Derivation of Faraday's law using Lorentz force?

For a stationary loop of wire with a time varying magnetic field $\vec{B}(t)$ through it. Is it possible to derive Faraday's law, using only the Lorentz force

$$\vec{F}=q \vec v\times \vec B \tag {1}$$

in some way? The way I would go about doing this is by finding the work done around on a pacticle as it made one full loop around the wire, and by defintion this is the induced emf in the wire. As follows:

$$\mathcal E =\oint{q(\vec v \times \vec B)\cdot d \vec l} \tag{2}$$

But I don't see how we can then turn this directly into flux, of in fact why it is not $0$ since $\vec v$ should, for a particle moving around the loop, be always perpenduclar to $d \vec l$ and thus the integral goes to 0. I assume that we need to introduce the electric field created by the time varying magnetic field for this to work. Is there any way of doing this without giving a circular argument? Or is there another way to derive Faraday's law for a stationary wire in a time varying magnetic field?

• My answer to your question is only partial, as the policy of this site is not to solve completely a question looking like a home exercise, and the rule is quite strict. So, I'll try to help you up to a point wherefrom I rely that you'll find your way on. – Sofia Feb 26 '15 at 21:58
• This question is of course not a home exercise...It involves deep analysing power. Hats off to Q S. – Ayan Biswas Apr 21 '16 at 7:49

With a stationary loop of wire and field $\vec B$ varying in time it doesn't seem that you'd be able to exploit the Lorentz force, because it's difficult to introduce the velocity $v$. But another configuration may be more helpful, see picture. A metallic rod (orange) crosses the magnetic lines (light-blue) with a constant velocity $v$ (blue) in the shown direction. It indeed moves the electrons in the rod as your formula $(1)$ says. The rod glides on two metallic tracks (light-gray) which are connected through another rod (light-gray) to form a closed circuit.

Although the field $B$ is constant, the magnetic flux $\Phi$ varies in time, because

$$\Phi_B = \int_{\scriptstyle_S} \mathbf{B} \cdot \text d\mathbf S \tag{i}$$

In our case is the surface $S$ is rectangular, so we can write the flux as $\Phi_B = B S = B L D$, where $D$ is the length of the rod, and $L$ the length of the metallic tracks on which the rod glides.

For the electromotive force we are interested in the flux derivative with time

$$\mathcal E = -\frac {\text d\Phi_B}{\text d t} = v B D. \tag{ii}$$

because $\text dL/dt = -v$.

Now, I believe that my formula $\text {(ii)}$ and your $(2)$ begin to resemble one another. What misses in my $\text {(ii)}$ is the charges $q$, and what has to appear in your $(2)$ is the rod length $D$, which probably is connected with the element of distance $\text d \vec {\ell}$. Also, since in our case the force $\vec F$ in your $(1)$ is along the rod, and the rod velocity $\vec v$ is perpendicular to $B$, the three vectors $\vec F$, $\vec v$ and $\text d \vec {\ell}$ are mutual perpendicular, s.t. the fact that in my $\text {(ii)}$, the vector product does not appear, is not a concern.

At this point I leave the issue to you. I suggest you to think what is the relationship between work $W$ and potential difference $\mathcal E$, and what is the connection between the integration over the loop in your $(2)$ and the quantity $D$ in my $\text {(ii)}$. Hint: in the grey bars no electromotive force is produced.

For a thin wire, with the charges staying inside the wire, the instantaneous change in magnetic flux is equal to the line integral of the Lorentz Force around the instantaneous circuit (which is actually different than the work done as a particle moves about a wire through time). However, the magnetic part of the Lorentz Force is entirely due to the motion of the circuit through the instantaneous $\vec{B}$ field, and the electric part of the Lorentz Force is 100% related to the instantaneous flux of $\partial \vec{B} /\partial t$ through an area instantaneously spanned by the circuit.

Thus if your circuit is stationary, your integral of $q\vec{vV}\times\vec{B}$ is going to give you zero since the velocity is either (thermally) random (so averages out over the many charges) or along the direction of the wire (why we are considering a thin wire), so this vector, $q\vec{v}\times\vec{B}$, is orthogonal to the wire.

Avoiding circularity is hard, since it is actually the circulating $\vec{E}$ that causes the change in $\vec{B}$. And there is a circulating $\vec{E}$ even inside the areas spanned by the circuit (far from the wires). if you want a completely rigorous derivation, you could consider the source of the $\vec{B}$ field, so you can see what causes the $\vec{B}$ to change. For instance, wrap a solenoid around your circuit. To get a current through your solenoid you need to apply a circulating electric field. And then the causality is clear.

But then it becomes a mutual induction problem.