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 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: 601 Thanks: 82 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 $f(R)=\frac{k}{R^2}$ In the first case the flux is 0 and in the second case the flux is $\ 2\pi k \$ per your equation.
 October 17th, 2012, 03:35 PM #3 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 601 Thanks: 82 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
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Re: vector field and flux

Quote:
 Originally Posted by jks 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.
thank u for your helps.I think it means the radius but not equal to 0, if R=0 then we don't have such circle any more ,it changes to a point, that is what i think both are solutions .

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