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

AB·dA = ΦB  is the flux of B through the area enclosed by the curve Γ.

∂/∂t∫AB·dA = ∂ΦB/∂t 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.

Γ E∙dr is the work done per unit charge in moving a test charge once around the curve Γ.

Faraday's law states that the absolute value or magnitude of the circulation of the electric field E around a closed loop is equal to the rate of change of the magnetic flux through the area enclosed by the loop.  The equation below expresses Faraday's law in mathematical form.

B/dt (through a fixed area) = -∫around loopE·dr (at a fixed time)

When the magnetic flux through the area enclosed by the loop changes, ∫around loopE·dr is not zero, the electric field E circulates.
E
∙dr is the work done per unit charge by the electric field in moving the charge a distance dr.
If the loop is an actual wire loop, then there is actual work done by the induced field on free charges.
around loopE·dr is the work per unit charge by the field in moving the charge once around the loop.
This is an induced emf, and it is measured in Volts.

B/dt (through a fixed area) = induced emf

If the curve Γ is the curve traced out by a wire loop with resistance R, then a current I = emf/R will flow in the wire.
The induced emf causes a current to flow without a potential difference due to separated charges.

The induced electric field is NOT a conservative field.  When you move a charge against the induced field once around the loop, you have to do work.  But your work is NOT stored as potential energy.  You cannot let the electric field do work to recover the energy you expended in moving the charge.  The induced electric field disappears as soon as the magnetic flux is no longer changing.  The work you do on a charge against the induced field is not locally stored.  The energy may be transported away in the form of an electromagnetic wave.  Electromagnetic waves carry energy through free space.

### What is the direction of the dynamic (induced) field?

There is an easy way to remember this direction.  The circulation of the induced 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 produce it.
This is called Lenz's law.

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

#### Example:

A magnet is moved quickly towards a wire loop as shown.
The flux through the wire loop is increasing in the downward direction.
A current starts flowing in the loop in the direction indicated by the arrow.
The magnetic field produced by this current points upward, it opposes the flux changes that produce it.

#### Problem:

Consider a flat square coil with N = 5 loops.
The coil is 20 cm on each side, and has a magnetic field of 0.3 T passing through it.
The plane of the coil is perpendicular to the magnetic field: the field points out of the page.
(a)  If nothing is changed, what is the induced emf?
(b)  The magnetic field increases uniformly from 0.3 T to 0.8 T in 1 s.  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 coil has 5 turns.  Each turn has area A = (0.2 m)2.  The initial magnetic flux through each turn of the the coil is Φ0 = B0A = 0.3*(0.2)2 Tm2 = 0.012 Tm2.
The final magnetic flux through each turn of the coil is Φf = BfA = 0.8*(0.2)2 Tm2 = 0.032 Tm2.
The total change of the flux through the coil N(Φf - Φ0), with N = 5.  The induced emf is
emf = -N∆Φ/∆t = -N(Φf - Φ0)/∆t = [-5*(0.032  -0.012)/1.0] V = -0.1 V.
(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 long 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 current produces 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 long coil is B = μ0(N/ℓ)I.
• The flux through the coil is NBA = μ0(N2/ℓ)IA.
• The change in flux per unit time is μ0(N2/ℓ)A ∆I/∆t = L*∆I/∆t, since I is the only quantity changing with time.
L = μ0(N2/ℓ)A is called the self inductance of the coil.  The units of inductance are Henry (H).  1 H = 1 Vs/A.
• The induced emf is emf = -L*dI/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 emf = -L*∆I/∆t.  The self inductance L depends only on the geometry of the circuit.

#### Problem:

A coil has an self inductance of 3 mH, and a current through it changes from 0.2 A to 1.5 A in a time of 0.2 s.  Find the magnitude of the average induced emf in the coil during this time.

Solution:
L = 3 mH, ∆I/∆t = (1.5 A - 0.2 A)/0.2 s = 6.5 A/s.
emf = -L*∆I/∆t = -(0.003 Vs/A)(6.5 A/s) = -0.0195 V.
The minus sign indicates that the induced emf opposes the flux changes that produced it.

#### Problem:

A 25 turn circular coil of wire has a diameter of 1 m.  It is placed with its axis along the direction of the Earth's magnetic field (magnitude 50 microT), and then, in 0.2 s, it is flipped 180o.  What is the average emf generated?

Solution:
emf = -∆ΦB/∆t. ΦB(initial) = NAB = 25*π*(0.5 m)250*10-6 T = 9.82*10-4Tm2.
ΦB(final)  = -ΦB(initial) since the coil is flipped.
|∆ΦB| = 2ΦB(initial).
|∆ΦB/∆t| = 2*(9.82*10-4 Tm2)/(0.2 s) = 9.82*10-3 V.

#### Problem:

A 0.5 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field.  Find the magnetic field strength needed to induce an average emf of 10,000 V.

Solution:
emf = -∆ΦB/∆t.  ΦB = NABcosθ changes from NAB to 0 in 4.17 ms, since θ changes from 0 to 90o in 4.17 ms.
|∆ΦB| = NAB = 500*π*(0.5 m)2*B = (393 m2) * B.
We want
|emf| = |∆ΦB/∆t| =  (393 m2)/(4.17*10-3 s) * B = (94174 m2/s)*B = 10000 V.
B = 0.1 Vs/m2 = 0.1 T.