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A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would be
- a)3.6 volt
- b)2.8 volt
- c)4.9 volt
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A square loop of wire 25 cm has a voltmeter (of infinite impedance) co...
Given,
The required induced voltage is given by
Hence, the maximum induced voltage
The required induced voltage is given by
Hence, the maximum induced voltage
Most Upvoted Answer
A square loop of wire 25 cm has a voltmeter (of infinite impedance) co...
Understanding the Problem
To find the voltage indicated by the voltmeter connected to a square loop of wire in an alternating magnetic field, we need to use Faraday's law of electromagnetic induction.
Given Parameters:
- Side length of the square loop: 25 cm (0.25 m)
- Frequency of the alternating field: 10 MHz (10^6 Hz)
- Maximum magnetic field intensity: 1 A/m
Calculating Area of the Loop:
- The area \(A\) of the square loop is calculated as:
A = side_length × side_length = 0.25 m × 0.25 m = 0.0625 m²
Magnetic Flux and Induced Voltage:
- The magnetic flux \(Φ\) through the loop is given by:
Φ = B × A
where B can be derived from the relationship between magnetic field intensity (H) and magnetic flux density (B):
B = μ₀ × H = (4π × 10^(-7) T·m/A) × 1 A/m = 4π × 10^(-7) T
- Thus, the flux becomes:
Φ = (4π × 10^(-7) T) × 0.0625 m²
Induced Voltage Calculation:
- The induced voltage \(V\) in the loop is given by:
V = -dΦ/dt
For a sinusoidal field, the voltage can be expressed as:
V = A × (dB/dt) × frequency
- The maximum change in magnetic flux density dB can be related to the field intensity, hence:
dB/dt = 2πfB_max = 2π(10^6 Hz)(4π × 10^(-7) T)
- Plugging the values into the induced voltage formula gives:
V = 0.0625 m² × 2π(10^6 Hz)(4π × 10^(-7) T) = 4.9 V
Conclusion:
- Therefore, the voltage indicated by the meter when the loop is placed in the alternating field is 4.9 volts, confirming that option 'C' is indeed the correct answer.
To find the voltage indicated by the voltmeter connected to a square loop of wire in an alternating magnetic field, we need to use Faraday's law of electromagnetic induction.
Given Parameters:
- Side length of the square loop: 25 cm (0.25 m)
- Frequency of the alternating field: 10 MHz (10^6 Hz)
- Maximum magnetic field intensity: 1 A/m
Calculating Area of the Loop:
- The area \(A\) of the square loop is calculated as:
A = side_length × side_length = 0.25 m × 0.25 m = 0.0625 m²
Magnetic Flux and Induced Voltage:
- The magnetic flux \(Φ\) through the loop is given by:
Φ = B × A
where B can be derived from the relationship between magnetic field intensity (H) and magnetic flux density (B):
B = μ₀ × H = (4π × 10^(-7) T·m/A) × 1 A/m = 4π × 10^(-7) T
- Thus, the flux becomes:
Φ = (4π × 10^(-7) T) × 0.0625 m²
Induced Voltage Calculation:
- The induced voltage \(V\) in the loop is given by:
V = -dΦ/dt
For a sinusoidal field, the voltage can be expressed as:
V = A × (dB/dt) × frequency
- The maximum change in magnetic flux density dB can be related to the field intensity, hence:
dB/dt = 2πfB_max = 2π(10^6 Hz)(4π × 10^(-7) T)
- Plugging the values into the induced voltage formula gives:
V = 0.0625 m² × 2π(10^6 Hz)(4π × 10^(-7) T) = 4.9 V
Conclusion:
- Therefore, the voltage indicated by the meter when the loop is placed in the alternating field is 4.9 volts, confirming that option 'C' is indeed the correct answer.
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A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer?
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A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer?.
A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer?.
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A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A square loop of wire 25 cm has a voltmeter (of infinite impedance) connected in series with one side. The plane of the ioop is perpendicular to the magnetic field and the frequency is 10 MHz. If the maximum intensity is 1 Amp/m, then the voltage indicated by the meter when the loop is placed in the alternating field would bea)3.6 voltb)2.8 voltc)4.9 voltd)none of theseCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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