Inductance - AP Physics C Electricity and Magnetism

A 500-turn inductor has an inductance of 25 mH. If a steady current of 4.0 A flows through it, what is the magnetic flux through a single turn of the inductor?
Answer: $2.0 \times 10^{-4}$ Wb Inductance is related to the total flux linkage ( $N\Phi_B$ ) by $L = \frac{N\Phi_B}{I}$ , where $\Phi_B$ is the flux through a single turn. Solving for $\Phi_B$ gives $\Phi_B = \frac{LI}{N}$ . Plugging in the values: $\Phi_B = \frac{(25 \times 10^{-3} \text{ H})(4.0 \text{ A})}{500} = \frac{0.10 \text{ H} \cdot \text{A}}{500} = 2.0 \times 10^{-4}$ Wb.
A long solenoid with $N$ turns, length ${\ell}$ , and cross-sectional area $A$ has an inductance $L$ . What would be the inductance of a solenoid with $N$ turns, length $2{\ell}$ , and cross-sectional area $A/2$ ?
Answer: $L/4$ The inductance of a long solenoid is given by $L = \frac{\mu_0 N^2 A}{\ell}$ . The new inductance, $L'$ will be $L' = \frac{\mu_0 N^2 (A/2)}{2\ell} = \frac{1}{4} \left( \frac{\mu_0 N^2 A}{\ell} \right) = \frac{L}{4}$ .
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?
Answer: The inductor induces a current in the same direction as the original current, which then decays through the resistor. 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.
The SI unit of inductance is the henry (H). Which of the following combinations of units is equivalent to the henry?
Answer: Volt-second per ampere ( $\text{V} \cdot \text{s/A}$ ) From the formula for induced EMF, ${\mathcal{E}} = -L \frac{dI}{dt}$ , we can solve for the units of inductance $L$ . The units are $[L] = \frac{[\mathcal{E}]}{[dI/dt]} = \frac{\text{Volts}}{\text{Amperes/Second}} = \frac{\text{V} \cdot \text{s}}{\text{A}}$ .
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?
Answer: 40 mV 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.
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...
Answer: $\frac{N^2 r^2}{\ell}$ 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 $\frac{N^2 r^2}{\ell}$ .
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...
Answer: increase slightly, because the magnetic permeability of paramagnetic materials is slightly greater than ${\mu_0}$ . 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.
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...
Answer: nine times the original inductance. 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 $(3)^2 = 9$ .
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...
Answer: $4 U_L$ 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?
Answer: 9.0 J 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 inductance of a long, air-filled solenoid is $L_0$ . If its length and total number of turns are both doubled, while keeping its cross-sectional area constant, what is its new inductance $L'$ ?
Answer: $2L_0$ The inductance of a long solenoid is given by the formula $L = \frac{\mu_0 N^2 A}{\ell}$ . The new parameters are $N' = 2N$ and ${\ell}' = 2{\ell}$ . The new inductance is $L' = \frac{\mu_0 (N')^2 A}{{\ell}'} = \frac{\mu_0 (2N)^2 A}{2{\ell}} = \frac{\mu_0 4N^2 A}{2{\ell}} = 2 \left( \frac{\mu_0 N^2 A}{\ell} \right) = 2L_0$ .
An ideal solenoid contains a uniform magnetic field $B$ when it carries a current $I$ . The energy density (energy per unit volume) of the magnetic field inside the solenoid is proportional to which of the following quantities?
Answer: $B^2$ The energy stored in an inductor is $U_L = \frac{1}{2}LI^2$ . For a solenoid, $L=\frac{\mu_0N^2A}{\ell}$ and $B=\mu_0nI=\mu_0\frac{N}{\ell}I$ , so $I=\frac{B\ell}{\mu_0N}$ . Substituting these into the energy equation gives $U_L=\frac{1}{2} \left( \frac{\mu_0N^2A}{\ell} \right) \left( \frac{B\ell}{\mu_0N} \right)^2 = \frac{1}{2\mu_0}B^2(A\ell)$ . The volume of the solenoid is $V=A\ell$ . The energy density is $u_B = U_L/V = \frac{B^2}{2\mu_0}$ . Thus, the energy density is proportional to $B^2$ .
A constant potential difference of 12 V is maintained across a 4.0 mH inductor. What must be the rate of change of current through the inductor?
Answer: 3000 A/s A potential difference across an ideal inductor is due to a changing current. The relationship is $|\mathcal{E}| = L |\frac{dI}{dt}|$ (ignoring internal resistance). To maintain a constant potential difference, the current must be changing at a constant rate. Rearranging for the rate of change: $|\frac{dI}{dt}| = \frac{|\mathcal{E}|}{L} = \frac{12 \text{ V}}{4.0 \times 10^{-3} \text{ H}} = 3000$ A/s.
What is the ratio of the inductance of solenoid B to the inductance of solenoid A, $L_B / L_A$ ?
Answer: $1$ The inductance of a solenoid is given by $L = \frac{\mu_0 N^2 A}{\ell}$ . For solenoid A, $L_A = \frac{\mu_0 N^2 A}{\ell}$ . For solenoid B, $L_B = \frac{\mu_0 (2N)^2 A}{4\ell} = \frac{\mu_0 4N^2 A}{4\ell} = \frac{\mu_0 N^2 A}{\ell}$ . Therefore, $L_B = L_A$ , and the ratio $L_B / L_A = 1$ .
Immediately after a switch is closed to connect an unpowered series RL circuit to a DC battery, the inductor behaves most like...
Answer: an open circuit with infinite resistance, allowing essentially zero initial current to flow. At the instant the switch is closed ( $t=0^+$ , the current begins to change from zero. The inductor opposes this change by inducing a back EMF equal in magnitude to the battery's EMF. This opposition is so strong initially that it prevents any significant current from flowing, making the inductor behave like an open circuit or a break in the wire.
Inductance is a property of an electrical conductor that measures its tendency to...
Answer: oppose a change in the electric current flowing through it by inducing an electromotive force. 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.
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?
Answer: The inductor dissipates no energy, while the resistor causes a potential drop and dissipates energy continuously. 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.
A circuit contains an ideal inductor and a switch connected to a battery. When the switch is suddenly closed, the inductor generates a 'back EMF'. This induced EMF is a direct consequence of...
Answer: the magnetic flux created by the rising current, which itself induces an EMF that opposes this change in flux. According to Faraday's law of induction and Lenz's law, as the current begins to flow and increase, it creates a changing magnetic flux through the inductor's coils. This changing flux induces an EMF that opposes the change, which in this case is the increase in current. This opposition is the source of the back EMF.

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