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Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by?
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Work Done in Pulling a Loop in a Magnetic Field
When a square loop of side 'a' is pulled out of a constant perpendicular magnetic field 'B' with a constant velocity 'v', the work done can be calculated considering the induced electromotive force (emf) due to the motion of the loop.
Induced EMF Calculation
- As the loop is pulled out, it cuts through magnetic field lines.
- According to Faraday's law of electromagnetic induction, the induced emf (ε) is given by:
ε = -dΦ/dt
- Where Φ is the magnetic flux through the loop. The magnetic flux is given by:
Φ = B * A
- Here, A is the area of the loop inside the magnetic field, which changes as the loop is pulled out. If x is the length of the loop still inside the field, the area A = a * x.
Changing Area and EMF
- As the loop is pulled out, x decreases at a rate of 'v':
dx/dt = -v
- Therefore, the change in area over time is:
dA/dt = a * (-v)
- Substituting this into the flux equation gives:
ε = -B * (dA/dt) = B * a * v
Current Induced and Resistance
- The induced current (I) in the loop can be calculated using Ohm’s Law:
I = ε/R
- Where R is the resistance of the loop.
Work Done Calculation
- The power (P) delivered to the circuit is given by:
P = I * ε
- Therefore, the work done (W) over time 't' is:
W = P * t = I * ε * t
- As a result, substituting the expressions for I and ε leads to the conclusion that the work done in pulling the loop out is directly proportional to the magnetic field strength, the area of the loop, the velocity, and the resistance of the loop.
When a square loop of side 'a' is pulled out of a constant perpendicular magnetic field 'B' with a constant velocity 'v', the work done can be calculated considering the induced electromotive force (emf) due to the motion of the loop.
Induced EMF Calculation
- As the loop is pulled out, it cuts through magnetic field lines.
- According to Faraday's law of electromagnetic induction, the induced emf (ε) is given by:
ε = -dΦ/dt
- Where Φ is the magnetic flux through the loop. The magnetic flux is given by:
Φ = B * A
- Here, A is the area of the loop inside the magnetic field, which changes as the loop is pulled out. If x is the length of the loop still inside the field, the area A = a * x.
Changing Area and EMF
- As the loop is pulled out, x decreases at a rate of 'v':
dx/dt = -v
- Therefore, the change in area over time is:
dA/dt = a * (-v)
- Substituting this into the flux equation gives:
ε = -B * (dA/dt) = B * a * v
Current Induced and Resistance
- The induced current (I) in the loop can be calculated using Ohm’s Law:
I = ε/R
- Where R is the resistance of the loop.
Work Done Calculation
- The power (P) delivered to the circuit is given by:
P = I * ε
- Therefore, the work done (W) over time 't' is:
W = P * t = I * ε * t
- As a result, substituting the expressions for I and ε leads to the conclusion that the work done in pulling the loop out is directly proportional to the magnetic field strength, the area of the loop, the velocity, and the resistance of the loop.
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Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by?
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Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by?.
Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Square root of side a is pulled out of constant perpendicular magnetic field b with constant velocity v if velocity is parallel to one of its sides and resistance of loop is are the work done in process given by?.
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