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Induced Electric Field

by zorro
Tags: electric, field, induced
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zorro
#1
Apr4-11, 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=-kπR2
E*x*cos(180-θ)=-kπR2, where x is horizontal distance of the particle from the initial position.
Substituting x for dtanθ and solving, I get W= qkπR2 which is incorrect.

The answer given is qkπR2/4
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aim1732
#2
Apr4-11, 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.
zorro
#3
Apr4-11, 01:22 PM
P: 1,394
E=kπR2/xcosθ
Which integration are you talking about?

aim1732
#4
Apr4-11, 02:46 PM
P: 431
Induced Electric Field

E*x*cos(180-θ)=-kπR2
Ambiguous since Maxwell's Law is to be applied only to fixed mathematical loops.

Anyways since the angle force vector makes with the displacement is varying you will have to integrate to get the work.
vela
#5
Apr4-11, 03:01 PM
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Explain how you went from
Quote Quote by Abdul Quadeer View Post
E.dl=-kπR2
to
E*x*cos(180-θ)=-kπR2, where x is horizontal distance of the particle from the initial position.
As aim1732 has suggested, you're not evaluating the LHS correctly. You can use Faraday's law to calculate E as a function of the displacement r. Once you have that, you can then find the force on the charge as a function of the displacement r and integrate it to find the work done.
zorro
#6
Apr4-11, 04:17 PM
P: 1,394
Ambiguous since Maxwell's Law is to be applied only to fixed mathematical loops.
But the particle is not moving along a loop here. Sorry I don't understand your point.


Explain how you went from.....
At any instant, let the particle be at P distant 'x' from the initial point. E at that point has the direction as shown.



dl vector is directed along the path travelled by the particle. I just used the dot product of two vectors.
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vela
#7
Apr4-11, 04:41 PM
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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 -kπR2 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.
zorro
#8
Apr5-11, 12:27 AM
P: 1,394
Ok.
What is the correct expression?
aim1732
#9
Apr5-11, 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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