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A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil is
- a)0.1 V
- b)80 KV
- c)7.96 V
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
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A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil...
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A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil...
Given data:
Field strength, H = 5 x 10^4/p A-turns/m
Number of turns, N = 50
Area, A = 10^-2 m^2
Time, t = 0.1 s
Formula used:
The induced emf in a coil is given by the formula,
E = -NdΦ/dt
where N is the number of turns in the coil, Φ is the magnetic flux passing through the coil, and dt is the time for which the magnetic flux changes.
Calculation:
Magnetic flux density, B = μH
where μ is the permeability of the medium, which is taken as μ = μ0 for air, and μ0 = 4π x 10^-7 T-m/A
B = μH = 4π x 10^-7 x 5 x 10^4/p = 2π/p x 10^-2 T
Magnetic flux passing through the coil, Φ = NBA
where B is the magnetic flux density, A is the area of the coil
Φ = NBA = 2π/p x 10^-2 x 50 = π/p x 10^-1 Wb
The rate of change of magnetic flux, dΦ/dt = Φ/t = π/p x 10^-1/0.1 = π/p x V/s
The induced emf in the coil, E = -NdΦ/dt
E = -50 x π/p x 10^-1/0.1 = -5π/p V
Since the emf is induced in the opposite direction to the change in magnetic flux, we take the magnitude of the emf.
E = 5π/p V = 1.6 V/p V (approx.)
Therefore, the induced emf in the coil is 0.1 V (approx.), which is closest to option (a).
Field strength, H = 5 x 10^4/p A-turns/m
Number of turns, N = 50
Area, A = 10^-2 m^2
Time, t = 0.1 s
Formula used:
The induced emf in a coil is given by the formula,
E = -NdΦ/dt
where N is the number of turns in the coil, Φ is the magnetic flux passing through the coil, and dt is the time for which the magnetic flux changes.
Calculation:
Magnetic flux density, B = μH
where μ is the permeability of the medium, which is taken as μ = μ0 for air, and μ0 = 4π x 10^-7 T-m/A
B = μH = 4π x 10^-7 x 5 x 10^4/p = 2π/p x 10^-2 T
Magnetic flux passing through the coil, Φ = NBA
where B is the magnetic flux density, A is the area of the coil
Φ = NBA = 2π/p x 10^-2 x 50 = π/p x 10^-1 Wb
The rate of change of magnetic flux, dΦ/dt = Φ/t = π/p x 10^-1/0.1 = π/p x V/s
The induced emf in the coil, E = -NdΦ/dt
E = -50 x π/p x 10^-1/0.1 = -5π/p V
Since the emf is induced in the opposite direction to the change in magnetic flux, we take the magnitude of the emf.
E = 5π/p V = 1.6 V/p V (approx.)
Therefore, the induced emf in the coil is 0.1 V (approx.), which is closest to option (a).
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A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer?.
A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A field of 5 x 104/p ampere-turns/metre acts at right angles to a coil of 50 turns of area 10-2 m2. The coil is removed from the field in 0.1 second. Then the induced emf in the coil isa)0.1 Vb)80 KVc)7.96 Vd)none of the aboveCorrect answer is option 'A'. Can you explain this answer?.
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