How do you calculate induced EMF in an open loop with changing magnetic field?

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SUMMARY

The discussion focuses on calculating induced electromotive force (EMF) in an open loop subjected to a changing magnetic field. It confirms that while Faraday's law may not directly apply due to the lack of an enclosed area, the induced EMF can be determined by solving Maxwell's Equations to find the electric field (##\vec{E}##). The voltage across the open loop can be computed using the integral formula ##V=-\int_C \vec{E}.\vec{\hat{l}} dl##. Additionally, it highlights that current can flow through the wire due to the capacitive effects at the ends, represented by the formula for capacitive reactance, ##X_c = 1/(omega*C)##.

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  • Understanding of Faraday's Law of Induction
  • Familiarity with Maxwell's Equations
  • Knowledge of electric fields and their calculations
  • Basic concepts of capacitance and capacitive reactance
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  • Study Maxwell's Equations in detail to understand their application in electromagnetic theory
  • Learn how to calculate electric fields generated by changing magnetic fields
  • Explore the integral calculation of voltage across open loops in electromagnetic contexts
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Let's say you have an open loop (like a section of a circle) in a changing magnetic field. I think there would be an induced EMF, but no current. What I can't figure out, though, is how to calculate the induced EMF. Using Faraday's law doesn't seem to help, as there's no enclosed area.
 
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Suppose there is EMF (voltage difference) between the ends. What do you think will happen? Would current flow through the wire?
 
scottdave said:
Suppose there is EMF (voltage difference) between the ends. What do you think will happen? Would current flow through the wire?
No, as there's no path for current to flow.
 
You would need to find the electric field (##\vec{E}##) due to changing magnetic field (by solving Maxwell's Equations), and then compute the voltage across the open loop ##C## by integrating the electric field along it ##V=-\int_C \vec{E}.\vec{\hat{l}} dl##
 
There is a capacitance due to the ends of the wire. Current will flow, due to Xc = 1/(omega*C)
 

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