If a rectangle loop would move inside a magnetic field, would the induced ϵ be zero?
Why would it? Is it due to the induced ϵ canceling out? Being in series and in opposition?
If it we're a single wire it would have an induced ϵ = −vBL.
Why would induced EMF be zero
electromagnetism
Related Solutions
Any time there's a change in the flux in the coil for whatever reason there will be an induced EMF in the coils.
A loop of wire, with the ends not touching nor connected to anything else, is this:
simulate this circuit – Schematic created using CircuitLab
If you look at the schematic, it even makes intuitive sense, graphically. As you can see at the gap between the capacitor "plates", the "wires" are not touching. That the plates (which are really just the wires that make up your loop) are really small and far apart doesn't make it not a capacitor: it just makes the capacitance very low, some picofarads at most.
Normally it's so low we can ignore it, but in this case, there's nothing else, so it's pretty key. When the magnetic flux through the loop changes, an EMF is induced just as you'd expect. Some current does flow, according to the governing laws of a capacitor:
$$ I(t) = C\frac{\mathrm{d}V(t)}{\mathrm{d}t} $$
Of course, with C = 1pF, the current will be very low indeed, and if the EMF isn't changing (\$\mathrm{d}V(t)/\mathrm{d}t = 0\$), then there is no current.
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Best Answer
Yes, it would be zero. According to Faraday's Law the induced EMF would be zero because by moving the loop in that way the magnetic flux would not change.
For an EMF to be induced the magnetic flux must change. This could happen if you spinned the loop so that the angle between \$ \vec{B} \$ and the normal to the loop would change.
But if the magnetic field itself was nonuniform then by moving the loop through areas with different intensities you could induce an EMF.
Edit:
A single wire moving in a magnetic field could have a current, which is explained by Lorentz force: because the wire is being moved so are its charges with velocity \$ \vec{v} \$, and so a force is applied on those charges, hence the current.
The same principle applies to the loop in the picture. Since \$ \vec{v} \$ is to the left and \$ \vec{B} \$ points out of the screen, \$ \vec{v}\times\vec{B} \$ will point down. Thus the current should have that direction (the electrons move in the opposite direction, i.e. up). But now notice both the wire on the left and on the right have currents that cancel out (they have the same absolute value but point opposite ways). So the resulting current is null.