# Two questions about the nature of induced emf

I have two questions about magnetic induction (I am in university level introductory E and M so maybe my questions will be answered over the next few years):

In the linear generator (shown below), I understand that a changing magnetic flux causes an induced emf, which causes a current to flow, and therefore a magnetic force opposing the hand pulling the wire. The magnitude of the opposing force is $\frac{B^2L^2v}{R}$.

Then the book discusses power output, by assuming constant speed (ie. the force of the hand is also the above so that v stays constant and no net force). You get a power = Fv from the hand, and you get a power output from the resistance. These turn out to be equal to $\frac{(BLv)^2}{R}$, so the book concludes that conservation of energy is at hand. But doesn't the opposing magnetic force also have a power output? We've only equated the force of the hand and the resistance of the loop but haven't said anything about the power caused by the magnetic force opposing the motion.

My second question is this: The book imagines a large region of uniform magnetic field that is changing with time. Then an imaginary loop of radius R is drawn in space. By using Faraday's law and some symmetry we work out that the induced electric field at any point on that loop must be $E=\frac{R}{2}\frac{dB}{dT}$, directed tangentially to the loop. However, if we draw another identical loop tangent to that first one, we get the same magnitude but in the opposite direction. Which emf is correct?

Please accept this poorly GIMPed image: • Both emfs are correct, you are running trough the loop in the opposite direction, i.e. your area vector points in the opposite direction. With regards of the power required to move the wire, the book neglects the changes in the magnetization of the magnet's material due to the induced current. That's an allowed simplification for the purposes of the example. If you ever sign up for the design of electric machines in a company that manufactures them, then you will not be allowed to neglect those energy terms. – CuriousOne Jun 18 '15 at 0:44
• Both E vectors point clockwise though, so why would the area vectors be opposite? – Faraz Masroor Jun 18 '15 at 1:25
• If you are merely translating the loop, then the sign of the will be the same. The only way you get the opposite sign is by switching the direction which you use to measure the potential. I hope you are aware that emf depends on the loop, it's not a property of the field alone? You are dealing with a non-conservative field here. – CuriousOne Jun 18 '15 at 5:48

First, keep in mind that many textbooks give incorrect definitons. An EMF is not work done per unit charge. An EMF depends on your loop, and it is the force per unit charge dotted with the instantaneous direction of the wire. Whereas work would require you dot with the direction the charge is going.

To avoid confusion, call the direction of a little bit of thin wire, the direction of the wire. And call the direction of motion of the entire loop the direction of the loop.

For a moving loop, charge must have a component of velocity orthogonal to the wire direction to stay in the loop. For a stationary loop, the charge can go entirely in the direction of the wire. So when you computed work for the stationary loop, it equalled the EMF. But for a moving loop it doesn't.

Next, now that we know about work and EMF we have to consider which forces are responsible for an EMF. They could be electric forces, or they could be magnetic forces.

When the wire is thin and stationary (the loop doesn't move), then the velocity of the charges has to be in the direction of the wire and hence the magnetic force is orthogonal to the direction of the wire, so contributes zero to the EMF, so the EMF is entirely electric. That's what happens to a stationary loop in a non conservative electric field, which is associated with the rate the magentic field is changing in time (but is independent of the current value of the magnetic field).

Now, if the loop is moving, then the charges can have a component of velocity orthogonal to the instantaneous direction of the wire, and hence when you cross that with the instantaneous magnetic field you get a force that does have a component in the direction of the wire. Hence the magnetic force contributes to the EMF so now the EMF could have electric and magnetic forces contribute.

If you want to track how the hand is involved, the hand grabs the solid loop and pulls it, and since there are charges in a solid matrix they are bound to each other and that drags the whole collection of charges bound in that matrix, which I'll call matrix charges (which isnt a standard term, but I worried that bound charges could be confused with another term that is standard). And the loop has more charges than just matrix charges, it has mobile charges too. Imagine four protons on the corners of a square that stay square corner shaped and four electrons that move along the square but are only sometimes at the corners. The matrix charges are like the four protons at the corners and the mobile charges are like the four electrons moving along the sides the of the square. A real wire has more protons, a large number, and the large number of protons have a large number of electrons that stay with them, so only a few of the many electrons are cruising around.

The matrix charges do exert forces on the mobile charges to drag them, and the forces the mobile charges exert on the matrix charges is an additional force the hand needs to overcome in order to move the wire.

A similar situation happens in mechanics (specifically in statics) when you hold one end of a spring and place a mass on the other end of the spring. The spring needs to exert a force on the mass equal and opposite to the weight of the mass to get zero motion of the mass. So the spring feels an equal and opposite force. So the spring feels it's own weight, plus a force equal to the weight of the hanging mass. So when you exert a force on the top of the spring; if you want to exert enough force to keep the spring from falling you need to exert a force equal and opposite to the combined weight of the spring and the mass.

So if you want to have a particular motion of the loop, any forces by your hand on the matrix charges in the loop will have to overcome the force the mobile charges exert on the matrix charges. And those forces are opposite (not really, you'd have to ignore radiation and such) to the force the matrix charges exert on the mobile charges.

If you worry this seems overly complicated, I bring up all these forces only because you wanted to know what is causing what.

Matrix charges do exert forces on charges. So let's start with a stationary example. For instance, when current flows around a corner the mobile charges feel a force from the matrix charges to accelerate around that corner. You have actual surface charge imblances on the outsides of the wire and particularly at the bend of the wire that exert forces on electrons that point in the direction to make them turn the corner, and that have a magnitude such that when you go at the drift velocity you get the force required to turn (you actually also get some from the surface charge to overcome the resistance, but the resistance is just another force from the matrix charges). So the matrix charges exert a force on the mobile charges so they can turn the corner and the equal and opposite force the matrix charges feel would cause them to go the opposite direction but the material is strained in that direction, giving a stress that counters that force. This is like if you magically attached one electron each to each end of a spring. The spring would be stretched a bit so the spring force countered the electrostatic force. In this case the matrix expands a bit so each bit is a feeling stress pushing itself inwards, but they feel an outwards force from the mobile charges, so in the end the mobile charges feel that force inwards and those they can make the turn.

So it's normal for a matrix charge to exert a force on a mobile charge, that's how current turns a corner, and it's how resistance affects current in a resistor too.

Now let's say you have a current carrying loop that is moving. The motion of the mobile charges has a component of velocity in the instantaneous direction of the wire and a component in the direction the loop is going. The motion in the instantaneous direction of the wire crosses with the instantaneous magnetic field to give a magnetic force that is orthogonal to the wire, so the matrix charges oppose that force on the mobile charges, and so the matrix charges feel a force from the mobile charges equal to that force. But it is strained to then feel a force that opposes the force from the mobile charges.

Now, since the loop is moving, the mobile charges also have a component of velocity orthogonal to the wire. This produces a magnetic force that does have a component in the direction of the wire. This is also opposed by the matrix charges. But from the resistive forces. Which means you need to have a current that is high enough to have large enough resistive forces to overcome this magnetic force. Note that some of the matrix charges have the opposite charge, but the force on the matrix charges again is countered by the stress caused by the strain of the solid. The wire expands and compresses to take the shape it needs to exert forces that counter all the net forces from everything else, from the external magnetic field on the matrix charges, from the adjunct electric and magnetic fields due to the mobile charges and the force those adjunct fields exert on the matrix charges. From whatever.

There are also electric forces, and they exert forces on the mobile charges. So the current needs to be high enough that the resistive forces from the matrix charges balance the electric force on the mobile charges.

In short. The mobile charges feel electric forces and magnetic forces. They feel forces from the matrix charges, some are resistive forces based on the current, some are forces that keep the current flowing inside the wire (like the forces that help current turn a corner). The solid feels forces from the mobile charges, and from the electric and magnetic fields, but it is defined (strained) in a way to give stresses on itself to oppose those forces (since the loop is moving with a steady velocity). So imagine a deformed solid wire of protons with a fluid of moving charges inside it, with the deformation of the wire exerting a force on each different part that opposes the net force of everything else on that part of the wire. Except the wire also has many electrons in the solid matrix too, it has everything except the mobile charges.

Now we have a pretty accurate picture (except we are still ignoring radiation and such). Let's compare with Faraday's law, which relates the electric EMF to the current position of the loop and the amount of flux of $\frac{\partial \vec B}{\partial t}$ through the instantaneous position of the loop. And you can compute that the magnetic EMF for a thin wire, with the charges constrained to stay in the wire (and no magnetic monopoles) is equal to the different between the time change of magnetic flux and the flux of the time rate of magnetic flux.

It's like a product rule. You have a magnetic field at two times $\vec B_1$ and $\vec B_2$ and you have two loops with surfaces $S_1$ and $S_2$. And so you can look at how the magnetic flux changes.

$$\frac{\iint \vec B_2\cdot \mathrm d \vec S_2-\iint \vec B_1\cdot \mathrm d \vec S_1}{t_2-t_1}$$ and it approximately equals $$\frac{\iint \left( \vec B_2-\vec B_1\right)\cdot \mathrm d \vec S_1}{t_2-t_1}+\frac{\iint \vec B_1\cdot \mathrm d \vec S_2-\iint \vec B_1\cdot \mathrm d \vec S_1}{t_2-t_1}$$.

And like the product rule you get an exact correspondence when the $\Delta t$ gets small. The change in the flux equals the combination of the flux in the loop now due to the change of the field, plus the change in flux of the field now due to the change in the loop.

Leave the magnetic field constant and change the loop and you get a magnetic EMF.

Leave the loop constant and change the magnetic field and you get an electric EMF.

Move a loop through a changing field and you get an EMF due to electric and magnetic forces.

Even though magnetic fields don't do work, since an EMF isn't about work. It's about the force exerted on the mobile charges in the direction of the wire, and hence about the current level needed to have the resistance force be large enough to overcome the EMF.

As for your second question you can't effectively use Faraday's law to find an electric field if you don't know the direction of the electric field. In this case if you had a giant circular disk region of increasing magnetic fields you expect the electric fields to go in circles around the center. But if you zoomed in to a random small section of the disk and only looked at that small section of the disk, you wouldn't know the direction (are you on the left of center of the disk, or the right of center of the disk. Hence you can't effectively use Faraday's law to find the electric field from your zoomed in picture.

If you had an actual solenoid and you knew the current in the solenoid as a function of time, you could compute the electric and magnetic fields everywhere inside the solenoid. But if you want to be completely accurate you'd eventually be taking into account things like the time delay.

• "Now it turns out that the force opposing the hand is first countered by a hall voltage across the wire that cones from an actual charge imbalance that developed between that leading edge and the trailing edge of the moving element." Are you now writing about the conductor starting from zero velocity and accelerating up to a final constant velocity? – Farcher Apr 28 '16 at 8:35
• @Farcher No. When moving at a constant velocity with a steady current there is a magnetic force from the instantaneous value of the magnetic field. This force is opposed by a Hall voltage. And it's the electric forces that do work. The EMF law requires electric and magnetic forces for the EMF, and there have to be forces that keep the current in the circuit. But I realize the answer was not clear, so I did a total rewrite. – Timaeus Apr 29 '16 at 16:05