Calculate Torque of Eddy Current Brake for Rotating Disk with Magnetic Field

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SUMMARY

The discussion focuses on calculating the torque of an eddy current brake acting on a rotating disk subjected to a perpendicular magnetic field B. The torque is derived using the equations for electromotive force (EMF) and resistance, leading to the expression τ = -B²a³ωσb. The user attempts to integrate the force F = -B²a²ωσb over the radius of the disk, indicating a need for revision in their integral handling. Key variables include conductivity σ, thickness b, angular velocity ω, and area a.

PREREQUISITES
  • Understanding of electromagnetism principles, specifically Faraday's law of induction.
  • Familiarity with the concepts of torque and rotational dynamics.
  • Knowledge of resistance in conductive materials, particularly the formula R = L/(σA).
  • Basic calculus skills for performing integrals in physics applications.
NEXT STEPS
  • Study the derivation of Faraday's law of induction in detail.
  • Learn about the dynamics of rotating systems and the role of torque in angular motion.
  • Explore the applications of eddy current brakes in engineering contexts.
  • Practice integration techniques relevant to physics problems involving forces and torques.
USEFUL FOR

Students in physics or engineering disciplines, particularly those studying electromagnetism and rotational dynamics, as well as professionals working with eddy current braking systems.

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Homework Statement



Refer to Figure attached.

A disk of conductivity σ and thickness b rotates around an axis through its center with a magnetic field B applied perpendicular to the plane of the disk over a small area a2. Find the approximate torque tending to slow the disk at the instant its angular velocity is ω.

Homework Equations


The Attempt at a Solution



[itex]\xi = -B \frac{dA}{dt}[/itex]

[itex]= -B a \frac{da}{dt}[/itex]

(Right here ^ Can I do this?)

[itex]= -B a v[/itex]

[itex]= -B a \omega dr[/itex]

[itex]\Rightarrow \int^{r+a/2}_{r-a/2} -B a \omega dr[/itex]

[itex]= - B a^2 \omega[/itex]

And

[itex]R = \frac{L}{\sigma A} = \frac{a}{\sigma ab} = \frac{1}{\sigma b}[/itex]

-----
-----

Thus

[itex]F = i a \times B[/itex]

[itex]= \frac{\xi}{R} a \times B[/itex]

[itex]F = - B^2 a^2 \omega \sigma b[/itex]

-----
-----

[itex]\tau = \int F dr[/itex]

[itex]\tau = - B^2 a^2 \omega \sigma b \int^{r+a/2}_{r-a/2} dr[/itex]

[itex]\tau = - B^2 a^3 \omega \sigma b[/itex]
 

Attachments

  • eddybrake.png
    eddybrake.png
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Last edited:
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I think I've handled my integrals poorly...
will revise.
 

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