{1}{2}L'I^2 = \frac{1}{2}(16L)I^2 = 16U$. 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 ($N^2$ for turns, $r^2$ for area), so geometric changes can produce surprisingly large effects on stored energy.
8. 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 $\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$ and neglect end effects)
   Answer: $7.9 \times 10^{-5} \, \text{H}$ 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: $L = \mu_0 n^2 A \ell$, where $n$ is turns per unit length, $A$ is cross-sectional area, and $\ell$ is coil length. However, for a single-layer circular coil (like this problem), we use the approximation: $L = \frac{\mu_0 N^2 A}{2R}$, where $N$ is total turns, $A = \pi r^2$ is the coil's area, and $R$ is the radius. Let's calculate: $N = 200$, $r = 0.05 \text{ m}$, so $A = \pi (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2$. $L = \frac{(4\pi \times 10^{-7})(200)^2(7.85 \times 10^{-3})}{2(0.05)}$ $L = \frac{(4\pi \times 10^{-7})(40000)(7.85 \times 10^{-3})}{0.1} = 7.9 \times 10^{-5} \text{ H}$ This confirms answer A is correct. Answer B ($1.6 \times 10^{-4} \text{ H}$) results from using the wrong geometric factor—likely forgetting the factor of 2 in the denominator. Answer C ($3.1 \times 10^{-4} \text{ H}$) comes from incorrectly using the coil circumference instead of area. Answer D ($6.3 \times 10^{-4} \text{ H}$) 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.
9. An inductor stores $0.25 \text{ J}$ of energy when the current through it is $2.0 \text{ A}$. If this inductor is connected in series with a $100 \text{ Ω}$ resistor and a $12 \text{ V}$ battery, what is the time constant of the resulting RL circuit?
   Answer: $1.25 \times 10^{-3} \text{ s}$ 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 $U = \frac{1}{2}LI^2$. With $U = 0.25 \text{ J}$ and $I = 2.0 \text{ A}$: $0.25 = \frac{1}{2}L(2.0)^2$ $0.25 = 2L$ $L = 0.125 \text{ H}$ The time constant of an RL circuit is $\tau = \frac{L}{R}$. With $L = 0.125 \text{ H}$ and $R = 100 \text{ Ω}$: $\tau = \frac{0.125}{100} = 1.25 \times 10^{-3} \text{ s}$ This confirms answer D is correct. Answer A ($1.00 \times 10^{-2} \text{ s}$) suggests using $L = 1.0 \text{ H}$, which would result from incorrectly solving the energy equation. Answer B ($2.50 \times 10^{-3} \text{ s}$) comes from using $L = 0.25 \text{ H}$, likely from setting $L = U$ directly without proper calculation. Answer C ($5.00 \times 10^{-3} \text{ s}$) results from using $L = 0.5 \text{ H}$, possibly from omitting the factor of $\frac{1}{2}$ 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.
10. An inductor with self-inductance 0.50 H carries a current of 2.0 A. How much magnetic energy is stored in the inductor?
    Answer: $1.0 \, \text{J}$ 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 $U = \frac{1}{2}LI^2$, where $L$ is the self-inductance and $I$ is the current. This formula is analogous to the kinetic energy formula $KE = \frac{1}{2}mv^2$, with inductance playing the role of "magnetic mass." Substituting the given values: $U = \frac{1}{2}(0.50 \, \text{H})(2.0 \, \text{A})^2 = \frac{1}{2}(0.50)(4.0) = 1.0 \, \text{J}$. This confirms answer B. Looking at the wrong answers: A) $0.50 \, \text{J}$ results from forgetting to square the current, calculating $\frac{1}{2}LI$ instead of $\frac{1}{2}LI^2$. C) $1.5 \, \text{J}$ comes from incorrectly using $\frac{3}{4}LI^2$ instead of the correct $\frac{1}{2}$ factor. D) $2.0 \, \text{J}$ results from omitting the $\frac{1}{2}$ factor entirely, calculating just $LI^2$. Study tip: Remember that energy formulas in physics often contain a $\frac{1}{2}$ 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.
11. 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?
    Answer: The inductance increases by a factor of 2 For a solenoid, $L = \mu_0 n^2 A \ell$ where $n = N/\ell$ is the turn density. When length is halved: $n$ doubles (from $N/\ell$ to $N/(\ell/2) = 2N/\ell$), so $n^2$ increases by factor of 4. However, $\ell$ decreases by factor of 2. Net change: $L_{new} = \mu_0 (2n)^2 A (\ell/2) = \mu_0 (4n^2) A (\ell/2) = 2\mu_0 n^2 A \ell = 2L_{original}$. Choice A incorrectly focuses only on length decrease. Choice B assumes inductance is independent of geometry. Choice D incorrectly considers only the $n^2$ increase while ignoring the length decrease.
12. A solenoid with 500 turns and a cross-sectional area of $2.0 \times 10^{-3} \, \text{m}^2$ 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? ($\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$)
    Answer: $9.4 \times 10^{-3} \, \text{V}$ 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 $|\mathcal{E}| = L \left|\frac{dI}{dt}\right|$, where $L$ is the self-inductance. For a solenoid, the inductance is $L = \frac{\mu_0 N^2 A}{l}$, where $N$ is the number of turns, $A$ is the cross-sectional area, and $l$ is the length. First, calculate the inductance: $L = \frac{(4\pi \times 10^{-7})(500)^2(2.0 \times 10^{-3})}{0.20} = \frac{(4\pi \times 10^{-7})(250,000)(2.0 \times 10^{-3})}{0.20} = 3.14 \times 10^{-3} \text{ H}$ Then find the EMF: $|\mathcal{E}| = (3.14 \times 10^{-3})(3.0) = 9.4 \times 10^{-3} \text{ V}$ This confirms answer A is correct. Answer B ($1.9 \times 10^{-2} \, \text{V}$) likely results from doubling the correct answer, perhaps from incorrectly including a factor of 2 in the inductance formula. Answer C ($3.8 \times 10^{-2} \, \text{V}$) represents roughly four times the correct value, possibly from squaring an incorrect factor. Answer D ($7.5 \times 10^{-2} \, \text{V}$) 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.
13. A toroidal inductor has a rectangular cross-section with inner radius $a = 2.0 \text{ cm}$, outer radius $b = 3.0 \text{ cm}$, and height $h = 1.5 \text{ cm}$. If the toroid has 200 turns, what is its self-inductance? (Use $\mu_0 = 4\pi \times 10^{-7} \text{ H/m}$)
    Answer: $5.1 \times 10^{-5} \text{ H}$ For a toroidal inductor, $L = \frac{\mu_0 N^2 h}{2\pi} \ln(b/a)$. Substituting values: $L = \frac{(4\pi \times 10^{-7})(200)^2(0.015)}{2\pi} \ln(3.0/2.0) = \frac{(4\pi \times 10^{-7})(40000)(0.015)}{2\pi} \ln(1.5) = (2 \times 10^{-7})(600) \ln(1.5) = 1.2 \times 10^{-4} \ln(1.5) \approx 1.2 \times 10^{-4}(0.405) \approx 5.1 \times 10^{-5} \text{ H}$. Choice B omits the logarithmic factor. Choice C uses an incorrect formula with linear dimensions. Choice D incorrectly uses $(b-a)$ instead of $\ln(b/a)$.
14. An RL circuit consists of a $25 \text{ mH}$ inductor and a $50 \text{ Ω}$ resistor connected in series with a battery. At $t = 0$, 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?
    Answer: $1 - e^{-2} \approx 0.86$ In an RL circuit, current grows as $I(t) = I_f(1 - e^{-t/\tau})$ where $\tau = L/R$ is the time constant and $I_f$ is the final current. The time constant is $\tau = (25 \times 10^{-3})/(50) = 5.0 \times 10^{-4} \text{ s}$. At $t = 2\tau$, the fraction is $(1 - e^{-2\tau/\tau}) = 1 - e^{-2} \approx 0.86$. Choice B gives the fraction at $t = \tau$, not $2\tau$. 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.
15. 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?
    Answer: $5.0 \times 10^{-2} \, \text{H}$ 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: $\mathcal{E} = -M \frac{dI}{dt}$, where $M$ is the mutual inductance, and $\frac{dI}{dt}$ is the rate of current change. First, calculate the rate of current change in coil A: $\frac{dI}{dt} = \frac{6.0 \, \text{A} - 2.0 \, \text{A}}{0.10 \, \text{s}} = \frac{4.0 \, \text{A}}{0.10 \, \text{s}} = 40 \, \text{A/s}$ Now solve for mutual inductance: $M = \frac{|\mathcal{E}|}{|\frac{dI}{dt}|} = \frac{0.20 \, \text{V}}{40 \, \text{A/s}} = 0.005 \, \text{H} = 5.0 \times 10^{-2} \, \text{H}$ Looking at the wrong answers: Choice A ($5.0 \times 10^{-3} \, \text{H}$) results from incorrectly using the total current change (4.0 A) instead of the rate of change (40 A/s). Choice B ($2.0 \times 10^{-2} \, \text{H}$) comes from using only the initial current (2.0 A) in place of the current change rate. Choice C ($4.0 \times 10^{-2} \, \text{H}$) results from using the current change (4.0 A) directly instead of dividing by time to get the rate. The answer is D: $5.0 \times 10^{-2} \, \text{H}$. Study tip: Always remember that Faraday's law involves the rate of change, not just the change itself. Calculate $\frac{dI}{dt}$ first, then solve for inductance.
16. 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? ($\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$)
    Answer: $0.13 \, \text{H}$ 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: $L = \frac{\mu_0 \mu_r N^2 h}{2\pi} \ln\left(\frac{r_2}{r_1}\right)$ where $N$ is the number of turns, $h$ is the height, $r_2$ is the outer radius, $r_1$ is the inner radius, and $\mu_r$ is the relative permeability. Substituting the given values: $L = \frac{(4\pi \times 10^{-7})(2000)(800)^2(0.02)}{2\pi} \ln\left(\frac{0.06}{0.04}\right)$ $L = \frac{(4\pi \times 10^{-7})(2000)(640,000)(0.02)}{2\pi} \ln(1.5)$ $L = (4 \times 10^{-7})(2000)(640,000)(0.02) \times 0.405$ $L = 0.13 \, \text{H}$ 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.

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