Electromotive force (EMF) induced in a rectangular loop

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SUMMARY

The discussion centers on the calculation of electromotive force (EMF) induced in a rectangular loop, specifically using Faraday's Law. The relationship between induced EMF and magnetic flux is defined by the equation Emfinduced = -N(dΦB/dt). When the loop rotates at a constant speed, the formula simplifies to Emfinduced = -N(ΔΦ/Δt). A participant successfully calculated the induced EMF to be 1.708 x 10-3 V using these principles.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Knowledge of magnetic flux calculation (ΦB = ∫B·dA)
  • Familiarity with the concept of induced EMF in rotating loops
  • Basic proficiency in calculus for differentiation and limits
NEXT STEPS
  • Study the derivation and applications of Faraday's Law in different contexts
  • Explore the effects of varying magnetic fields on induced EMF
  • Learn about Lenz's Law and its implications for induced currents
  • Investigate practical applications of EMF in electrical engineering and physics
USEFUL FOR

Students of physics, electrical engineers, and anyone interested in the principles of electromagnetic induction and its applications in technology.

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Homework Statement
A rectangular single loop of wire of length 0.035m and width 0.015m is positioned at 45° to a uniform magnetic field of 2.3T. The loop is rotated through 45° so that its plane is parallel to the magnetic field. Determine the electromotive force (e.m.f.) induced in the loop if the rotation takes 0.5s
Relevant Equations
emf = BAN cosθ
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Hi, so you have calculated the flux \Phi_B = \int \vec B \cdot d\vec A, but what is the relationship between Emf induced and the flux? Have you heard of Faraday's Law, which states that the emf induced is given by Emf_{induced} = -N\frac{d\Phi_{B}}{dt}. If you are assuming that this rotates at a constant speed, then you can change the formula to Emf_{induced} = -N\frac{\Delta \Phi}{\Delta t}.

Hope that is some help. I can answer more questions, but I am trying not to give the answer away.
 
Hello, thank you for your message! After some thought and looking around in textbooks etc. I believe that by using Emf_{induced} = -N\frac{\Delta \Phi}{\Delta t} I have calculated the Emf to be -1\frac{0 - (8.54\text{ x }10^{-4})}{0.5} = 1.708 \text{ x } 10^{-3} \text{ V}

I hope this is correct?
Thank you
 

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