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Induced Electric Field 
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#1
Apr411, 05:16 AM

P: 1,394

1. The problem statement, all variables and given/known data
In a cylindrical region of radius R, there exists a time varying magnetic field B such that dB/dt=k(>0) . A charged particle having charge q is placed at the point P at a distance d (> R) from its centre O. Now, the particle is moved in the direction perpendicular to OP (see figure) by an external agent upto infinity so that there is no gain in kinetic energy of the charged particle. Show that the work done by the external agent is independent of d and find it. 3. The attempt at a solution Let the angle between the line joining the particle to the point O make an angle θ with the vertical. ∫E.dl=kπR^{2} E*x*cos(180θ)=kπR^{2}, where x is horizontal distance of the particle from the initial position. Substituting x for dtanθ and solving, I get W= qkπR^{2} which is incorrect. The answer given is qkπR^{2}/4 


#2
Apr411, 01:04 PM

P: 431

What's your calculated E? How come I see no integration over cycle when you are clearly using Maxwell's Law? That's a blatant error,sorry.



#3
Apr411, 01:22 PM

P: 1,394

E=kπR^{2}/xcosθ
Which integration are you talking about? 


#4
Apr411, 02:46 PM

P: 431

Induced Electric Field
Anyways since the angle force vector makes with the displacement is varying you will have to integrate to get the work. 


#5
Apr411, 03:01 PM

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PF Gold
P: 11,774

Explain how you went from



#6
Apr411, 04:17 PM

P: 1,394

dl vector is directed along the path travelled by the particle. I just used the dot product of two vectors. 


#7
Apr411, 04:41 PM

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PF Gold
P: 11,774

Faraday's Law says
[tex]\oint_{\partial S} \mathbf{E}\cdot d\mathbf{l} = \frac{\partial \Phi_{B,S}}{\partial t}[/tex] where S is a surface and ∂S is the closed boundary of that surface. The expression kπR^{2} is equal to the rate of change of flux, but your calculation of the line integral is wrong. The path the particle travels has nothing to do with the boundary of S. 


#8
Apr511, 12:27 AM

P: 1,394

Ok.
What is the correct expression? 


#9
Apr511, 12:33 PM

P: 431

First separately derive the electric field at a distance r from the centre of vortex. Then write out the differential work in terms of r and angle b/w displacement and force. Eliminate angle and integrate under proper limits..
By the way the existence of a vortex has absolutely nothing to do with the whether there is a charged particle at that point or not. 


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