Deriving wave eqn. from Faraday & A.M.

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SUMMARY

The discussion focuses on deriving the wave equation for the magnetic field, B, from Faraday's Law and the Ampere-Maxwell Law. The equation is expressed as ∇²B = (1/c²) ∂²B/∂t². The user attempts to manipulate the equations, specifically using curl operations, and encounters confusion regarding the transition from -ε₀μ₀ d²B/dt² to -ε₀μ₀ dB/dt. The key equations referenced include curl E = -dB/dt and curl B = ε₀μ₀ dE/dt.

PREREQUISITES
  • Understanding of Faraday's Law and Ampere-Maxwell Law
  • Familiarity with vector calculus operations, specifically curl and divergence
  • Knowledge of electromagnetic theory, particularly Maxwell's equations
  • Basic understanding of wave equations in physics
NEXT STEPS
  • Study the derivation of the wave equation from Maxwell's equations
  • Learn about vector calculus identities, particularly curl and divergence
  • Explore the physical implications of electromagnetic waves in a vacuum
  • Investigate the role of ε₀ and μ₀ in electromagnetic theory
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Students and professionals in physics, particularly those focusing on electromagnetism and wave propagation, as well as educators teaching advanced physics concepts.

Roodles01
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Homework Statement


I'm asked to start from Faradays Law & Ampere-Maxwell law then derive the equation for the magnetic field, B, in the form;

del2B = 1/c2 * d2B/dt2



Homework Equations


Faraday: curl E = -dB/dt
A.M : curl B = ε0μ0 dE/dt


The Attempt at a Solution


taking curl of both sides of AM

rhs
curl curlB = grad div B - del2B
no monopole says divB = 0
so = - del2B

lhs
curl curl B = ε0μ0 curl dE/dt
rearrange
= ε0μ0 d/dt curl E
Faraday says curl E = -dB/dt

in the next step of working out I don't get why this goes to
-ε0μ0 d2B/dt2
instead of
-ε0μ0 dB/dt

Addled, I must be [Yoda]
 

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-\nabla ^2 \mathbf{B} = \epsilon_0 \mu_0 \frac{\partial}{\partial t}\nabla \times E

Try the substitution for \nabla \times E again.
 
Thank you.
Nose - Face - well known phrase.
 

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