Question 1

AC generator: Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm², how many turns do you need?

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Question 2

AC generator: A rectangular coil having N turns and measuring 15 cm by 25 cm is rotating in a uniform 1.6-T magnetic field with a frequency of 75 Hz. The rotation axis is perpendicular to the direction of the field. If the coil develops a sinusoidal emf of maximum value 56.9 V, what is the value of N?

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Question 3

LR circuits: A series circuit consists of a 0.55-H inductor with internal resistance of 8.0 Ω connected in series with a 4.0-Ω resistor, an open switch, and an ideal 12-V battery.
(a) When the switch is closed, what is the initial current through the 4.0-Ω resistor?
(b) What is the current through the 4.0-Ω resistor a very long time after the switch is closed?

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Question 4

Self-inductance: What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10^-4 m²? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The formula for self-inductance of a solenoid is L = μN²Aℓ, where μ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and ℓ is the length of the solenoid. Plugging in the given values, we get:
L = (4π × 10^-7 T ∙ m/A) × (100 turns)^2 × (1.00 × 10^-4 m^2) × (0.300 m)
L = 4.19 × 10^-6 H = 4.19 µH

Therefore, the answer is C) 4.19 µH.

Question 5

LR circuits: A series LR circuit contains an emf source of  having no internal resistance, a resistor, a  inductor having no appreciable resistance, and a switch. If the emf across the inductor is  of its maximum value  after the switch is closed, what is the resistance of the resistor?

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Question 6

LR circuits: What resistance should be added in series with a 3.0-H inductor to complete an LR circuit with a time constant of  ?

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Question 7

AC generator: The coil in a 50-Hz ac generator has 283 turns, each having an area of 3.0 × 10⁻² m² and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?

Accepted Answer

The peak output voltage of an AC generator is given by:V = N * A * B * w * sin(wt)where:N is the number of turns in the coilA is the area of each turnB is the magnetic field strengthw is the angular frequency (2 * pi * f)t is the timeIn this problem, we have:N = 283A = 3.0 × 10^-2 m^2B = 0.12 Tf = 50 Hzw = 2 * pi * f = 314 rad/sPlugging these values into the equation, we get:V = 283 * 3.0 × 10^-2 m^2 * 0.12 T * 314 rad/s * sin(314t)The peak output voltage is the maximum value of V, which occurs when sin(314t) = 1. Therefore, the peak output voltage is:V = 283 * 3.0 × 10^-2 m^2 * 0.12 T * 314 rad/s = 120 VTherefore, the correct answer is A) 120 V.

Question 8

Energy in an inductor: At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is increasing at a rate of 3.0 J/s, how fast is the current changing?

Accepted Answer

The energy stored in an inductor is given by $ E = \frac{1}{2} L I^2 $. Differentiating with respect to time, $ \frac{dE}{dt} = L I \frac{dI}{dt} $. Given $ \frac{dE}{dt} = 3.0 \, \text{J/s} $, $ L = 5.0 \, \text{H} $, and $ I = 3.0 \, \text{A} $, we solve for $ \frac{dI}{dt} $: $ 3.0 = 5.0 \times 3.0 \times \frac{dI}{dt} $, leading to $ \frac{dI}{dt} = 0.20 \, \text{A/s} $.

Question 9

AC generator: You are designing a generator to have a maximum emf of 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030 m², what should be the frequency of the generator in a uniform magnetic field of 0.030 T?

Accepted Answer

The max emf induced in the coil is given by:
$E_{max} = NAB\omega$
where N is the number of turns, A is the cross-sectional area of the coil, B is the magnetic field strength, and $\omega$ is the angular frequency.

Rearranging, we get:
$\omega = \frac{E_{max}}{NAB}$

Substituting the given values, we get:
$\omega = \frac{8.0	ext{ V}}{(200)(0.030	ext{ m}^2)(0.030	ext{ T})} \approx 7.1	ext{ Hz}$

Therefore, the answer is A) 7.1 Hz.

Question 10

Mutual inductance: The mutual inductance between two coils is 10.0 mH. The current in the first coil changes uniformly from 2.70 A to 5.00 A in 0.160 s. If the second coil has a resistance of 0.600 Ω, what is the magnitude of the induced current in the second coil?

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Question 11

LR circuits: A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Accepted Answer

The current in an LR circuit is given by:$$i(t) = \frac{V}{R} (1 - e^{-t/	au})$$where:* V is the voltage of the power supply* R is the resistance of the resistor* L is the inductance of the inductor* $	au$ is the time constant of the circuit, which is equal to L/RIn this problem, we have:* V = 15 V* R = 60 Ω* L = 45 mH = 0.045 H* t = 7.0 ms = 0.007 sSo, the time constant is:$$	au = \frac{L}{R} = \frac{0.045 H}{60 \Omega} = 0.00075 s$$And the current at t = 7.0 ms is:$$i(0.007 s) = \frac{15 V}{60 \Omega} (1 - e^{-0.007 s/0.00075 s}) = 0.25 A = 250 mA$$Therefore, the correct answer is A.

Question 12

Energy in an inductor: An insulated wire of diameter 1.0 mm and negligible resistance is wrapped tightly around a cylindrical core of radius 5.0 cm and length 30 cm to build a solenoid. What is the energy stored in this solenoid when a current I = 0.20 A flows through it? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Question 13

Self-inductance: At what rate would the current in a 100-mH inductor have to change to induce an emf of 1000 V in the inductor?

Accepted Answer

The formula for emf induced in an inductor is given by 
emf = - L(dI/dt)
where L is the inductance, I is the current and (dI/dt) is the rate of change of current. 
Rearranging the formula, we get 
(dI/dt) = - emf/L = -1000/0.1 = -10000 A/s
Since the question asks for the magnitude of the rate of change of current, we take the absolute value which is 10000 A/s. So the answer is D.

Question 14

Energy density: What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 A? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

The energy density ($u$) in a magnetic field can be calculated using the formula $u = \frac{B^2}{2\mu_0}$, where $B$ is the magnetic field strength and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}$). The magnetic field ($B$) around a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$, where $I$ is the current and $r$ is the distance from the wire. Substituting $I = 12 \, \text{A}$ and $r = 0.25 \, \text{m}$, and then calculating $u$, results in an energy density of $3.7 \times 10^{-5} \, \text{J/m}^3$, which corresponds to choice B.

Question 15

Energy in an inductor: A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (μ₀ = 4π × 10^-7 T ∙ m/A)

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Question 16

Energy density: How much energy is stored in a room 3.0 m by 4.0 m by 2.4 m due to the earth's magnetic field with a strength of 5.0 × 10^-5 T? (μ₀ = 4π × 10^-7 T ∙ m/A)

Accepted Answer

First, calculate the magnetic field energy density using:
$u = \frac{B^2}{2\mu_0}$ = $\frac{(5.0 	imes 10^{-5}	ext{ T})^2}{2(4\pi 	imes 10^{-7}	ext{ T} \cdot 	ext{m/A})}$ = $1.57 	imes 10^{-2} 	ext{ J/m}^3$
Then, calculate the volume of the room: 3.0 m $	imes$ 4.0 m $	imes$ 2.4 m = 28.8 m$^3$
Finally, calculate the total energy stored:
$E = uV = (1.57 	imes 10^{-2}	ext{ J/m}^3)(28.8	ext{ m}^3) = 0.450 mJ \approx 29	ext{ mJ}$

Question 17

AC generator: Wire is wound on a square frame, 30 cm by 30 cm, to form a coil of 7 turns. The frame is mounted on a horizontal shaft through its center (perpendicular to the plane of the diagram), as shown in the figure. The coil is in clockwise rotation, with a period of 0.060 s. A uniform, horizontal, magnetic field of magnitude 0.40 T is present. At a given instant, the plane of the coil forms a 60° angle with the horizontal, as shown. At that instant, what is the magnitude of the emf induced in the coil?

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Question 18

LR circuits: A series LR circuit consists of a 2.0-H inductor with negligible internal resistance, a 100-ohm resistor, an open switch, and a 9.0-V ideal power source. After the switch is closed, what is the maximum power delivered by the power supply?

Accepted Answer

The maximum power delivered by the power supply occurs when the circuit reaches its steady state, where the current through the circuit is constant. In a series LR circuit, the current reaches its maximum value (I_max) when the inductor has reached its maximum current (I_Lmax), which occurs when the voltage across the resistor is equal to the voltage across the inductor (V_R = V_L).

We can use Ohm's law to find I_max: 
V = IR + VL 
9.0 V = I(100 Ω) + (2.0 H)(dI/dt)

At the maximum current, dI/dt = 0, so we have: 
I_max = V/R = 9.0 V/100 Ω = 0.090 A

The maximum power delivered by the power supply is given by: 
P_max = V^2/R = (9.0 V)^2/100 Ω = 0.81 W

Therefore, the correct answer is (E) 0.81 W.

Question 19

Self-inductance: In the figure, the current in a solenoid having no appreciable resistance is flowing from b to a and is decreasing at a rate of 9.6 A/s. The self-induced emf in the solenoid is found to be 8.4 V.  (a) What is the self-inductance of the solenoid?
(b) Which point, a or b is at higher potential?

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Question 20

Mutual inductance: A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoid at its center. The coil's resistance is 9.9 ohms. The mutual inductance of the coil and solenoid is  At a given instant, the current in the solenoid is  and is decreasing at the rate of  At the given instant, what is the magnitude of the induced current in the coil?

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Quiz 24: Capacitance and Dielectrics

AC generator: Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need?

Self-inductance: What is the self-inductance of a solenoid 30.0 cm long having 100 turns of wire and a cross-sectional area of 1.00 × 10-4 m2? (μ0 = 4π × 10-7 T ∙ m/A)

LR circuits: A series LR circuit contains an emf source of having no internal resistance, a resistor, a inductor having no appreciable resistance, and a switch. If the emf across the inductor is of its maximum value after the switch is closed, what is the resistance of the resistor?

LR circuits: What resistance should be added in series with a 3.0-H inductor to complete an LR circuit with a time constant of ?

AC generator: The coil in a 50-Hz ac generator has 283 turns, each having an area of 3.0 × 10−2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?