1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Faraday's law of electromagnetic induction -- Motional EMF

  1. Jun 1, 2017 #1
    There is magnetic field in B = B ##\hat z## in a region from x=0 to x=l.
    There is a metal rectangular wire loop with length l and width w in x- y plane with coordinates of four corners as (0,0),(0,w),(l,0),(l,w). This loop is moved with velocity v=v##\hat x##.
    Now according to Faraday's law of electromagnetic induction,
    there will be an emf and as a result there will be a current in the loop.
    emf ≡ ξ = ##\frac{-dΦ}{dt} = Bwv##
    This ξ is potential difference across which two points?
    I think this is the potential difference across that two points which is just going out of the magnetic field region because in a small region around these two points the Φ is changing.
    Now, applying Lenz's law,
    since here the flux is decreasing, current will flow in a direction which will oppose this change in flux.
    To increase the flux, magnetic field due to this current should be in ##\hat z## direction and for this the current should move in the anti - clockwise direction.

    Now at time t, the two points which are going out of the magnetic field region is (l-vt,0),(l-vt,w).
    The current is going from (l-vt,0) to (l-vt,w). Since the current flows from higher potential to lower potential,the potential at(l-vt,0) is higher than that at (l-vt,w).

    Now it is said that to verify the presence of ξ, we need a loop (so that we can check the current.),but ξ exists independently of the presence of the loop. So, even if there is no loop, there will be ξ.

    Now, the problem is : this ξ wil be across which two points?
    And if there is no loop here, then the flux is not changing.
    In the presence of loop, the flux over the area covered by the loop is changing.

    From the presence of ξ, it is induced that there is electric field and this electric field is known as induced electric field Ein.
    But magnetic field is not changing here. So, what is creating Ein?
    Now the relation between V and E is given as
    V(b)-V(a) = -∫abE.dl
    But the relation between ξ and Ein is given as
    ##ξ =\oint \vec{ E _{in}}\cdot d \vec l ##

    From where do we get this relation? Is it empirical?
    Note that ξ is defined only for a circular path and there is absence of negative sign on R.H.S.,too. Can anyone please explain this?
    Can we say that ξ is defined only in the presence of a loop?

    And so, the following question cannot be answered:
    This ξ is potential difference across which two points?

    But in general, it is said that Ein exists independently of the presence of loop.
    So, if in a region R,I change B, there will be Ein. Will this Ein be
    only in this region R or will it exist in the whole space (of course, decreasing in magnitude as the distance from this region increases)?

    Now , if in this region ,I place a test charge q stationary for sometime , so that magnetic force acting on it is 0, then this charge will experience a force F = q Ein.
    Is this correct?
    Last edited: Jun 1, 2017
  2. jcsd
  3. Jun 2, 2017 #2


    User Avatar
    Homework Helper

    Uhm no! Without a loop, there is no induced emf.

    An emf can be defined by: ##\xi = \oint_C \vec E . \vec{dA}## where C is the loop. Since it is a contour integral, you see that a loop is required.
  4. Jun 2, 2017 #3


    User Avatar
    Homework Helper
    Gold Member

    Across the resistance of the loop. You can model the loop as a conductor loop with a resistance R.
    But the "induced" emf is along the moving length perpendicular to the field. It is non-conservative.
    It is called 'motional emf'. The v×B force on the electrons causes thus emf.
  5. Jun 2, 2017 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    It also follows from the completed (only correct!) integral version of Faraday's law for moving surfaces ##A## with boundary ##\partial A##:
    $$\int_{\partial A} \mathrm{d} \vec{x} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
    Here ##\vec{v}=\vec{v}(t,\vec{x})## is the velocity field of the surface (here needed of course only along the boundary).

    I don't understand, why many textbooks carelessly through a wrong integral Faraday law at their readers instead of giving the really nice correct derivation, which you even find in the Wikipedia:

  6. Jun 2, 2017 #5
    Griffith, electrodynamics

    In page no. 304,
    Griffith defines e.m.f as ξ = ##\oint f.dl ##,
    and he says that e.m.f. is the net effect of the mechanism which drives the current in the circuit.
    And after some calculation it turns out that this e.m.f. is equal to the potential difference across the battery.
    But conceptually e.m.f. and potential difference are two different things.

    In page 305, just above the equation (7.11),he says the following
    If the entire loop is pulled to the right with speed v,
    the charges in segment ab experience a magnetic force whose vertical component
    qvB drives current around the loop, in the clockwise direction.
    This means that it is the magnetic force which pushes the charges up, right?
    And so the magnetic force is doing the work.

    In equation (7.11) he derives e.m.f. as line integral of magnetic force.

    Then, he says that magnetic force does no work. It is the person who is pulling the loop is doing the work on the charge.

    Then, he says that the emf is equal to this work done per unit charge. But, the emf and the work done per unit charge are conceptually different.
    Now, my question is : Is work done per unit charge always equal to the e.m.f. ? I am this because calculating e.m.f. is easier than calculating work done per unit charge.

    ed 309.png While in page 30 9, he uses emf as potential difference for calculating current.

    So, can anyone please tell me the physical significance of e.m.f.?
    I think practically, in any circuit , I can calculate e.m.f. and then take this emf as work done per unit charge i.e. potential difference for further calculation.
    Is this correct?

    Attached Files:

  7. Jun 4, 2017 #6


    User Avatar
    Homework Helper
    Gold Member

    The magnetic force exists only because of the velocity of the rod, which is the result of the person's physical effort. No effort, no velocity and no magnetic force. Hence, it's the mechanical force by the person that does the work here. Energy is not consumed unless there is current in the electrical circuit. Hence, no current, no work. The braking Bil force on the current carrying rod indicates that the energy is being consumed in the electric circuit and and the person must apply an equal and opposite force in order to keep the rod moving with the same velocity.
    Magnetic field acts as a medium for conversion between electrical and mechanical energy.
    Work done per unit charge in order to move it from a to b is the potential difference between a and b. Work done per unit charge in order to move it from +ve terminal of the battery to its negative terminal is the emf.
  8. Jun 4, 2017 #7
    What is it?
  9. Jun 4, 2017 #8


    User Avatar
    Homework Helper
    Gold Member

    The force acting on the rod when a current starts flowing through it. Have you studied Flemming's left hand rule?
  10. Jun 4, 2017 #9
    No, I have heard of it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted