What is the difference between emf induced in Faraday's law and the potential difference due to electric field?
If we thake a conducting circular loop in a changing magnetic field then we have an emf induced in it.
But in electric circuits the potential difference between the two points of a conducting wire is zero.
Also explain how this emf causes the electric current to flow.
The difference between EMF induced in Faraday’s law and the potential difference due to electric field
electric-circuitselectromagnetic-inductionelectromagnetismpotentialvoltage
Related Solutions
- 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.
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!).
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.
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.
You're right. When you have a conservative $E$ field, you can define an electric potential $V$. And when you don't, you can't define such a potential (just as you can't define a potential energy $U$ for a nonconservative force $F$ - nonconservative forces still do work - but there is no associated potential energy function for such forces).
General Remarks
Electric potential is measured in volts and defined by $$V(x) = -\int_{\mathcal{O}}^x \vec{E}\cdot d\vec{l} \tag{1}$$ A voltage is defined by (in the textbooks) as a difference in potential $$\Delta V = V_b - V_a = -\int_a^b \vec{E}\cdot d\vec{l} \tag{2}$$
However this is too restrictive. In general, anything that takes the form $$\int \frac{\vec{F}}{q}\cdot d\vec{l} \tag{3}$$ can be called voltage and is measured in volts. So yes $\text{(2)}$ above is a voltage and even $\text{(1)}$ is a voltage if you like (as $V(x)$ is secretly $V(x) - V(\mathcal{O}) = V(x) - 0$ and therefore a difference in potential). But note that $\vec{F}$ can be anything. It doesn't have to be a conservative electric force. Again, anything that takes the form of $\text{(3)}$ is measured in volts and can be called a voltage. EMF takes the form $\text{(3)}$ [I explain this further down]. So too does potential difference. This is one reason why EMF and potential difference get mixed up: they take similar forms and hence both are measured in volts and can be called voltage (or induced voltage or whatever). And actually, this is great. If you are doing anything in the lab or talking to engineers, it doesn't really matter whether voltage means potential difference or EMF. It does, but all we really care about is energy. Voltage (equation $\text{(3)}$)is energy. EMF and potential difference/potential are energy. Energy is energy, whether it be EMF or potential difference. Voltage is just a general term for energy. If you are ever in a situation where you don't know whether to say "the EMF is 5 volts" or the "potential difference is 5 volts", just say "5 volts" or "the voltage/induced voltage/whatever is 5 volts" and you are safe.
What is EMF
In order to get current to flow around a circuit, you need some force pushing charges around the wire. Let's call this the driving force $\vec{F}$. EMF $\mathcal{E}$ is defined as
$$ \mathcal{E} = \oint \frac{\vec{F}}{q}\cdot d\vec{l} \tag{4}$$ where the integration is taken around the loop. There are two main forces that drive current around a circuit: a "source" force from say a battery and a conservative electric field which pushes charges around the wire. $\vec{F} = \vec{F}_s + q\vec{E}$. Therefore, $\text{(4)}$ can be written
$$\mathcal{E} = \int_a^b \frac{\vec{F}_s}{q}\cdot d\vec{l} $$ as $\vec{F}_s$ is usually confined to a section of the loop and $E$ is conservative so it integrates to 0 (started where we left off - once around the loop). $\vec{F}_s$ can be anything. It can be a chemical force, some temperature gradient thing, pressure on a crystal, a magnetic force, a nonconservative $E$ field, etc. So consider a battery. A conservative electric field goes from the positive terminal, around the loop, to the negative terminal, as well as from the positive terminal to the negative terminal inside the battery. Using the last equation, assuming the battery is ideal so that the chemical force is equal and opposite to the electric force,
$$ \mathcal{E} = \int_a^b \frac{\vec{F}_{\text{chemical}}}{q}\cdot d\vec{l} = -\int_a^b \vec{E}\cdot d\vec{l} = V$$ The EMF of the battery is equal to the potential difference across its terminals. But this does not mean that EMF is potential difference. It just happens to be so in this case. Most simple circuits turn out to be this way but realize again that EMF and potential difference are totally different. In the first place, you can't have a potential difference without an EMF generating that separation of charge in the battery. EMF generates a potential difference that happens to match the numerical value of the EMF (which you can think of as energy conservation). Then if you have a resistor connected to this battery, current $I = V/R$. I could also say $I = \mathcal{E}/R$ as they are numerically equivalent. But it's more appropriate in my opinion to use $I = V/R$ as the energy drop is coming as electric potential energy in a conservative $E$ field (which exists throughout the wire doing the pushing). Here we begin to see, as in the next section, that All circuits require an EMF to function.
Let's Look at your example
There is no such thing as an electric potential in your example. There's an $E$ field, but it's not conservative. Therefore, don't say potential. There is an EMF however. You can say there's a voltage or an induced voltage if you like (from the above discussion). But there's definitely not a potential difference/potential present. For this specific example, the EMF is given by
$$ \mathcal{E} = -\frac{d\phi}{dt} = \oint \vec{E} \cdot d\vec{l}$$ where the driving force is that nonconservative $E$ field. The current as you say is $I = \mathcal{E}/R$. Here again we see a true instance of the following: all circuits require an EMF. The idea that all points in a conductor are at the same potential is equivalent to saying that there is no $E$ field in a conductor. Note that this idea of there being no $E$ field in a conductor only holds for electrostatics + no time varying external magnetic fields (having a time varying $B$ prevents electrostatics anyways so saying "+ no time varying $B$" was redundant). Having an $E$ field in a conductor is completely fine. Turn on an $E$ field in a conductor. There is definitely an $E$ field in the conductor until electrostatics is reached. This is why conducting wires in simple circuits can have $E$ fields in them. This $E$ field is essential for driving current around even though it's through a conductor. It's just that the conductor can never reach electrostatics when it's part of a circuit. It desperately tries, but the battery prevents the wire from coming to a static situation. And with time-varying $B$ fields, nothing wrong with having an $E$ field in a conductor. When you stop varying $B$, you'll stop changing the $E$ field and things will reach statics. While you are changing the $B$ field, $E$ is changing with time. The conductor is trying to reach statics, but can never do so. So there will always be an $E$ present and hence the conductor won't be an equipotential (albeit, in simple circuits, you can take the wires to be equipotentials because $E$ is so so tiny).
Too Much Theory, What to know about EMF
From equation $\text{(4)}$, because of the closed line integral, EMF does not care about conservative forces while electric potential crucially depends on a conservative E field. EMF and potential are both instances of equation $\text{(3)}$, and therefore both tell you about energy. Potential is energy in a conservative E field. EMF is energy added to your circuit through "nonconservative" driving forces. In order for circuits to work, you need to pump energy into them so that charges will flow back down to low energy, making a circuit. EMF tells you how much energy driving forces give to a unit charge in one trip around the loop. Conservative forces don't give any net energy to a charge after one complete loop (started where you stopped). "Nonconservative" forces will give you some nonzero value to equation $\text{(4)}$. EMF tells you how much energy was added by driving forces and hence how much energy must be dropped by dissipative/"friction" forces in one trip around the loop. An EMF of 5 volts means 5 volts must be dropped by every unit of charge. If you have a battery providing 2 volts of EMF and a changing magnetic field providing 6 volts of EMF, 8 volts must be dropped
[by the way, If you ever see an example circuit out there with both a battery and a changing flux enclosed by the loop, more than likely their derived equations have wrong explanations. Right equation. Wrong explanation (which is basically just as bad as not knowing what you are doing). What they do is say $-d\phi/dt = \oint \vec{E}\cdot d\vec{l}$. This is Faraday's Law, true in anycase. But what actually does that $\vec{E}$ mean? It's the net $E$ field on your loop. In the case of a battery and a changing flux, that $E$ has both a conservative and a nonconservative component. The conservative component integrates to zero, leaving only the nonconservative $E$ providing the $-d\phi/dt$. Therefore, if you have this simple battery + changing flux + resistor circuit, $-d\phi/dt = I_{\phi}R$ where $I_{\phi}R$ is the integral of nonconservative $E$ around the loop. Now we can add a constant to each side of the equation. Since EMF battery $V_0 = I_0R$, we can say $V_0 - d\phi/dt = (I_{\phi} + I_0)R = IR$. Or if you want, you can write out $-d\phi/dt = I_{\phi}R + \oint \vec{E}_{\text{conserv}} \cdot d\vec{l} = I_{\phi}R - V_0 + I_0R$].
Best Answer
Yes this issue often causes confusion. The basic thing you need to think about here is the electric field. In the first instance, remove any conducting wire and just suppose there is a region of space where there is a magnetic field, uniform spatially (i.e. uniform direction and size), but changing with time. Let's say it is increasing. In this scenario there is also an electric field in that region of space. The electric field in this situation runs in circular loops around the magnetic field lines.
Ok so far so good: we have a changing magnetic field, and, in that same region of space, also an electric field running in circular loops.
Now suppose you put a conducting wire in a loop in that same region, following the direction of the electric field lines, but for the moment do not close the circuit. That is, you have a loop of wire but with a gap in it so that it does not close. What will happen?
The electrons in the wire will be pushed by the electric field, and they will move, such that they start to pile up on one side of the gap. This imbalance in the charge distribution in the wire causes a counter-balancing electric field. The electrons keep moving until this counter-balancing electric field (caused by the electrons) is equal and opposite to the one caused by the changing magnetic field. Thus when the system settles down the net electric field inside the conducting wire is zero (I am assuming the rate of change of the $B$ field here is constant).
At this point there is a build-up of negative electric charge on one side of the gap in the wire loop, and a corresponding positive charge on the other side of the gap. Also there is potential difference across that gap: it is equal to the e.m.f. which you can calculate using Faraday's law. So if you were to now connect a resistor or a light bulb or something like that across the gap, then a current would flow.
When a resistor is connected across the gap, there is a potential difference across the resistor and an electric field inside the resistor. There is no electric field inside the conducting wire (if we assume zero resistance of the wire), and the electric field just outside it is also affected by the distribution of charge in the wire.
On electric potential difference
In the above I did not mention the concept of potential difference until near the end. This is because in electromagnetism it is best to regard the fields and the charges as the main idea, and then concepts such as potential difference come in as useful tools to help in calculating and in getting insight.
Electric potential comes into its own, as a concept, in static conditions, because then we can find a function $V(x,y,z)$ such that the electric field can be written as $$ {\bf E} = - {\bf \nabla} V. $$ (This is a standard shorthand notation for a gradient. In terms of components it means: $$ E_x = - \frac{\partial V}{\partial x},\\ E_y = - \frac{\partial V}{\partial y},\\ E_z = - \frac{\partial V}{\partial z}.) $$
In non-static cases, as for example in the presence of a changing magnetic field, things are not so simple, because now the electric field is such that it can point around a loop, and this means there is no function $V$, having a single value at each location, such that $\bf E$ is its gradient. However we can still investigate quantities such as $$ -\int_{P_1}^{P_2} {\bf E} \cdot d{\bf l} $$ where $P_1$ and $P_2$ are two points in space (or possibly the same point) and the integral is taken along a path running between those points. In a static problem, this integral would give the potential difference between $P_2$ and $P_1$. In a non-static problem, we may choose to give another name to the result of this integral. It is often called 'electromotive force' or 'emf' (this is arguably not a very clear name but is adopted for historical reasons). But this is just a name. You could call it a potential difference if you like, as long as you realise that you are really talking about the integral of electric field along a given path. What you know is that if a charge were to move along that path, then the net energy given to the charge by the electric field is minus the amount given by this integral. This can be deduced because the force on such a charge would be $$ {\bf f} = q ( {\bf E} + {\bf v} \times {\bf B} ) $$ and therefore the work done when the charge moves through a displacement $d{\bf r}$ is $$ {\bf f} \cdot d{\bf r} = q {\bf E} \cdot d{\bf r}. $$ Here the magnetic field term does not contribute because for a charge moving along a path described by $\bf r$ we have that $\bf v$ and $d{\bf r}$ are parallel so the magnetic force is perpendicular to $d{\bf r}$.
The main point of this final section of my answer is to say that 'potential difference' and 'emf' are different words for essentially the same thing, namely the integral of electric field along a path. The reason to have two terms is that the first one (potential difference) draws attention to a useful property of static fields, and the second one draws attention to the fact that the situation under consideration is not static so we have to proceed a little more carefully in our reasoning. In particular, for a non-static case we should not assume that there is any function $V(x,y,z)$ (having a single value at each point $(x,y,z)$) which gives the electric field as its gradient.