Amplitude of the induced voltage across the coil

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SUMMARY

The discussion focuses on calculating the amplitude of the induced voltage across a 5-turn coil with a cross-sectional area of 5 cm² rotating at 1200 RPM in a magnetic field of 10 mT. The magnetic flux (φ) is calculated as 2.5 × 10^-3 Wb using the formula φ = N.B.A. Two methods for calculating the induced voltage (V) are presented: one using the sinusoidal function V(t) = N.B.ω.A.sin(ωt), resulting in an amplitude of 31.415 volts, and another using the time derivative of magnetic flux, yielding 4.17 × 10^-3 V. The concern raised is the proper incorporation of the rotation speed (1200 RPM) in the calculations.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the formula for magnetic flux (φ = N.B.A)
  • Knowledge of angular velocity and its conversion from RPM to radians per second
  • Ability to differentiate sinusoidal functions for voltage calculations
NEXT STEPS
  • Learn how to convert RPM to angular velocity (ω) in radians per second
  • Study the principles of Faraday's Law of Electromagnetic Induction
  • Explore the differentiation of sinusoidal functions in the context of induced voltage
  • Investigate the effects of varying magnetic field strength on induced voltage
USEFUL FOR

Electrical engineers, physics students, and anyone involved in the study of electromagnetic systems and induction principles will benefit from this discussion.

lee.perrin@gm
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If anyone can advise that would be great.

1. The plane of a 5 turn coil of 5mm² cross sectional area is rotating a 1200 r.p.m in a magnetic field of 10mT.

Q. Calculate the amplitude of the induced voltage across the coil

Data:

5 turns
c = 1200 rpm
A = 5cm²
B = 10 mT

2. φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

V = dφ/dt or N.A. dB/dt

The equation thought to have been used is the following but the Length is not vissible to me.

V/L = c x B c being the speed
B field
L lenth of wire

3. I have tried two avenues please advise

a: V(t) = -N•d(B•A)/dt = -N•B•dA/dt = N•B•ω•A•sin(ωt)
V(t) =(5)(10x10^-3)(125.66370599999999)(5)sin(ωt) = 31.415 sin(ωt) volts.

Or

b: And using the following, saying t = 60 sec

Then φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

And V = dφ/dt or N.A. dB/dt = 4.17x10^-3 V

My concern here is that 1200rpm was not used.
 
Last edited:
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lee.perrin@gm said:
If anyone can advise that would be great.

1. The plane of a 5 turn coil of 5mm² cross sectional area is rotating a 1200 r.p.m in a magnetic field of 10mT.

Q. Calculate the amplitude of the induced voltage across the coil

Data:

5 turns
c = 1200 rpm
A = 5cm²
B = 10 mT

2. φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

V = dφ/dt or N.A. dB/dt

The equation thought to have been used is the following but the Length is not vissible to me.

V/L = c x B c being the speed
B field
L lenth of wire

3. I have tried two avenues please advise

a: V(t) = -N•d(B•A)/dt = -N•B•dA/dt = N•B•ω•A•sin(ωt)
V(t) =(5)(10x10^-3)(125.66370599999999)(5)sin(ωt) = 31.415 sin(ωt) volts.

Or

b: And using the following, saying t = 60 sec

Then φ = N.B.A φ = 5×10×10^-3 ×5 × 10^-2
φ = 2.5 ×10^-3

And V = dφ/dt or N.A. dB/dt = 4.17x10^-3 V

My concern here is that 1200rpm was not used.

You should write out the function φ(t) (hint: it is a sinusoidal function), then differentiate it to find V(t)...
 

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