The TRUTH about EMF

Well, the article isn't about anything new in Physics, what ever I say here is as old as electricity & magnetism is. The question is whether EMF [The Electro Motive Force (emf), is the external work  per unit charge to produce an electric potential difference across two open-circuited terminals.] is CONSERVATIVE OR NON CONSERVATIVE ?

Obviously, conservative and I will not object you.Well what if it's not conservative in a specific situation, How can one explain this paradox?  Who can explain it? Read and find out...

Before you begin, you should have a sound knowledge about  the difference between a conservative and non conservative forces,Kirchhoff 's Laws,Ohm's law,Faraday's laws.

So lets begin, is emf a conservative or non conservative force?

Still Yes, I don't really think so.

We will use here all the known laws like Kirchhoff's law, Faraday's laws, Ohm's law etc. to derive our conclusions regarding the conservative or non conservative nature.

CASE -1

Circuit Diagram Illustrating the CASE - 1

Consider a simple circuit as shown:  100% ideal (no voltage drop because of connecting wires & voltmeters)

We assume the following,

R₁ = 100Ω,  R₂ = 900Ω

EMF of the cell = 1 V, 

A & D are 2 points on the circuit 

V_D – V_A  Denotes the Potential drop across R₂

_{VA – VD} Denotes the Potential drop across R_{1 & Cell}

We get,

_{Effective Resistance,}  R = R₁ + R₂ = 1000Ω

_{Using Ohm’s Law,  V = IR,  [V = 1v]}

                                => I = 0.001 A

Using Kirchhoff’s Laws, (Loop rule- algebraic sum of all the potential drop & emf in a closed loop is zero) (direction on loop assumed in clockwise direction i.e. from A to D to A)

V_D – V_A = -IR₂             (the reading given by the voltmeter on right) = -0.9 v

V_A – V_D = E + (-IR₁) (the reading given by the  voltmeter on left) = 1- 0.1 = 0.9 v

Hence, V_D – V_D         (obtained by adding above 2 equations)       = 0.9 - 0.9 = 0  

This is exactly what Kirchhoff’s loop rule says, E + (-IR₁) + (-IR₂) = 0  => 1 - 0.1 - 0.9 = 0

* Note that the sign of IR₁ & IR₂ is negative, 

  which corresponds to the potential drop.

  It’s negative because we took clockwise 

  direction for applying Loop rule        *

We thus conclude, the potential or the emf along the closed loop is zero,

which means emf is conservative,as the total work done in a closed loop is zero

is a definite integral over a closed loop .i.e. it depends only on the initial and final values. which again conforms that emf is conservative according to Kirchhoff’s Laws.

 Now let’s go to the second half of the drama,

CASE-2 

The cell's role is now going to be played by a current carrying solenoid ( or a bar magnet) . This solenoid when moved in & out of the loop (which comprises of the 2 resistance,see fig) induces a current in the circuit of whose direction is given by Lenz’s law.

The New circuit diagram looks something like this,

Circuit illustraing the CASE -2 

The Central circle represents the solenoid moving in and out .

The rest everything unchanged.

R₁ = 100Ω,  R₂ = 900Ω

A & D are 2 points on the circuit 

V_D – V_A  Denotes the Potential drop across R₂ 

_{VA – VD} Denotes the Potential drop across R₁ 

_{*REMEMBER, THERE IS NO CELL THIS TIME IN THE CIRCUIT *}

We assume that at an instant of time the value of induced current in the wire due to solenoid to be 0.001 Ampere (same as earlier).

   V_D – V_A = - IR₂  (the reading given by the voltmeter on right) = -0.9 v

   V_A – V_D = (-IR₁) (the reading given by the voltmeter on left)  = -0.1 v

   Hence, V_D – V_D   (obtained by adding above 2 equations) = -0.9 - 0.1 = +1 v 

Well Surprising ?????  the total potential drop is not ZERO.....

_{A violation of Kirchhoff’s Laws…let me see...yes the algebraic sum of potential drop is NOT ZERO. Does it mean, emf in this case is non conservative…is it path dependent?}

_{Yes this emf is non conservative in this situation.}

.…….. WHO CAN EXPLAIN THIS CONTRADICTION NOW?

Faraday CAN …….This is not a violation of Faraday’s Laws. This is in accordance with faraday’s laws which states that: 

Where

Is the electromotive force

Φ_B is the magnetic flux

The equation says a changing magnetic field induces an emf in the loop. The changing magnetic field is brought by the help of the solenoid...and the loop is the wire.

The EMF induced in an electric circuit always acts in such a direction that the current it drives around the circuit opposes the change in magnetic flux which produces the EMF (Lenz’s Law).

But how is this paradox explained, well according to Faraday’s laws, when ever there is a rate of change of the flux linked with the coil, an EMF is induced in the coil (here the coil, being the connecting wire of the circuit) or loop and the magnitude of that emf is here the 1 volt which we obtained just a few minutes before [V_D – V_D is the closed loop].

But the  question still remains, is emf conservative (which we see in the 1^st case) or non conservative (which we see in the 2nd case)?

THE ANSWER YOU HAVE BEEN WAITING FOR : 

Induced electric field is produced not by static charges (the cell here) but by changing magnetic flux (the solenoid here). Although electric fields produced in either way exert SAME FORCE on charged particles, there is an important difference between them (which we normally forget or unware of). The simplest evidence of this difference is that the field lines of induced electric fields form CLOSED LOOPS, but field lines produced by static charges (the cell here) NEVER DO SO but must start on positive charges and end on negative charges.

In a more formal sense, we can say that:

ELECTRIC POTENTIAL (emf) HAS ONLY MEANING FOR ELECTRIC FIELDS THAT ARE PRODUCED BY STATIC CHARGES (the cell here); IT HAS NO MEANING FOR ELECTRIC FIELDS PRODUCED BY INDUCTION.

is zero, i.e. conservative only for emf produced by static charges. However if changing magnetic flux is present, the integral is not zero but is:

Assigning electric potential (emf) to induced electric fields leads us to a contradiction… we must conclude that EMF HAS NO MEANING FOR ELECTRIC FIELDS ASSOCIATED WITH INDUCTION or in another sense… we can say that…INDUCED EMF IS NON CONSERVATIVE AND STATIC EMF IS CONSERVATIVE. 

Thank You for your valuable patience for reading it, hoping it would be valuable for someone.

With this I conclude

THANK YOU

RAHUL M                                                                                                         

 CREDITS: Friends, Prof Walter Lewin  (MIT Phy DEPT.), Prof P.V Thomas  (Girideepam,HOD, Phy DEPT.), 

Books: Fundamentals of Physics By Halliday Resnick & Walker, 

               NCERT PHYSICS Part 1 (CLASS 12)

P.S :   I have uploaded the same article (old edition) online using www.docs.com .You can download it for FREE 

           THE DIRECT LINK (MS WORD 2007 FILE) : http://docs.com/A00B

           THE DIRECT LINK (PDF FORMAT FILE) : http://docs.com/A00M