# Induced EMF of a rectangular loop should be zero?

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

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:

**EDIT**:

Examples of induced $epsilon$ canceling out:

A –

B –

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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## Answers

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 possibleto change the magnetic flux through it. This is what we observe in superconductors.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 $mathcal E = -frac{dPhi_B}{dt}$, 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.

The EMF through the loop is equal to $-frac{dPhi_B}{dt}$ 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?

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.