Faraday's law

Charges do not have to be at rest or move with constant speed as steady currents.  If we do not restrict ourselves to steady state conditions, two additional terms appear in Maxwell's equations.

(1)
(2)
(3)
(4)

Let us first take a closer look at equation 2.  It is called Faraday's law of induction.

FB = òB×dA is the flux of B through the area enclosed by the curve G.
  is the partial derivative of this flux with respect to time.

Taking the partial derivative means taking the derivative of the flux with respect to time while keeping the area fixed.

is the work done per unit charge in moving a test charge once around the curve G.

In electrostatics this work is zero, because the electrostatic field is a conservative field.  But when we have a changing magnetic field, and the flux of the magnetic field through the area enclosed by the curve G changes, then this work is no longer zero.  In electrodynamics the electric field consist of a static field and a dynamic field, and the dynamic field is not conservative.  The dynamic field is produced by a changing magnetic field.  It encircles any area through which the magnetic flux is changing.  The strength of the dynamic field is proportional to the rate of change of the magnetic flux.

The circulation of the dynamic field is equal to an emf.  It is measured in volts.

If the curve G is the curve traced out by a wire loop with resistance R, then the dynamic field will do work on the electrons in the wire and a current    will flow in the wire.

What is the direction of the dynamic field?

For equations such as , which relate the circulation of some vector field around a curve G to the flux of some other vector field through the area enclosed by G, you can always use the right-hand rule to resolve the directional aspects.  If the fingers of your right hand curl into the direction of ds, i.e. into the direction of the integration, then your right thumb points out the direction for which dA is positive.

For the direction of the dynamic field the following rule therefore applies:

Let n be the normal to the area A through which the flux is increasing.  If the thumb of your right hand points in the direction of n, then the fingers of your right hand curl opposite to the direction of the induced field.  They do not curl into the direction but opposite to the direction of the dynamic field because of the minus sign in equation 2.

There is an easier way to remember the direction of the induced field.  The circulation of the dynamic field is equal to an emf.  Any current flowing as the result of that emf produces a magnetic field that opposes the flux changes that produced it.  This is called Lenz's law.

An induced emf acts as to oppose the change in flux that produced it.

Example:
A magnet is moved quickly towards a wire loop as shown.

lenz.gif (27873 bytes)

The flux through the wire loop is increasing.  A current starts flowing in the loop.  The magnetic field produced by this current opposes the flux changes that produced it.

Problem:

Consider a flat square coil with N = 5 loops.  The coil is 20cm on each side, and has a magnetic field of 0.3T passing through it.  The plane of the coil is perpendicular to the magnetic field: the field points out of the page.

Image1270.gif (1916 bytes)

(a)  If nothing is changed, what is the induced emf?

(b)  The magnetic field increases uniformly from 0.3T to 0.8T in 1s.  While the change is taking place, what is the induced emf in the coil?

(c)  While the magnetic field is changing, the emf induced in the coil causes a current to flow.  Does the current flow clockwise or counter-clockwise around the coil?

Solution:
(a)  An emf is induced by a changing magnetic flux.  If nothing changes, the induced emf is zero.
(b)  The initial magnetic flux through the coil is F0 = B0A = 0.3(0.2)2Tm2 = 0.012Tm2 .
The final magnetic flux through the coil is Ff = BfA = 0.8(0.2)2Tm2 = 0.032Tm2.
The coil has 5 turns. The induced emf therefore is
e = -NDF/Dt = -N(Ff - F0)/Dt = -5(0.032-0.012)V/1.0 = -0.1V
(c)  While the magnetic field is changing, the magnetic flux increased out of the page.  According to Lenz's law, the emf induced in the loop by this changing flux produces a current that sets up a field opposing the change.  The field set up by the current in the coil, points into the page, opposite to the direction of the increase in flux.  To produce a field into the page, the current must flow clockwise around the loop according to the right hand rule.

Self induction

If a coil of wire of cross sectional area A and length l with N turns is connected or disconnected from a battery, the changing magnetic flux through the coil produces an induced emf.  The induced currents produce a magnetic field, which opposes the change in the magnetic flux.  The magnitude of the induced emf can be calculated using Faraday's law.

The magnetic field inside the coil is B = m0(N/l)I.
The flux through the coil is NBA = m0(N2/l)IA.
The change in flux per unit time is m0(N2/l)AdI/dt = LdI/dt.
L = m0(N2/l)A is called the self inductance of the coil.  The units of inductance are Henry (H).  1H = 1Vs/A.
The induced emf is e = -LdI/dt, where the minus sign is a consequence of Lenz's law.

The induced emf is proportional to the rate of change of the current in the coil.  It can be several times the power supply voltage.  When a switch in a circuit carrying a large current is opened, reducing the current to zero in a very short time interval, this can result in a spark.  All circuits have self inductance, and we always have e = -LdI/dt.  The self inductance L depends only on the geometry of the circuit.

Problems:
A coil has an inductance of 2mH, and a current through it changes from 0.2A to 1.5A in a time of 0.2s.  Find the magnitude of the average induced emf in the coil during this time.
Solution:
L = 3mH, dI/dt = (1.5-0.2)A/0.2s = 6.5A/s.
e = -LdI/dt = -(0.3Vs/A)(6.5A/s) = -0.0195V.
The minus sign indicates that the induced emf opposes the flux changes that produced it.
A 25 turn circular coil of wire has a diameter of 1m.  It is placed with its axis along the direction of the Earth's magnetic field (magnitude 50mT), and then, in 0.2s, it is flipped 180o.  What is the average emf generated?
Solution:
e = -dF/dt. F = NAB = 25p(0.5m)250´10-6T = 9.82´10-4Tm2.
(dF/dt) = 2(9.82´10-4Tm2)/0.2s = 9.82´10-3V.

Exercise  (You can earn up to 5 points extra credit by completing this exercise.)

Link:

Induction exercises

Motional emf

A neutral, straight, conducting wire contains equal amounts of positive and negative charges.  However, the electrons are free to move inside the wire, while the positive nuclei are not.  

If a straight conducting wire is placed in a plane perpendicular to a uniform magnetic field, and is moving in a direction perpendicular to the field, then each charge q in the wire experiences a magnetic force of magnitude F = qvB.  The negatively charged electrons will accelerate in response to this force.  Since they cannot leave the wire, negative charge will accumulate on one end of the wire, while positive charge will be left behind on the other end.  The separated charges produce an electric field, which exerts a force on the other charges in the wire.  This electric force opposes the magnetic force.  Once the electric force is strong enough to cancel the magnetic force, electrons will no longer accelerate, and their net motion will stop due to the resistance of the wire.  We then have qvB = qE.  The electric field in the wire is equal to E = vB.

If we place the wire on a conducting rail, a current will start to flow in the circuit formed by the rail and the wire.  

The emf driving the current is equal to vB times the length d of the section of wire connecting the rails. (The work done per unit charge (voltage) is vBd, when a charge moves from one end of the moving wire to the other end.)  The current flowing in the circuit will be I = vBd/R, where R is the resistance of the circuit.

Problem:

In the figure below, assume that R = 6W, d = 1.2m, and a uniform 2.5T magnetic field is directed into the page.  At what speed should the bar be moved to produce 0.5A in the resistor?

Solution:
From above, I = vBd/R.  Therefore v = IR/Bd = 0.5A×6W/(2.5T×1.2m) = 1m/s

Link:

Motional EMF

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Induced emf

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