Register to reply 
Solving for Magnetic Fields from time varying Electric fields 
Share this thread: 
#1
Jun2312, 12:10 AM

P: 104

Hi all, new to the forums.
Anyways I had a day off and decided to try something quirky with maxwells equations. Using the AmpereMaxwell equation, I tried to solve for the magnetic field that would be created by a <0,0,t> Electric field. (No current and no magnetic materials) ∇ x B = εμ(∂E/∂t) ∇ x B = εμ<0,0,1> (Pardon the bad notation) ∇ x B=< ∂Bz/∂y∂By/∂z , ∂Bx/∂z∂Bz/∂x , ∂By/∂x∂Bx/∂y > Equating the components, we have: ∂Bz/∂y=∂By/∂z ∂Bx/∂z=∂Bz/∂x ∂By/∂x∂Bx/∂y=1 Here's the problem, How do I solve this mess? 


#2
Jun2312, 12:25 AM

P: 834

That the source of the magnetic field the timevarying electric field and not an actual current makes no difference to Maxwell's equations. Because [itex]\nabla \cdot B = 0[/itex] still, you can remove the derivative through its inverse, the freespace Green's function [itex]r/4\pir^3[/itex]. What you get is
[tex]B(r) \propto \int_V \frac{\hat z \times (r  r')}{4 \pi r  r'^3} \; d^3r'[/tex] This result (when the proper proportionality constants are in) is the BiotSavart law. The problem, however, is that if this electric field is timevarying everywhere, then I doubt this integral converges. You'll likely have to confine yourself to a 2D plane and invoke symmetry, or even simplify the source electric field to being nonzero only at a specific point, unless you're explicitly interested in the case of E being nonzero throughout a particular region. 


#3
Jun2312, 12:29 AM

P: 104

Sorry I'm not familiar with the method you're referring to here, could you elaborate?



#4
Jun2312, 01:04 AM

P: 834

Solving for Magnetic Fields from time varying Electric fields
You're not familiar with the use of Green's functions to solve differential equations? I'm not terribly surprised, I guess; the topic is sometimes treated like it's esoteric. Here's the basic idea:
Many differential equations are of the form [itex]dA/dx = J[/itex]. It's first order, and it's simple. The Green's function for [itex]d/dx[/itex] is defined such that [itex]dG/dx = \delta(x)[/itex]the delta function. This isn't immediately useful until you use it with the fundamental theorem of calculus, which can be written as [tex]\left. A(x) \right_a^b = \int_a^b \frac{dA}{dx} \; dx = \int_a^b J(x) \; dx[/tex] Now, using the fundamental theorem above (with [itex]dA/dx = J[/itex]), just for fun, consider [tex]\left. A(x') G(xx') \right_{x'=a}^b = \int_a^b \frac{dA(x')}{dx'} G(xx') + A(x') \frac{dG(xx')}{dx'} \; dx' = \int_a^b J(x') G(xx')  A(x') \delta(xx') \; dx'[/tex] Now, if we let [itex]a \to \infty[/itex] and [itex]b \to \infty[/itex], we cover the whole real line. We generally require the solution for [itex]G(x)[/itex] to be wellbehaved there, so the term on the lefthand side goes to zero. Remember also that when you integrate over a delta function, it forces the argument to zero. This leads to [itex]0 = \int_{\infty}^\infty J(x') G(xx') \; dx'  A(x) \implies A(x) = \int_{\infty}^\infty J(x') G(xx') \; dx'[/itex] So, if you know the form of the Green's function for a particular differential operator and the source that generates the field, you can solve for the field through this integration. Extending this to 3D isn't too hard, but it requires some mathematical formalism that may obscure the point. The Green's function for [itex]\nabla[/itex] is well knownyou use it every time you talk about the field from a point charge, because we model point charges as delta function sources anyway. 


Register to reply 
Related Discussions  
Time delay/magnetic fields/coloumb fields.  General Physics  18  
TimeVarying Magnetic Fields Q#2  Introductory Physics Homework  5  
TimeVarying Magnetic Fields  Introductory Physics Homework  4  
Electric fields, magnetic fields and Lorentz frames  Advanced Physics Homework  5  
Magnetic fields inducing electric fields  turning on lightbulb with magnetic field  Introductory Physics Homework  3 