October 17th, 2012, 12:00 PM  #1 
Member Joined: Oct 2012 Posts: 36 Thanks: 0  vector field and flux
v(x,y)=f(r)(x,y) is a vector field with r=(x^2+y^2)^(1/2) and f is continuously differentiable ,given v is consevative then i have found the scalar potential of v which is f(x,y)=?f(r)dr , and now C is the circle of radius R centred at the orign, and i found the counterclockwise flux is 2pif(R)R^2 ,now find all functions f so that this flux is independent of R, my answer to it is f is 0. can someone give me some hints or do my solutions get any wrong?? 
October 17th, 2012, 03:14 PM  #2 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 625 Thanks: 87 Math Focus: Electrical Engineering Applications  Re: vector field and flux
I think there are two solutions. One is f(R)=0 as you stated and the other is In the first case the flux is 0 and in the second case the flux is per your equation. 
October 17th, 2012, 03:35 PM  #3 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 625 Thanks: 87 Math Focus: Electrical Engineering Applications  Re: vector field and flux
Actually, I am kind of confused by your notation. Your last equation, which I keyed on, includes f(R), as if the field is a function of the radius of the circle. If you mean f(r) as in the distance from the origin, then I think f(r) = 0 is the only solution. I need to think about it some more. Any clarification you can provide would be helpful.

October 17th, 2012, 03:41 PM  #4  
Member Joined: Oct 2012 Posts: 36 Thanks: 0  Re: vector field and flux Quote:
 

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field, flux, vector 
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