I am confused what happens in a coil when flux linkage is changed. What I know about EMF is:
Potential difference between terminals of a battery before it is connected to a circuit.
Is this valid for induced EMF also?
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$\begingroup$ Welcome on Physics SE :) Please refrain from asking too many questions in one question and further, I am not sure whether you could not maybe go and read up a bit about electromagnetism and then maybe come back with a more specific question ... $\endgroup$– SanyaCommented Oct 17, 2016 at 15:37
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2 Answers
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It is noteworthy to mention an excerpt from Purcell:
The electromotive force was earlier defined as the work per unit charge involved in moving a charge around a circuit containing a voltaic cell. We now broaden the definition of emf to include any influence that causes charge to circulate around a closed path.
More specifically, Feynman writes:
[...] More specifically, emf is the tangential component of the force per unit charge , integrated along the wire once around the circuit. This quantity is equal, therefore, to the total work done on a single charge that travels one around the circuit.
So, emf is basically the work done by any means that circulate the electron around a closed curve.
In electrostatics, the curl of the electric field was zero since they were conservative and hence there is no non-zero circulation; but in general, the fields are not conservative.
Using Lorentz Force, we get \begin{align}\mathscr E_\textrm{induced} \equiv \textrm{Electromotive Force} &=\int_\Gamma (\mathbf E + \mathbf v\times \mathbf B)\,\mathrm d\mathbf s \\&=- \int_{S(t_0)}\, \left(\dfrac{\partial \mathbf B}{\partial t}\right)\bigg|_{t~=~t_0}\,\mathrm d\mathbf a +\int_{\Gamma}\,(\mathbf v\times \mathbf B)\cdot \mathrm d\mathbf s \\& = \left(-\int_{S(t_0)}\, \left(\dfrac{\partial \mathbf B}{\partial t}\right)\bigg|_{t~=~t_0}\,\mathrm d\mathbf a\right)+\left(-\dfrac{\mathrm d}{\mathrm dt}\, \int_{\mathrm dS}\mathbf B(t_0)\cdot \mathrm d\mathbf a\right) \\ &=-\dfrac{\mathrm d\Phi_\textrm{total}}{\mathrm dt}\;.\end{align}
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$\begingroup$ Then what is difference between emf and voltage? $\endgroup$– BeginnerCommented Oct 17, 2016 at 15:28
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$\begingroup$ Scalar Potential doesn't exist for a non-conservative field. $\endgroup$– user36790Commented Oct 17, 2016 at 15:51
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$\begingroup$ @Beginner It's a technical difference based on the source of the thing. As far as their influence on a charge is concerned, they are the same. $\endgroup$– garypCommented Oct 17, 2016 at 17:50
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$\begingroup$ Mafia, I downvoted this because it chucks out a couple of quotes followed by a lump of maths, and it doesn't actually explain anything. $\endgroup$ Commented Oct 18, 2016 at 12:55
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$\begingroup$ So, don't you think OP required the definition which Purcell was explaining his quote clearly followed by Feynman? $\endgroup$– user36790Commented Oct 18, 2016 at 12:57
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The definition of emf in a coil is the amount of work needed to move a unit positive charge around the coil. That is $\epsilon$ = $\int_C E.\vec {ds}$, where C denotes the coil contour. For an electrostatic E, $\epsilon$ is obviously zero as the integral around a closed loop vanishes, but for a non conservative electric field, $\epsilon$ is non zero.
