Inductance
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AP Physics C: Electricity and Magnetism › Inductance
An ideal air-core solenoid has $$N$$ turns, a length $${\ell}$$, and a cross-sectional area $$A$$. If the number of turns is tripled to $$3N$$ while the length and area remain constant, the new inductance will be...
one-ninth of the original inductance.
nine times the original inductance.
one-third of the original inductance.
three times the original inductance.
Explanation
The inductance of a long solenoid is given by $$L = \frac{\mu_0 N^2 A}{\ell}$$. Since the inductance $$L$$ is proportional to the square of the number of turns ($$N^2$$), tripling $$N$$ to $$3N$$ will change the inductance by a factor of $$(3)^2 = 9$$.
An ideal solenoid of length $${\ell}$$ and radius $$r$$ has $$N$$ turns of wire. Assuming the solenoid is long enough that end effects are negligible, its self-inductance is proportional to...
$$\frac{N r^2}{\ell}$$
$$\frac{N r}{\ell^2}$$
$$\frac{N^2 r}{\ell}$$
$$\frac{N^2 r^2}{\ell}$$
Explanation
The inductance of a long solenoid is $$L = \frac{\mu_0 N^2 A}{\ell}$$. The cross-sectional area is $$A = \pi r^2$$. Therefore, $$L = \frac{\mu_0 N^2 (\pi r^2)}{\ell}$$. Ignoring the constants $${\mu_0}$$ and $${\pi}$$, the inductance is proportional to $$\frac{N^2 r^2}{\ell}$$.
Which of the following statements best distinguishes the behavior of an ideal inductor from that of an ideal resistor in a DC circuit with a steady current?
The resistor causes a constant potential drop proportional to the current, while the inductor has zero potential drop.
The inductor dissipates no energy, while the resistor causes a potential drop and dissipates energy continuously.
Both dissipate energy as heat, but the inductor does so at a rate proportional to the current squared.
The inductor stores energy in its electric field, while the resistor stores energy in its magnetic field.
Explanation
In a DC circuit with a steady current ($$dI/dt = 0$$), an ideal inductor has no induced EMF ($$\mathcal{E} = -L(0) = 0$$) and behaves like a wire with zero resistance, thus dissipating no energy. An ideal resistor always has a potential drop $$V=IR$$ and dissipates power as heat ($$P=I^2R$$) as long as current flows. The correct answer reflects this difference in behavior under steady DC conditions.
An air-core solenoid has an inductance $$L_0$$. A rod of a paramagnetic material is then fully inserted into the solenoid's core. The inductance of the solenoid will...
remain unchanged, because paramagnetic materials do not affect the magnetic field or flux inside a solenoid.
decrease slightly, because the magnetic permeability of paramagnetic materials is slightly less than $${\mu_0}$$.
increase significantly, because paramagnetic materials have very high magnetic permeability.
increase slightly, because the magnetic permeability of paramagnetic materials is slightly greater than $${\mu_0}$$.
Explanation
The inductance of a solenoid is proportional to the magnetic permeability $${\mu}$$ of its core material ($$L = \frac{\mu N^2 A}{\ell}$$). Paramagnetic materials have a magnetic permeability $${\mu}$$ that is slightly greater than the permeability of free space, $${\mu_0}$$. Therefore, inserting a paramagnetic core will cause a slight increase in the inductance.
A steady current is flowing through a large inductor. The power supply is suddenly disconnected, and the inductor is simultaneously connected across a resistor. Which of the following occurs?
The inductor induces a current in the opposite direction to the original current, which then charges the resistor.
The magnetic field in the inductor reverses direction, causing an oscillating current to flow through the resistor.
The inductor induces a current in the same direction as the original current, which then decays through the resistor.
The current immediately drops to zero, and the stored magnetic energy is lost instantly.
Explanation
An inductor resists changes in current. When the power supply is removed, the current begins to decrease. To oppose this decrease, the inductor induces an EMF that drives a current in the same direction as the original current. This induced current flows through the resistor, and the energy stored in the inductor's magnetic field is dissipated as heat in the resistor, causing the current to decay over time.
Inductance is a property of an electrical conductor that measures its tendency to...
generate a constant magnetic field when a constant current is flowing through it.
resist the flow of a steady, direct current through it, converting electrical energy into thermal energy.
store electrical charge when a potential difference is applied across it, creating an electric field.
oppose a change in the electric current flowing through it by inducing an electromotive force.
Explanation
Inductance ($$L$$) is defined by the relationship where the induced electromotive force (EMF, $${\mathcal{E}}$$) is proportional to the rate of change of current ($$dI/dt$$), specifically $${\mathcal{E}} = -L(dI/dt)$$. This relationship shows that an inductor opposes changes in current.
The current $$I$$ in an inductor with inductance $$L = 5.0$$ mH is given by the function $$I(t) = 3.0t^2 - 4.0t$$, where $$I$$ is in amperes and $$t$$ is in seconds. What is the magnitude of the induced EMF in the inductor at time $$t=2.0$$ s?
30 mV
20 mV
40 mV
10 mV
Explanation
The induced EMF is given by $${\mathcal{E}} = -L \frac{dI}{dt}$$. First, find the derivative of the current: $$\frac{dI}{dt} = \frac{d}{dt}(3.0t^2 - 4.0t) = 6.0t - 4.0$$. At $$t=2.0$$ s, $$\frac{dI}{dt} = 6.0(2.0) - 4.0 = 12.0 - 4.0 = 8.0$$ A/s. The magnitude of the EMF is $$|{\mathcal{E}}| = |-(5.0 \times 10^{-3} \text{ H})(8.0 \text{ A/s})| = 40 \times 10^{-3} \text{ V} = 40$$ mV.
The energy stored in the magnetic field of an inductor is $$U_L$$. If the steady current flowing through the inductor is doubled, while its inductance remains constant, the new energy stored in the inductor will be...
$$4 U_L$$
$$2 U_L$$
$$\frac{1}{4} U_L$$
$$\frac{1}{2} U_L$$
Explanation
The energy stored in an inductor is given by the formula $$U_L = \frac{1}{2} L I^2$$. Since the energy $$U_L$$ is proportional to the square of the current ($$I^2$$), doubling the current from $$I$$ to $$2I$$ will change the stored energy by a factor of $$(2)^2 = 4$$.
An inductor with an inductance of 2.0 H carries a steady current of 3.0 A. How much energy is stored in the magnetic field of the inductor?
9.0 J
3.0 J
6.0 J
12.0 J
Explanation
The energy stored in an inductor is calculated using the formula $$U_L = \frac{1}{2} L I^2$$. Substituting the given values: $$U_L = \frac{1}{2} (2.0 \text{ H}) (3.0 \text{ A})^2 = \frac{1}{2} (2.0)(9.0) = 9.0$$ J.
The SI unit of inductance is the henry (H). Which of the following combinations of units is equivalent to the henry?
Volt-second per ampere ($$\text{V} \cdot \text{s/A}$$)
Volt per ampere ($$\text{V/A}$$)
Volt-ampere per second ($$\text{V} \cdot \text{A/s}$$)
Ampere-second per volt ($$\text{A} \cdot \text{s/V}$$)
Explanation
From the formula for induced EMF, $${\mathcal{E}} = -L \frac{dI}{dt}$$, we can solve for the units of inductance $$L$$. The units are $$[L] = \frac{[\mathcal{E}]}{[dI/dt]} = \frac{\text{Volts}}{\text{Amperes/Second}} = \frac{\text{V} \cdot \text{s}}{\text{A}}$$.