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A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring is
- a)120 V
- b)100 V .
- c)200 V
- d)150 V
Correct answer is option 'B'. Can you explain this answer?
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A 1.0 m long metallic rod is rotated with an angular frequency of 400 ...
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Given information:
- Length of metallic rod = 1.0 m
- Angular frequency of rotation of the rod = 400 rad/s
- Magnetic field strength = 0.5 T
- The axis of rotation is normal to the rod and passes through one of its ends
- The other end of the rod is in contact with a circular metallic ring
To find:
- The emf developed between the centre and the ring
Solution:
To calculate the emf developed, we can use Faraday's law of electromagnetic induction, which states that the emf induced in a closed circuit is proportional to the rate of change of magnetic flux through the circuit. Mathematically,
emf = -dΦ/dt
where emf is the induced electromotive force, Φ is the magnetic flux, and t is time.
Let's consider a small element of the rod of length dx at a distance x from the end in contact with the ring. The magnetic flux through this element is given by
dΦ = B*dx*cosθ
where B is the magnetic field strength, θ is the angle between the magnetic field and the element, and dx*cosθ is the component of the element perpendicular to the magnetic field.
The angle θ can be calculated as follows:
θ = π/2 - ωt
where ω is the angular frequency of rotation of the rod and t is time. At t = 0, the angle θ is π/2, and it decreases linearly with time.
The total magnetic flux through the rod is given by integrating dΦ over the length of the rod:
Φ = ∫dΦ = ∫B*dx*cosθ = B∫cos(π/2 - ωt)*dx
The limits of integration are 0 and 1 m, the length of the rod.
Φ = B∫cos(π/2 - ωt)*dx = B*sin(ωt)
The induced emf is then given by taking the derivative of Φ with respect to time:
emf = -dΦ/dt = -B*ω*cos(ωt)
The maximum value of emf occurs when cos(ωt) = -1, i.e., at t = (2n+1)π/2ω, where n is an integer. The maximum value of emf is then
emf(max) = B*ω
Substituting the given values, we get
emf = 0.5*400 = 200 V
However, the question asks for the emf between the centre and the ring, which is half of the total emf, i.e.,
emf = emf(max)/2 = 200/2 = 100 V
Therefore, the correct answer is option (b) 100 V.
- Length of metallic rod = 1.0 m
- Angular frequency of rotation of the rod = 400 rad/s
- Magnetic field strength = 0.5 T
- The axis of rotation is normal to the rod and passes through one of its ends
- The other end of the rod is in contact with a circular metallic ring
To find:
- The emf developed between the centre and the ring
Solution:
To calculate the emf developed, we can use Faraday's law of electromagnetic induction, which states that the emf induced in a closed circuit is proportional to the rate of change of magnetic flux through the circuit. Mathematically,
emf = -dΦ/dt
where emf is the induced electromotive force, Φ is the magnetic flux, and t is time.
Let's consider a small element of the rod of length dx at a distance x from the end in contact with the ring. The magnetic flux through this element is given by
dΦ = B*dx*cosθ
where B is the magnetic field strength, θ is the angle between the magnetic field and the element, and dx*cosθ is the component of the element perpendicular to the magnetic field.
The angle θ can be calculated as follows:
θ = π/2 - ωt
where ω is the angular frequency of rotation of the rod and t is time. At t = 0, the angle θ is π/2, and it decreases linearly with time.
The total magnetic flux through the rod is given by integrating dΦ over the length of the rod:
Φ = ∫dΦ = ∫B*dx*cosθ = B∫cos(π/2 - ωt)*dx
The limits of integration are 0 and 1 m, the length of the rod.
Φ = B∫cos(π/2 - ωt)*dx = B*sin(ωt)
The induced emf is then given by taking the derivative of Φ with respect to time:
emf = -dΦ/dt = -B*ω*cos(ωt)
The maximum value of emf occurs when cos(ωt) = -1, i.e., at t = (2n+1)π/2ω, where n is an integer. The maximum value of emf is then
emf(max) = B*ω
Substituting the given values, we get
emf = 0.5*400 = 200 V
However, the question asks for the emf between the centre and the ring, which is half of the total emf, i.e.,
emf = emf(max)/2 = 200/2 = 100 V
Therefore, the correct answer is option (b) 100 V.
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A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer?
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A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer?.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 1.0 m long metallic rod is rotated with an angular frequency of 400 rads−1about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Emf developed between the centre and the ring isa)120 Vb)100 V .c)200 Vd)150 VCorrect answer is option 'B'. Can you explain this answer?.
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