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Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise
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Figure 8 (below) shows a rectangular loop of wire of width L and resis...
**Problem Description**
Figure 8 shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?
**Solution**
To find the induced current in the loop, we need to apply Faraday's law of electromagnetic induction. According to Faraday's law, the induced EMF in a loop is equal to the rate of change of magnetic flux through the loop. Mathematically, we can write:
EMF = -dΦ/dt
where EMF is the induced electromotive force, Φ is the magnetic flux through the loop, and t is time.
**Magnetic Flux**
The magnetic flux through the loop can be calculated as the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field and the normal to the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the angle between the magnetic field and the normal to the loop is 90 degrees. Therefore, we can write:
Φ = B*A*cos(90) = B*A*0 = 0
where A is the area of the loop.
**Induced EMF**
Since the magnetic flux through the loop is constant (zero), the induced EMF will also be zero. However, we know that the loop is being pulled to the right at a constant speed v. This means that there is an external force acting on the loop, which is doing work against the resistance of the loop. This work is being converted into thermal energy, which means that there must be a current flowing in the loop.
**Induced Current**
The current induced in the loop can be calculated using Ohm's law, which relates the current to the voltage and resistance. In this case, the voltage is zero (since the induced EMF is zero), so the current will also be zero. However, we know that there must be a current flowing in the loop due to the external force. Therefore, we can conclude that there must be an external voltage source driving the current.
**Conclusion**
In conclusion, the induced EMF and current in the loop are both zero, but there is an external voltage source driving the current due to the applied force.
Figure 8 shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?
**Solution**
To find the induced current in the loop, we need to apply Faraday's law of electromagnetic induction. According to Faraday's law, the induced EMF in a loop is equal to the rate of change of magnetic flux through the loop. Mathematically, we can write:
EMF = -dΦ/dt
where EMF is the induced electromotive force, Φ is the magnetic flux through the loop, and t is time.
**Magnetic Flux**
The magnetic flux through the loop can be calculated as the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field and the normal to the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the angle between the magnetic field and the normal to the loop is 90 degrees. Therefore, we can write:
Φ = B*A*cos(90) = B*A*0 = 0
where A is the area of the loop.
**Induced EMF**
Since the magnetic flux through the loop is constant (zero), the induced EMF will also be zero. However, we know that the loop is being pulled to the right at a constant speed v. This means that there is an external force acting on the loop, which is doing work against the resistance of the loop. This work is being converted into thermal energy, which means that there must be a current flowing in the loop.
**Induced Current**
The current induced in the loop can be calculated using Ohm's law, which relates the current to the voltage and resistance. In this case, the voltage is zero (since the induced EMF is zero), so the current will also be zero. However, we know that there must be a current flowing in the loop due to the external force. Therefore, we can conclude that there must be an external voltage source driving the current.
**Conclusion**
In conclusion, the induced EMF and current in the loop are both zero, but there is an external voltage source driving the current due to the applied force.
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Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.?
Question Description
Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.?.
Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? for Mechanical Engineering 2025 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? covers all topics & solutions for Mechanical Engineering 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.?.
Solutions for Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Here you can find the meaning of Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? defined & explained in the simplest way possible. Besides giving the explanation of
Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.?, a detailed solution for Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? has been provided alongside types of Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? theory, EduRev gives you an
ample number of questions to practice Figure 8 (below) shows a rectangular loop of wire of width L and resistance R. One end of the loop is in a uniform magnetic field of strength B directed perpendicular into the plane of the page. The loop is, pulled to the right at a constant speed v. What are the magnitude and direction of the induced current in the loop?A. I = BLvR and clockwiseB. I = BLv/R and clockwiseC. I = BLvR and counterclockwiseD. I = BLv/R and counterclockwise Related: Lecture 20 - Classical Size Effects, Perpendicular Direction - Mechanical Engg.? tests, examples and also practice Mechanical Engineering tests.
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