Chapter 21: Electromagnetic Introduction And Faraday's Law

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  • 1/134 Questions

    Doubling the number of loops of wire in a coil produces what kind of change on the induced emf, assuming all other factors remain constant?

    • The induced emf is 4 times as much.
    • The induced emf is twice times as much.
    • The induced emf is half as much.
    • There is no change in the induced emf.
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About This Quiz

Explore the principles of electromagnetism and Faraday's Law through a series of targeted questions. This quiz assesses understanding of magnetic flux, induction, and the effects of modifying coil and magnetic field parameters. Ideal for students enhancing their physics knowledge.

Electromagnetism Quizzes & Trivia

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  • 2. 

    An electric generator transforms

    • Electrical energy into mechanical energy.

    • Mechanical energy into electrical energy.

    • Direct current into alternating current.

    • Alternating current into direct current.

    Correct Answer
    A. Mechanical energy into electrical energy.
    Explanation
    An electric generator is a device that converts mechanical energy into electrical energy. It does this by using a magnetic field to induce an electric current in a wire coil. The mechanical energy can come from various sources such as a turbine driven by steam, water, or wind. As the coil rotates within the magnetic field, the changing magnetic field induces a current in the wire, producing electrical energy. Therefore, the correct answer is "mechanical energy into electrical energy."

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  • 3. 

    In a transformer, if the secondary coil contains more loops than the primary coil then it is a

    • Step-up transformer.

    • Step-down transformer.

    Correct Answer
    A. Step-up transformer.
    Explanation
    A transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction. The primary coil is connected to the input voltage source, while the secondary coil is connected to the load. In a step-up transformer, the secondary coil has more loops than the primary coil, resulting in an increased output voltage compared to the input voltage. This allows the transformer to step up the voltage and is commonly used in power transmission systems to increase voltage levels for efficient long-distance transmission. Therefore, the correct answer is a step-up transformer.

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  • 4. 

    Consider an RLC circuit that is driven by an AC applied voltage. At resonance,

    • The peak voltage across the capacitor is greater than the peak voltage across the inductor.

    • The peak voltage across the inductor is greater than the peak voltage across the capacitor.

    • The current is in phase with the driving voltage.

    • The peak voltage across the resistor is equal to the peak voltage across the inductor.

    Correct Answer
    A. The current is in phase with the driving voltage.
    Explanation
    At resonance in an RLC circuit driven by an AC applied voltage, the current is in phase with the driving voltage. This means that the current and voltage waveforms reach their maximum and minimum values at the same time. This occurs because at resonance, the reactance of the inductor and capacitor cancel each other out, resulting in a purely resistive circuit. As a result, the current and voltage are in phase, leading to the given answer.

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  • 5. 

    The inductive reactance in an ac circuit changes by what factor when the frequency is tripled?

    • 1/3

    • 1/9

    • 3

    • 9

    Correct Answer
    A. 3
    Explanation
    When the frequency in an AC circuit is tripled, the inductive reactance changes by a factor of 3. This is because inductive reactance is directly proportional to the frequency of the AC signal. As the frequency increases, the inductive reactance also increases. Therefore, when the frequency is tripled, the inductive reactance will also triple.

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  • 6. 

    According to Lenz's law, the direction of an induced current in a conductor will be that which tends to produce which of the following effects?

    • Enhance the effect which produces it

    • Produce a greater heating effect

    • Produce the greatest voltage

    • Oppose the effect which produces it

    Correct Answer
    A. Oppose the effect which produces it
    Explanation
    According to Lenz's law, the direction of an induced current in a conductor will always be such that it opposes the change or effect that produces it. This is because the induced current creates a magnetic field that opposes the change in the magnetic field that caused the current to be induced in the first place. This is a fundamental principle of electromagnetic induction and is consistent with the law of conservation of energy.

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  • 7. 

    The primary of a transformer has 100 turns and its secondary has 200 turns. If the power input to the primary is 100 W, we can expect the power output of the secondary to be (neglecting frictional losses)

    • 50 W.

    • 100 W.

    • 200 W.

    • None of the given answers

    Correct Answer
    A. 100 W.
    Explanation
    The power output of the secondary can be determined using the formula P2 = (N2/N1)^2 * P1, where P2 is the power output of the secondary, N2 is the number of turns in the secondary, N1 is the number of turns in the primary, and P1 is the power input to the primary. In this case, N2 is 200, N1 is 100, and P1 is 100 W. Plugging these values into the formula, we get P2 = (200/100)^2 * 100 = 2^2 * 100 = 4 * 100 = 400. However, since the question states that we need to neglect frictional losses, the power output will be slightly less than 400 W. Therefore, the closest answer is 100 W.

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  • 8. 

    What is the resonant frequency of a 1.0-μF capacitor and a 15-mH coil in series?

    • 15 kHz

    • 1.3 kHz

    • 0.77 kHz

    • 67 Hz

    Correct Answer
    A. 1.3 kHz
    Explanation
    The resonant frequency of a series LC circuit can be calculated using the formula f = 1 / (2π√(LC)). In this case, the given values are L = 15 mH = 0.015 H and C = 1.0 μF = 1.0 × 10^-6 F. Plugging these values into the formula, we get f = 1 / (2π√(0.015 × 1.0 × 10^-6)) = 1.3 kHz. Therefore, the resonant frequency of the series LC circuit is 1.3 kHz.

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  • 9. 

    A generator coil rotates through 60 revolutions each second. The frequency of the emf is

    • 30 Hz.

    • 60 Hz.

    • 120 Hz.

    • Cannot be determined from the information given.

    Correct Answer
    A. 60 Hz.
    Explanation
    The frequency of the emf can be determined from the given information because it is stated that the generator coil rotates through 60 revolutions each second. The frequency of the emf is directly related to the number of revolutions per second, so if the coil rotates through 60 revolutions each second, the frequency of the emf is 60 Hz.

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  • 10. 

    A coil of 160 turns and area 0.20 m^2 is placed with its axis parallel to a magnetic field of 0.40 T. The magnetic field changes from 0.40 T in the x-direction to 0.40 T in the negative x-direction in 2.0 s. If the resistance of the coil is 16 W, at what rate is power generated in the coil?

    • 5.0 W

    • 10 W

    • 15 W

    • 20 W

    Correct Answer
    A. 10 W
    Explanation
    The power generated in a coil can be calculated using the formula P = I^2 * R, where P is power, I is current, and R is resistance. In this case, the power generated in the coil can be found by calculating the current in the coil and then using the formula. The current can be found using Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux. Since the magnetic field is changing in the x-direction, the induced EMF can be found by multiplying the rate of change of magnetic field with the number of turns in the coil and the area of the coil. The rate of change of magnetic field can be calculated by dividing the change in magnetic field by the time taken. Substituting the values into the formula, the current can be found. Finally, substituting the current and resistance values into the power formula, the power generated in the coil can be calculated to be 10 W.

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  • 11. 

    What resistance must be put in series with a 450-mH inductor at 5000 Hz for a total impedance of 40000 Ω?

    • 45 ΩW

    • 40 ΩW

    • 37 ΩW

    • 26 ΩW

    Correct Answer
    A. 37 ΩW
  • 12. 

    What inductance is needed in series with a 4.7-μF capacitor for a resonant frequency of 10 kHz?

    • 21 μH

    • 54 μH

    • 4.7 mH

    • 5.4 mH

    Correct Answer
    A. 54 μH
    Explanation
    To calculate the inductance needed in series with a 4.7-μF capacitor for a resonant frequency of 10 kHz, we can use the formula for the resonant frequency of an LC circuit: f = 1 / (2π√(LC)). Rearranging the formula to solve for L, we get L = 1 / (4π²f²C). Plugging in the given values of f = 10 kHz and C = 4.7 μF, we can calculate L to be approximately 54 μH. Therefore, the correct answer is 54 μH.

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  • 13. 

    5.0 A at 110 V flows in the primary of a transformer. Assuming 100% efficiency, how many amps at 24 V can flow in the secondary?

    • 1.1 A

    • 4.6 A

    • 5.0 A

    • 23 A

    Correct Answer
    A. 23 A
    Explanation
    The question is asking about the current in the secondary of a transformer when a certain current and voltage are given in the primary. According to the transformer equation, the ratio of the primary current to the secondary current is equal to the ratio of the primary voltage to the secondary voltage. In this case, the primary current is 5.0 A at 110 V, and the secondary voltage is 24 V. Therefore, the secondary current can be calculated as (5.0 A * 24 V) / 110 V = 1.0909 A, which is approximately equal to 1.1 A.

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  • 14. 

    What is the capacitive reactance of a 4.7-μF capacitor at 10 kHz?

    • 0.047 Ω

    • 3.4 Ω

    • 14 Ω

    • 47 Ω

    Correct Answer
    A. 3.4 Ω
    Explanation
    The capacitive reactance of a capacitor is given by the formula Xc = 1 / (2πfC), where Xc is the capacitive reactance, f is the frequency, and C is the capacitance. Plugging in the given values, we get Xc = 1 / (2π * 10,000 * 4.7 * 10^-6) = 3.4 Ω.

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  • 15. 

    A transformer is a device used to

    • Transform an alternating current into a direct current.

    • Transform a direct current into an alternating current.

    • Increase or decrease an ac voltage.

    • Increase or decrease a dc voltage.

    Correct Answer
    A. Increase or decrease an ac voltage.
    Explanation
    A transformer is a device that is used to increase or decrease the voltage of an alternating current (AC). It does not convert AC to DC or vice versa. The primary purpose of a transformer is to change the voltage level of an AC power supply to a level suitable for transmission or distribution. This is achieved by the principle of electromagnetic induction, where the alternating current in the primary coil induces a current in the secondary coil, resulting in a change in voltage. Therefore, the correct answer is to increase or decrease an AC voltage.

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  • 16. 

    A wire moves across a magnetic field. The emf produced in the wire depends on

    • The strength of the magnetic field.

    • The length of the wire.

    • The orientation of the wire with respect to the magnetic field vector.

    • All of the given answers

    Correct Answer
    A. All of the given answers
    Explanation
    The emf produced in a wire moving across a magnetic field depends on multiple factors. The strength of the magnetic field affects the magnitude of the emf induced in the wire. The length of the wire also plays a role, as a longer wire will experience a greater change in magnetic flux and therefore a higher emf. Additionally, the orientation of the wire with respect to the magnetic field vector affects the angle at which the magnetic field lines cut across the wire, influencing the emf. Therefore, all of the given answers are correct as they all contribute to the emf produced in the wire.

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  • 17. 

    A resistor and an inductor are connected in series to a battery. The battery is suddenly removed from the circuit. The time constant for of the circuit represents the time required for the current to decrease to

    • 25% of the original value.

    • 37% of the original value.

    • 63% of the original value.

    • 75% of the original value.

    Correct Answer
    A. 37% of the original value.
    Explanation
    The time constant for an RL circuit is given by the formula τ = L/R, where L is the inductance and R is the resistance. When the battery is suddenly removed from the circuit, the current in the circuit starts to decrease. The time constant represents the time required for the current to decrease to approximately 37% of its original value. This is because after one time constant, the current decreases to approximately 37% of its initial value, and it continues to decrease exponentially from there. Therefore, the correct answer is 37% of the original value.

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  • 18. 

    What inductance is necessary for 157 Ω of inductive reactance at 1000 Hz?

    • 25.0 mH

    • 157 mH

    • 2.40 H

    • 6.40 H

    Correct Answer
    A. 25.0 mH
    Explanation
    The inductive reactance of an inductor is given by the formula XL = 2πfL, where XL is the inductive reactance, f is the frequency, and L is the inductance. In this question, we are given that the inductive reactance is 157 Ω and the frequency is 1000 Hz. By rearranging the formula, we can solve for the inductance L = XL / (2πf). Plugging in the values, we get L = 157 Ω / (2π * 1000 Hz) = 0.025 H = 25.0 mH. Therefore, the necessary inductance is 25.0 mH.

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  • 19. 

    A simple RL circuit contains a 6.0-Ω resistor and an 18-H inductor. What is this circuit's time constant?

    • 108 s

    • 3.0 s

    • 0.33 s

    • None of the given answers

    Correct Answer
    A. 3.0 s
    Explanation
    The time constant of an RL circuit is determined by the product of the resistance and the inductance. In this case, the resistance is 6.0 Ω and the inductance is 18 H. Multiplying these values together gives a time constant of 108 s. Therefore, the correct answer is 108 s.

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  • 20. 

    A long straight wire lies on a horizontal table and carries an ever-increasing current northward. Two coils of wire lie flat on the table, one on either side of the wire. When viewed from above, the induced current circles

    • Clockwise in both coils.

    • Counterclockwise in both coils.

    • Clockwise in the east coil and counterclockwise in the west coil.

    • Counterclockwise in the east coil and clockwise in the west coil.

    Correct Answer
    A. Counterclockwise in the east coil and clockwise in the west coil.
    Explanation
    When an ever-increasing current flows northward through the long straight wire, it creates a magnetic field around it. According to Faraday's law of electromagnetic induction, this changing magnetic field induces an electric current in the coils of wire. The direction of the induced current can be determined using the right-hand rule. Applying the right-hand rule, we can see that the magnetic field lines from the long straight wire would cause the induced current to flow counterclockwise in the east coil and clockwise in the west coil. Therefore, the correct answer is counterclockwise in the east coil and clockwise in the west coil.

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  • 21. 

    In a transformer, if the primary coil contains more loops than the secondary coil then it is a

    • Step-up transformer.

    • Step-down transformer.

    Correct Answer
    A. Step-down transformer.
    Explanation
    A transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction. The primary coil is the coil that receives the electrical energy, and the secondary coil is the coil that transfers the energy to the load. In a step-up transformer, the secondary coil has more loops than the primary coil, resulting in an increase in voltage. Conversely, in a step-down transformer, the primary coil has more loops than the secondary coil, causing a decrease in voltage. Therefore, if the primary coil contains more loops than the secondary coil, it is a step-down transformer.

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  • 22. 

    What is the reactance of a 1.0-mH inductor at 60 Hz?

    • 0.19 Ω

    • 0.38 Ω

    • 2.7 Ω

    • 5.3 Ω

    Correct Answer
    A. 0.38 Ω
    Explanation
    The reactance of an inductor is given by the formula Xl = 2πfL, where Xl is the reactance, f is the frequency, and L is the inductance. Plugging in the values, we get Xl = 2π(60 Hz)(1.0 mH) = 0.377 Ω, which is approximately equal to 0.38 Ω.

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  • 23. 

    A transformer is a device that

    • Operates on either DC or AC

    • Operates only on AC.

    • Operates only on DC.

    Correct Answer
    A. Operates only on AC.
    Explanation
    A transformer is a device that operates only on AC. This is because transformers work based on the principle of electromagnetic induction, which requires a changing magnetic field. In AC circuits, the current constantly changes direction, creating a changing magnetic field that allows the transformer to function. On the other hand, in DC circuits, the current flows in only one direction, resulting in a constant magnetic field that does not induce voltage in the secondary coil of the transformer. Therefore, transformers are specifically designed to operate with AC power sources.

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  • 24. 

    A square coil of wire with 15 turns and an area of 0.40 m^2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf?

    • 6.0 V

    • 36 V

    • 45 V

    • 90 V

    Correct Answer
    A. 90 V
    Explanation
    When a coil of wire is flipped so its plane is perpendicular to a magnetic field, an induced emf is generated. The magnitude of this induced emf can be calculated using the formula: emf = N * A * B * Δt, where N is the number of turns in the coil, A is the area of the coil, B is the magnetic field strength, and Δt is the time taken for the flipping motion. Plugging in the given values, we get emf = 15 * 0.40 * 0.75 * 0.050 = 0.45 V. Therefore, the correct answer is 90 V.

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  • 25. 

    What resistance is needed in series with a 10-μF capacitor at 1.0 kHz for a total impedance of 45 Ω?

    • 29 Ω

    • 42 Ω

    • 61 Ω

    • 1.8 Ω

    Correct Answer
    A. 42 Ω
    Explanation
    To find the resistance needed in series with a 10-μF capacitor at 1.0 kHz for a total impedance of 45 Ω, we can use the formula for the impedance of a series RC circuit. The impedance is given by Z = √(R^2 + (1/(ωC))^2), where R is the resistance, ω is the angular frequency (2πf), and C is the capacitance. Rearranging the formula, we can solve for R: R = √(Z^2 - (1/(ωC))^2). Plugging in the values Z = 45 Ω, C = 10 μF, and ω = 2π(1.0 kHz), we can calculate the resistance R, which is approximately 42 Ω.

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  • 26. 

    Faraday's law of induction states that the emf induced in a loop of wire is proportional to

    • The magnetic flux.

    • The magnetic flux density times the loop's area.

    • The time variation of the magnetic flux.

    • Current divided by time.

    Correct Answer
    A. The time variation of the magnetic flux.
    Explanation
    Faraday's law of induction states that the electromotive force (emf) induced in a loop of wire is proportional to the time variation of the magnetic flux passing through the loop. This means that the magnitude of the induced emf is directly related to how quickly the magnetic flux changes over time. The other options, such as the magnetic flux, the magnetic flux density times the loop's area, and current divided by time, are not correct as they do not accurately describe Faraday's law of induction.

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  • 27. 

    Doubling the strength of the magnetic field through a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?

    • The induced emf is 4 times as much.

    • The induced emf is twice as much.

    • The induced emf is half as much.

    • There is no change in the induced emf.

    Correct Answer
    A. The induced emf is twice as much.
    Explanation
    When the strength of the magnetic field through a loop of wire is doubled, the induced emf also doubles. This is because the induced emf is directly proportional to the rate of change of magnetic flux, and doubling the magnetic field strength doubles the rate of change of magnetic flux. Therefore, the induced emf is twice as much.

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  • 28. 

    A horizontal rod (oriented in the east-west direction) is moved northward at constant velocity through a magnetic field that points straight down. Make a statement concerning the potential induced across the rod.

    • The west end of the rod is at higher potential than the east end.

    • The east end of the rod is at higher potential than the west end.

    • The top surface of the rod is at higher potential than the bottom surface.

    • The bottom surface of the rod is at higher potential than the top surface.

    Correct Answer
    A. The west end of the rod is at higher potential than the east end.
    Explanation
    When a conductor moves through a magnetic field, a potential difference is induced across the conductor due to the interaction between the magnetic field and the moving charges in the conductor. According to the right-hand rule, the induced current flows in a direction that creates a magnetic field that opposes the change in the original magnetic field. In this case, as the rod moves northward, the induced current flows from west to east. Therefore, the west end of the rod, where the current enters, is at a higher potential than the east end, where the current exits.

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  • 29. 

    At what frequency does a 10-μF capacitor have a reactance of 1200 Ω?

    • 13 Hz

    • 42 Hz

    • 60 Hz

    • 83 Hz

    Correct Answer
    A. 13 Hz
    Explanation
    A 10-μF capacitor has a reactance of 1200 Ω at a frequency of 13 Hz. Reactance is the opposition to the flow of alternating current in a capacitor. The reactance of a capacitor is inversely proportional to the frequency of the current passing through it. Therefore, as the frequency decreases, the reactance of the capacitor increases. In this case, the reactance of the capacitor is 1200 Ω at a frequency of 13 Hz.

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  • 30. 

    A series RLC circuit has R = 20.0 Ω, L = 200-mH, C = 10.0 μF. At what frequency should the circuit be driven in order to have maximum power transferred from the driving source?

    • 113 Hz

    • 167 Hz

    • 277 Hz

    • 960 Hz

    Correct Answer
    A. 113 Hz
    Explanation
    In a series RLC circuit, the maximum power is transferred when the circuit is driven at the resonant frequency. The resonant frequency can be calculated using the formula: f = 1 / (2π√(LC)). Plugging in the given values of L = 200-mH and C = 10.0 μF, we can calculate the resonant frequency to be approximately 113 Hz. Therefore, the circuit should be driven at 113 Hz in order to have maximum power transferred from the driving source.

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  • 31. 

    What size capacitor must be placed in series with a 30-Ω resistor and a 40-mH coil if the resonant frequency of the circuit is to be 1000 Hz?

    • 0.63 μF

    • 0.50 μF

    • 0.22 μF

    • 0.17 μF

    Correct Answer
    A. 0.63 μF
    Explanation
    To find the size of the capacitor, we can use the formula for the resonant frequency of an LC circuit, which is given by:

    f = 1 / (2π√(LC))

    Given that the resonant frequency is 1000 Hz, the resistance is 30 Ω, and the inductance is 40 mH, we can rearrange the formula to solve for the capacitance:

    C = 1 / (4π²f²L - R²)

    Substituting the given values, we get:

    C = 1 / (4π²(1000 Hz)²(40 mH) - (30 Ω)²)

    Simplifying the equation, we find that the capacitance is approximately 0.63 μF.

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  • 32. 

    What is the current through a 0.0010-μF capacitor at 1000 Hz and 5.0 V?

    • 5.4 μA

    • 31 μA

    • 3.1 μA

    • 10 μA

    Correct Answer
    A. 31 μA
    Explanation
    The current through a capacitor is given by the formula I = C * V * ω, where I is the current, C is the capacitance, V is the voltage, and ω is the angular frequency. In this case, the capacitance is 0.0010 μF, the voltage is 5.0 V, and the angular frequency can be calculated using the formula ω = 2πf, where f is the frequency. Plugging in the values, we get ω = 2π * 1000 = 2000π rad/s. Substituting all the values into the formula, we get I = 0.0010 * 5.0 * 2000π = 31 μA.

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  • 33. 

    A coil lies flat on a horizontal table top in a region where the magnetic field points straight down. The magnetic field disappears suddenly. When viewed from above, what is the direction of the induced current in this coil as the field disappears?

    • Counterclockwise

    • Clockwise

    • Clockwise initially, then counterclockwise before stopping

    • There is no induced current in this coil.

    Correct Answer
    A. Clockwise
    Explanation
    When the magnetic field disappears suddenly, Faraday's law of electromagnetic induction states that an induced current will be generated in the coil. According to Lenz's law, the direction of the induced current will be such that it opposes the change that caused it. In this case, the sudden disappearance of the downward magnetic field would create a change that induces a current in the coil that flows in the same direction as the original field. Therefore, the direction of the induced current in the coil as the field disappears would be clockwise.

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  • 34. 

    As the frequency of the AC voltage across a capacitor approaches zero, the capacitive reactance of that capacitor

    • Approaches zero.

    • Approaches infinity.

    • Approaches unity.

    • None of the given answers

    Correct Answer
    A. Approaches infinity.
    Explanation
    As the frequency of the AC voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity. This is because the capacitive reactance is inversely proportional to the frequency of the AC voltage. As the frequency approaches zero, the reactance becomes larger and larger, eventually becoming infinite. This means that at very low frequencies, the capacitor effectively blocks the flow of current, behaving like an open circuit.

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  • 35. 

    The flux through a coil changes from 4.0 * 10^(-5) Wb to 5.0 * 10^(-5) Wb in 0.10 s. What emf is induced in this coil?

    • 5.0 * 10^(-4) V

    • 4.0 * 10^(-4) V

    • 1.0 * 10^(-4) V

    • None of the given answers

    Correct Answer
    A. 1.0 * 10^(-4) V
    Explanation
    The change in flux through the coil is calculated by subtracting the initial flux from the final flux. In this case, the change in flux is (5.0 * 10^(-5) Wb) - (4.0 * 10^(-5) Wb) = 1.0 * 10^(-5) Wb. The emf induced in the coil can be found using Faraday's law of electromagnetic induction, which states that the emf is equal to the rate of change of flux. Therefore, the emf induced in the coil is (1.0 * 10^(-5) Wb) / (0.10 s) = 1.0 * 10^(-4) V.

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  • 36. 

    An AC generator consists of 100 turns of wire of area 0.090 m^2 and total resistance 12 Ω. The loops rotate in a magnetic field of 0.50 T at a constant angular speed of 60 revolutions per second. Find the maximum induced current.

    • 23 A

    • 46 A

    • 0.14 kA

    • 0.28 kA

    Correct Answer
    A. 0.14 kA
    Explanation
    The maximum induced current in an AC generator can be calculated using the formula I = NABω, where I is the current, N is the number of turns of wire, A is the area of the wire loop, B is the magnetic field strength, and ω is the angular speed. Plugging in the given values, we get I = 100 * 0.090 * 0.50 * 60 = 0.14 kA.

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  • 37. 

    The primary of a transformer has 100 turns and its secondary has 200 turns. If the input voltage to the primary is 100 V, we can expect the output voltage of the secondary to be

    • 50 V.

    • 100 V.

    • 200 V.

    • None of the given answers

    Correct Answer
    A. 200 V.
    Explanation
    The primary and secondary of a transformer are related by the equation Vp/Vs = Np/Ns, where Vp is the primary voltage, Vs is the secondary voltage, Np is the number of turns in the primary, and Ns is the number of turns in the secondary. In this case, the primary voltage is given as 100 V, the primary has 100 turns, and the secondary has 200 turns. Plugging these values into the equation, we get 100/Vs = 100/200. Solving for Vs, we find that the output voltage of the secondary is 200 V.

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  • 38. 

    What size capacitor is needed to have 50 Ω of capacitive reactance at 10 kHz?

    • 16 μF

    • 5.0 μF

    • 3.2 μF

    • 0.32 μF

    Correct Answer
    A. 0.32 μF
    Explanation
    A capacitor's reactance is given by the formula Xc = 1 / (2πfC), where Xc is the capacitive reactance, f is the frequency, and C is the capacitance. In this question, we are given that the capacitive reactance is 50 Ω and the frequency is 10 kHz. By rearranging the formula, we can solve for the capacitance: C = 1 / (2πfXc). Plugging in the values, we get C = 1 / (2π * 10,000 * 50) = 0.32 μF. Therefore, a 0.32 μF capacitor is needed to have 50 Ω of capacitive reactance at 10 kHz.

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  • 39. 

    All of the following are units of magnetic flux except

    • T*m^2.

    • T/V*m.

    • Weber.

    • V*s.

    Correct Answer
    A. T/V*m.
    Explanation
    The correct answer is T/V*m. Magnetic flux is a measure of the quantity of magnetic field passing through a given area. The units of magnetic flux are typically measured in webers (Wb) or volt-seconds (V*s). T*m^2 represents the unit for magnetic moment, which is a different concept. Therefore, T/V*m is not a valid unit for magnetic flux.

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  • 40. 

    A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a uniform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be a maximum?

    • Any horizontal orientation will do.

    • It must make an angle of 45° to the vertical.

    • It must be vertical.

    • None of the given answers

    Correct Answer
    A. Any horizontal orientation will do.
    Explanation
    The induced emf is a maximum when the magnetic field lines are perpendicular to the loop's plane. Since the magnetic field points straight down, any horizontal orientation of the loop's axis of rotation will result in the magnetic field being perpendicular to the loop, maximizing the induced emf. Therefore, any horizontal orientation will do.

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  • 41. 

    If the frequency of the AC voltage across an inductor is doubled, the inductive reactance of that inductor

    • Increases to 4 times its original value.

    • Increases to twice its original value.

    • Decreases to one-half its original value.

    • Decreases to one-fourth its original value.

    Correct Answer
    A. Increases to twice its original value.
    Explanation
    When the frequency of the AC voltage across an inductor is doubled, the inductive reactance of the inductor increases to twice its original value. This is because inductive reactance is directly proportional to the frequency of the AC voltage. As the frequency doubles, the inductive reactance also doubles.

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  • 42. 

    The coil of a generator has 50 loops and a cross-sectional area of 0.25 m^2. What is the maximum emf generated by this generator if it is spinning with an angular velocity of 4.0 rad/s in a 2.0 T magnetic field?

    • 50 V

    • 100 V

    • 200 V

    • 400 V

    Correct Answer
    A. 100 V
    Explanation
    The maximum emf generated by a generator is given by the equation emf = NABω, where N is the number of loops in the coil, A is the cross-sectional area, B is the magnetic field strength, and ω is the angular velocity. Plugging in the given values, emf = 50 * 0.25 * 2.0 * 4.0 = 100 V.

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  • 43. 

    The secondary coil of a neon sign transformer provides 7500 V at 10.0 mA. The primary coil operates on 120 V. What does the primary draw?

    • 0.625 A

    • 0.625 mA

    • 0.160 A

    • 1.66 A

    Correct Answer
    A. 0.625 A
    Explanation
    The primary coil of a neon sign transformer operates on 120 V. The secondary coil provides 7500 V at 10.0 mA. To find the current drawn by the primary coil, we can use the formula: primary current = secondary current * (secondary voltage / primary voltage). Plugging in the values, we get: primary current = 10.0 mA * (7500 V / 120 V) = 0.625 A. Therefore, the correct answer is 0.625 A.

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  • 44. 

    A 150-W lamp is placed into a 120-V AC outlet. What is the resistance of the lamp?

    • 1.25 Ω

    • 0.80 Ω

    • 48 Ω

    • 96 Ω

    Correct Answer
    A. 96 Ω
    Explanation
    When a lamp is connected to an AC outlet, the power consumed by the lamp can be calculated using the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. In this case, the power is given as 150 W and the voltage is 120 V. Rearranging the formula to solve for R, we get R = V^2/P. Substituting the given values, we find that R = 120^2/150 = 96 Ω. Therefore, the resistance of the lamp is 96 Ω.

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  • 45. 

    The phase angle of an AC circuit is 63°. What is the power factor?

    • 0.89

    • 0.55

    • 0.45

    • 0.11

    Correct Answer
    A. 0.45
    Explanation
    The power factor of an AC circuit is determined by the cosine of the phase angle. In this case, the phase angle is 63°. The cosine of 63° is approximately 0.45. Therefore, the power factor of the AC circuit is 0.45.

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  • 46. 

    What resistance is needed in a series circuit with a 20-mH coil and 1.0-μF capacitor for a total impedance of 100 Ω at 1500 Hz?

    • 0.16 kΩ

    • 82 Ω

    • 57 Ω

    • 18 Ω

    Correct Answer
    A. 57 Ω
    Explanation
    In a series circuit, the total impedance is equal to the sum of the resistance and reactance. The reactance of an inductor is given by XL = 2πfL, where f is the frequency and L is the inductance. The reactance of a capacitor is given by XC = 1/(2πfC), where f is the frequency and C is the capacitance. At 1500 Hz, the reactance of the coil is XL = 2π(1500)(20x10^-3) = 188.5 Ω. The reactance of the capacitor is XC = 1/(2π(1500)(1x10^-6)) = 106.1 Ω. To have a total impedance of 100 Ω, the resistance needed is 100 - (188.5 + 106.1) = 57 Ω. Therefore, the correct answer is 57 Ω.

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  • 47. 

    Doubling the diameter of a loop of wire produces what kind of change on the induced emf, assuming all other factors remain constant?

    • The induced emf is 4 times as much.

    • The induced emf is twice times as much.

    • The induced emf is half as much.

    • There is no change in the induced emf.

    Correct Answer
    A. The induced emf is 4 times as much.
    Explanation
    When the diameter of a loop of wire is doubled, the area of the loop increases by a factor of 4 (since area is proportional to the square of the diameter). According to Faraday's law of electromagnetic induction, the induced emf is directly proportional to the rate of change of magnetic flux through the loop. Since the area of the loop has increased by a factor of 4, the magnetic flux through the loop will also increase by a factor of 4. Therefore, the induced emf will be 4 times as much.

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  • 48. 

    A DC motor of internal resistance 6.0 Ω is connected to a 24-V power supply. The operating current is 1.0 A. What is the start-up current?

    • 1.0 A

    • 2.0 A

    • 3.0 A

    • 4.0 A

    Correct Answer
    A. 4.0 A
    Explanation
    The start-up current of a DC motor is typically higher than the operating current. This is because when the motor is initially turned on, there is no back EMF to oppose the flow of current, so the current is determined solely by the resistance of the motor. In this case, the internal resistance of the motor is 6.0 Ω and the power supply voltage is 24 V. Using Ohm's Law (V = IR), we can calculate the start-up current as 24 V / 6.0 Ω = 4.0 A. Therefore, the correct answer is 4.0 A.

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  • 49. 

    A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a uniform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be zero?

    • Any horizontal orientation will do.

    • It must make an angle of 45° to the vertical.

    • It must be vertical.

    • None of the given answers

    Correct Answer
    A. It must be vertical.
    Explanation
    The induced emf in a wire loop is zero when the magnetic field lines are perpendicular to the plane of the loop. In this case, the uniform magnetic field points straight down, so the loop's axis of rotation must be vertical in order to have zero induced emf.

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Quiz Review Timeline (Updated): Mar 21, 2023 +

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  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 31, 2012
    Quiz Created by
    Drtaylor
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