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anantchowdhary
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Using Faraday's law for a closed loop we say that the EMI is -d(phi)/dt .Now this emf is induced in the closed loop...So between which two points is the PD equal to -d(phi)/dt
Thanks
Thanks
anantchowdhary said:Using Faraday's law for a closed loop we say that the EMI is -d(phi)/dt .Now this emf is induced in the closed loop...So between which two points is the PD equal to -d(phi)/dt
anantchowdhary said:hehe...
i just don't understand the points betwwen which the emf is calculated...
tiny-tim said:Does http://members.lycos.nl/AmazingArt/E/artist7.html" help?
anantchowdhary said:um...but while solving problems in case of say a fixed square semi-loop(duno what to call it) and a rod that is free to slide...we take the EMF between the two end points of the rod...now isn't this contradictory to your statement?:O
anantchowdhary said:If we add the Potential drop across two 'infinitesmally' close points,then shoudlnt the PD across the two ends be HUGE?this is as the EMF across those close points is equal to d(phI)/dt
sorry..i know this is a bit irritating
tiny-tim said:Electric PD isn't "conservative", like gravitational PD.
anantchowdhary said:Well...i have a new doubt
Please see this (the figure for the problem)
http://img80.imageshack.us/my.php?image=picjt0.jpg
Suppose that the rod ab moves with a constant speed v.Now there is a uniform magnetic field B..in the loop..
So if we want to know the PD between a and b,as tiny_tim said
we can think of a battery between two infinitesimally close points...the current in the circuit will be Blv/R(where R is the resistance of ab).
Now with two different circuits we get different answers why?i mean...we assumed the emf to be induced between any two infinitesimally close points..
anantchowdhary said:im getting some of this...Could you please elaborate upon how electric field is non conservative here...and is potential difference taken as the integral including time or not...acc to me it is not over time...
thanks
anantchowdhary said:how is possible to create such an electric field using charges?
Yes there are two cases, one which is time-varying and one which is steady state. Both cases are covered by one of Maxwell's equations:anantchowdhary said:This i assume is only for time varying fields...but isn't PD calculated at an instant...by moving a hypothetical test charge...otherwise we can't solve problems by applying ohm's law
therefore
i suppose it is calculated over time also..am i correct?
I assume you mean two very close points both of which are along the length of ab and not a,b themselves. For one thing, I believe this picture may be a little misleading since considering a small segment dx of the ab rod does not take into account that it is a closed loop, which is required for emf to be generated. Think instead of the entire length ab, and the rate at which the magnetic flux through the loop is changed, given by Blv.anantchowdhary said:Please see this (the figure for the problem)
http://img80.imageshack.us/my.php?image=picjt0.jpg
Suppose that the rod ab moves with a constant speed v.Now there is a uniform magnetic field B..in the loop..
So if we want to know the PD between a and b,as tiny_tim said
we can think of a battery between two infinitesimally close points...the current in the circuit will be Blv/R(where R is the resistance of ab).
Now with two different circuits we get different answers why?i mean...we assumed the emf to be induced between any two infinitesimally close points..
As said above, the electric field is non-conservative if it is induced by a time-varying magnetic flux through a loop. The potential difference is defined as the rate at which magnetic flux through the loop is changed.anantchowdhary said:im getting some of this...Could you please elaborate upon how electric field is non conservative here...and is potential difference taken as the integral including time or not...acc to me it is not over time...
thanks
I'm not following you here. My understanding is that the electric field would be induced in the wire which causes a current flow which in turn induces a B-field opposing the increasing B-field through the loop. What do you mean by the electric field "propagating outwards" and that it isn't restricted to the wire?Gear300 said:Lets say you had a circular loop of wire with a magnetic field running through the loop. The magnitude of the magnetic field is constantly changing over time and so an electric field is propagating outward due to the changing magnetic flux. This allows for a voltage to be established throughout the loop along with a current.
If the electric field here was electrostatic, then the electric field would be dependent on position rather than path. But the electric field we're focusing on here is distributed across the wire and depends on the path the wire provides. Otherwise the closed loop integral provided would be 0 in value:
http://upload.wikimedia.org/math/7/6...aed959500d.png
Also note that the electric field isn't restricted to the wire...its propagating outward.
Defennder said:I'm not following you here. My understanding is that the electric field would be induced in the wire which causes a current flow which in turn induces a B-field opposing the increasing B-field through the loop. What do you mean by the electric field "propagating outwards" and that it isn't restricted to the wire?
What is a "non-stationary" loop? If the electric/magnetic flux is time-varying then this itself would induce a time varying magnetic/electric field.anantchowdhary said:Is an induced electric field present in a non stationary loop...but whose flux is changing..wrt time...?
Kirchoff's voltage law still applies here. Although the electric field is non-conversative, the drop in potential across any circuit loop is still 0. The potential difference across the inductor is assigned that value for which the loop pd is 0. See here for a clearer explanation:And how can we apply Kirchoff's Loop Law in an LR circuit as there is a non conservative electric field induced in the solenoid?
You don't use that equation when you're dealing with time-varying fields. Instead the more general circuit law version applies:and can anyone please confirm that is emf defined by
[tex]
\oint \mathbf{E} \cdot d\mathbf{r}
[/tex]
starting from ANY point and ending at the same point on the loop
even in a moving conductor
Thanks
Yes it's non-conservative. You're describing the setup of a sliding rod cutting the field lines of a constant magnetic fields, right?anantchowdhary said:the field inside a conductor which is non stationary but is immersed in uniform and constant magnetic field with changing flux is non conservative is it?
Defennder said:You don't use that equation when you're dealing with time-varying fields. Instead the more general circuit law version applies:
[tex]\sum^{n}_{r=1} V_r = 0[/tex] for any circuit loop.
anantchowdhary said:And how can we apply Kirchoff's Loop Law in an LR circuit as there is a non conservative electric field induced in the solenoid?
and can anyone please confirm that is emf defined by
∫E . dr
starting from ANY point and ending at the same point on the loop
even in a moving conductor
Defennder said:Kirchoff's voltage law still applies here. Although the electric field is non-conversative, the drop in potential across any circuit loop is still 0 …
Hi anantchowdhary!anantchowdhary said:Is there anything wrong in using
∫E . dr
∫E . dr represents the line integral of the electric field (E) along a path (dr) in a moving conductor. This calculation is used to determine the induced electromotive force (EMF) in the conductor.
No, it is not incorrect to use ∫E . dr in a stationary conductor. However, the value of the line integral will be zero since there is no induced EMF in a stationary conductor.
∫E . dr is a key component of Faraday's law of induction, which states that the induced EMF in a closed loop is equal to the negative of the rate of change of the magnetic flux through the loop. The line integral ∫E . dr is used to calculate the induced EMF in the conductor.
No, ∫E . dr cannot be used to directly calculate the current in a moving conductor. However, it can be used to calculate the induced EMF, which can then be used in Ohm's law (V=IR) to calculate the current.
Yes, there are a few limitations to using ∫E . dr in a moving conductor. This calculation assumes that the conductor is moving in a uniform magnetic field and that the path of the conductor is known. It also does not take into account any resistance in the conductor, so the calculated induced EMF may differ from the actual value in a real-world scenario.