Magnet Coils & EMF Homework Help

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SUMMARY

The discussion focuses on calculating the magnetic flux and induced electromotive force (emf) in a coil due to a magnetized needle. The magnetic field (B) is derived from the surface-current-induced field of the needle, with the initial magnetic flux (Φ0) calculated as πA²B, where B = μ0M. As the needle exits the coil, the flux decreases to zero, leading to an induced emf expressed as emf(t) = -N dΦ/dt. The work done in moving the needle is equated to the electrical energy dissipated, highlighting the relationship between work and the rate of decay of magnetic flux.

PREREQUISITES
  • Understanding of magnetic fields and flux
  • Familiarity with Faraday's law of electromagnetic induction
  • Knowledge of basic calculus for rate of change calculations
  • Concept of electrical energy dissipation in circuits
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  • Study Faraday's law of electromagnetic induction in detail
  • Explore the concept of magnetic flux and its applications in electromagnetism
  • Learn about the relationship between work and energy in electromagnetic systems
  • Investigate the properties of permanent magnets and their effects on coils
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Homework Statement


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Homework Equations

The Attempt at a Solution


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I got a Magnetic field by "Magnetized needle". I think it's just magnet.

But i don't know how to get a magnetic flux due to the variable "d" that should be erased.
 
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OK, I'm answering this now because no one else has chosen to. Which surprises me since I don't think this is a trivial problem. I hope it wasn't part of an introductory physics course. Or I'm missing something big. So the below is my take which may be subject to revision by others or even myself:

Part (a): You seem to be trying to compute the B field a distance d away from the coil center. Is that what is asked?

As I see it, the flux thru the coil is just the surface-current-induced B field of the permanent magnet (the needle) which, since the needle is assumed uniforly magnetized, would initially (t=0) be a constant Φ0 = πA2B with B = μ0M since the needle is half-way into the coil. But over a short time Δt as the needle leaves the plane of the coil, Φ → 0. Don't know any other way to look at it. So there is a step decrease in Φ from Φ0 to 0, and emf(t) = V(t) = -N dΦ/dt = NπA2μ0M / Δt.

Part (b): To get the work done in bringing the needle from its mid-point to where its end is in the coil plane you can just equate that work with the electrical energy dissipated = ∫V2/R dt from L/2v - Δt/2 to L/2v +Δt/2T. Unfortunately, doing this implies that the work is inversely proportinal to the rate of decay of Φ as the needle leaves the coil plane. So no hard-and-fast answer.
 

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