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Considering the shape of a rectangular loop in a changing magnetic field:

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The induced $\epsilon$ would be zero? Since a rectangular loop is a combination of wires in series to create such a shape. Each wire in this loop induces $\epsilon$ opposes the other, and they should each cancel out?

Here is the diagram adjusted with polarities:

enter image description here

EDIT:

Examples of induced $\epsilon$ canceling out:

A -

enter image description here

B -

enter image description here

Where there are two separate conductors that are wired in series together, each in the same magnetic field, that experience the same flux change over the same time period.

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If you are moving your rectangular loop in a constant magnetic field. You will have zero E.M.F. Each rod of the rectangle would have an E.M.F if it was moving independently in the magnetic field.

Any closed conducting loop will never have an induced current when moving in a uniform magnetic field.

The only other way to induce an E.M.F in the circuit (or drive a current) is by changing the magnetic field or by changing the area so as to cause a net change in flux.

$$\mathcal E = -\frac{d\phi_B}{dt}$$

A changing magnetic flux will induce a closed non-conservative electric field which drives the current.

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The induced ϵ would be zero?

Assuming the magnetic field is out of the page, the emf is simply

$$\mathcal E = -\frac{dB}{dt}xy$$

For an increasing $\vec B$, the induced electric field (and resulting current) is clockwise 'round the loop.

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The EMF through the loop is equal to $-\frac{d\Phi_B}{dt}$ i.e., the rate of change of magnetic flux. If the magnetic field is changing, then the EMF is non-zero. You can't really calculate the EMF for an open stationary conductor if only the $\vec{B}$ field is changing. You need to use Faraday's law. How do you find an EMF for a single straight piece of wire?

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  • $\begingroup$ Edited and added a few more examples where I think would have the induced-emf cancel out since they are wired in series. Could you please give me you're thoughts about them. $\endgroup$ – Pupil Dec 7 '14 at 22:33
  • $\begingroup$ EMF around a loop is equal to rate of change of flux through the loop. You can't talk about the change in flux through/near a wire. It has to be a closed loop. $\endgroup$ – lionelbrits Dec 7 '14 at 22:51
  • $\begingroup$ Well you have two "bars" in examples A & B, they experience a changing magnetic field, both connected together forming a circuit. This is a closed loop(somewhat). If those bars induced equal $\epsilon$ and are wired in series they should cancel each out,therefore, no current flow? $\endgroup$ – Pupil Dec 7 '14 at 23:11
  • $\begingroup$ A change in B may cause current to circulate in those bars, but does not affect the potential between one end and the other. I suggest that you carefully look at how emf is defined. The only loop you need to worry about is the big one. The flux through the loop depends on the number of field lines surrounded by the loop $\endgroup$ – lionelbrits Dec 7 '14 at 23:49
  • $\begingroup$ E.M.F does not require the circuit to be closed. A battery has E.M.F without being present in a closed circuit. When you move a conducting rod in a magnetic field, you get E.M.F due to charge separation. $\endgroup$ – Yashas Feb 13 '17 at 14:50
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You can only calculate the EMF induced from a close loop, instead of "a conductor" as you illustrate in your example A and B.

We know $\mathcal E = -\frac{d\Phi_B}{dt}$, where $\Phi_B=BA$.

EMF induced in a "conductor" makes no sense because we cannot find the surface $A$ to calculate the magnetic flux $\Phi_B$ through the conductor.

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  • $\begingroup$ E.M.F does not require the circuit to be closed. A battery has E.M.F without being present in a closed circuit. When you move a conducting rod in a magnetic field, you get E.M.F due to charge separation. $\endgroup$ – Yashas Feb 13 '17 at 14:51
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I think your problem comes from the fact that a 'wire' is an idealized conductor with zero resistance. A loop formed by such wires does inded not allow for any voltage drops to appear.

Any attempt to induce a voltage (such as changing the magnetic flux) would immediately create an induced current, which in turn would cancel any change in flux.

Conclusion: In an idealized loop such as you have presented, there would indeed not appear any emf, simply because it is not possible to change the magnetic flux through it. This is what we observe in superconductors.

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