# What force causes the induced EMF of a loop? And, the difference between a loop EMF to motional EMF?

If a loop of wire is placed perpendicular to a changing magnetic field, there is an induced EMF such that the induced current is flowing in a direction to create an opposing magnetic field like this simple diagram:

While a rod(or wire) moving in a uniform magnetic field experiences a magnetic force pushing the positive charges upwards to create an EMF like this diagram:

Questions:

1) The rod has a magnetic force due to it's motion in $B$ that causes the induce EMF, for a changing magnetic field perpendicular to a loop (like the first diagram) is there a magnetic force as well creating the induced EMF? If not, what is the force(s) causing the induced EMF?

2) For the motional EMF of a rod, by increasing the velocity how does it increase the induced EMF? The expression $\epsilon = -vBL$ is simple but imagining it is confusing. By having more charges? For a stronger E field?

3) If the rod is moving in a non-uniform magnetic field is it considered a varying B field? Is the formula for motional EMF valid? My approach is to calculate $B$ at certain points of $x$ and use the formula. Is this valid?

4) This relates to the first question, for this diagram:

A loop moving in a uniform $B$, has an induced EMF = 0, because of EMF's of each side of the wire cancels out like so:

Also they are in series, for me it's easier to imagine those wires as battery sources all of equal potential opposing one another. Why isn't this similar to the first diagram Of a varying $B_{\bot}$? The rectangular loop moving in a magnetic field has zero EMF, while a loop in a changing magnetic field has a non-zero EMF, why aren't the sides canceling out?

1. The case of a moving conductor and a stationary conductor is fundamentally different. When the conductor is stationary, a changing magnetic field produces an electric field everywhere in space, whose curl along any loop enclosing the varying magnetic field is non-zero, given by $$\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}$$. Using Stokes law, we easily find the emf to be the rate of change of flux through the loop. NOTE: it is the electric field produced in space that is the cause of emf(in this case) not any force.
1. The emf across an open ended conductor(suppose the ends are A and B) is ONLY due to the conservative electrostatic field produced by the separated charges, due to the magnetic force, given by the integral of the electrostatic field over the length of the conductor for A to B : $$\mathcal E=-\int \mathbf E_s\cdot d\mathbf l$$. As the conductor speeds up, due the greater magnetic force, greater charge is deposited at the ends, resulting in greater magnitude of the electrostatic field ($$\mathbf E_s$$) across the conductor, which in turn increases its integral over the length (which is the emf of course!).
2. Again, for an open-ended conductor, we take the electrostatic field, which is $$qvB/q = vB$$. Substituting In the formula and integrating, $$\mathcal E = -vBL$$. So, the formula remains valid.
3. You are again confusing the non-conservative electric field $$\mathbf E$$ with the conservative electrostatic field ($$\mathbf E_s$$). When the circuit is closed, the emf: $$\mathcal E=-\int \mathbf E\cdot d\mathbf l$$, where $$\mathbf E$$ is the total field( due to electric as well as electrostatic over the whole loop). The electrostatic part is obviously zero over the whole loop (because it is conservative), but the electric field is given by Maxwell’s Equation: $$\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}$$. For constant $$\mathbf B$$, $$\nabla\times\mathbf E=0$$, which from Stokes law suggests that its integral around the loop is also zero, producing zero emf. For varying magnetic field, the integral of electrostatic field vanishes anyway, but $$\nabla\times\mathbf E$$ is not zero, which renders the emf non zero in the integral. Do not bring in electrostatic field in emf in a closed circuit. You’ll confuse yourself.