Chapter 27: Electromagnetic Induction Farady’s Law Discovery of Farady’s law of induction

Farady’s Law Farady’s law of induction An emf in volts is induced in a circuit that is equal to the time rate of change of the total magnetic flux in webers threading (linking) the circuit: The flux through the circuit may be changed in several different ways B may be made more intense. The coil may be enlarged. The coil may be moved into a region of stronger field. The angle between the plane of the coil and B may change.

Farady’s Law Farady’s law of induction (cont’d)

Farady’s Law Induced electric field Consider work done in moving a test charge around the loop in one revolution of induced emf. Work done by emf : Work done by electric field: emf = 2prE for a circular loop For a circular current loop In general : Faraday’s law rewritten

Lenz’s Law Lenz’s Law Extend Farady’s law to solenoids with N turns: Direction of induced emf and Lenz’s law Extend Farady’s law to solenoids with N turns: Why the minus sign and what does it mean? Number of turns Lenz’s Law The sign of the induced emf is such that it tries to produce a current that would create a magnetic flux to cancel (oppose) the original flux change. or – the induced emf and induced current are in such a direction as to oppose the change that produces them!

Lenz’s Law Example 1 B due to induced current

Lenz’s Law The bar magnet moves towards loop. Example 2 The bar magnet moves towards loop. The flux through loop increases, and an emf induced in the loop produces current in the direction shown. B field due to induced current in the loop (indicated by the dashed lines) produces a flux opposing the increasing flux through the loop due to the motion of the magnet.

Motional Electromotive Force Origin of motional electromotive force I FE FB

Motional Electromotive Force Origin of motional electromotive force I (cont’d)

Motional Electromotive Force Origin of motional electromotive force II v: constant . B Bind

Motional Electromotive Force Origin of motional electromotive force II (cont’d)

Motional Electromotive Force Origin of motional electromotive force II (cont’d)

Motional Electromotive Force Origin of motional electromotive force II (cont’d)

Motional Electromotive Force Origin of motional electromotive force II (cont’d)

Motional Electromotive Force Origin of motional electromotive force III

Motional Electromotive Force Origin of motional electromotive force III (cont’d)

Motional Electromotive Force Origin of motional electromotive force III (cont’d)

Motional Electromotive Force A bar magnet and a loop (again) In this example, a magnet is being pushed towards a closed loop. The number of field lines linking the loop is evidently increasing. There is relative motion between the loop and the field lines and an observer at any point in the metal of the loop, or the charges in the loop, will see an E field Also we have

Motional Electromotive Force Example: A bar magnet and a loop (cont’d) loop For the example just considered, let us see what happens in a small interval dt. The relative displacement υloopdt causes a small area of B field to enter the loop. For a length dL of the loop the ddΦB passing inside is d[(dA)B] = dL υloopdt sinθ B. We can see this as Integrating this expression right round the circuit ( i.e. over dL) shows that this υ×B interpretation recovers Faraday’s law. You will also see that the sign of E is consistent with Lenz’s Law.

Motional Electromotive Force Example: A generator (alternator) top The armature of the generator is rotating in a uniform B field with angular velocity ω this can be treated as a simple case of the E = υ×B field. On the ends of the loop υ×B is perpendicular to the conductor so does not contribute to the emf. On the top υ×B is parallel to the conductor and has the value E = υB cos θ = ωRB cos ωt. The bottom conductor has the same value of E in the opposite direction but the same sense of circulation. B bottom

Eddy Current Eddy current: Examples

Eddy Current Eddy current : Examples (cont’d)

Eddy Current Eddy current prevention

The orange represents a magnetic field pointing into the screen and let say it is increasing at a steady rate like 100 gauss per sec. Then we put a copper ring In the field as shown below. What does Faradays Law say will happen? Current will flow in the ring. What will happen If there is no ring present? Now consider a hypothetical path Without any copper ring.There will be an induced Emf with electric field lines as shown above. In fact there will be many concentric circles everywhere in space. The red circuits have equal areas. Emf is the same in 1 and 2, less in 3 and 0 in 4. Note no current flows. Therefore, no thermal energy is dissipated

Example: A magnetic field is  to the board (screen) and uniform inside a radius R. The magnetic field is increasing at a steady rate. What is the magnitude of the induced field at a distance r from the center? R x r E is parallel to dl Field circulates around B field B Notice that there is no wire or loop of wire. To find E use Faraday’s Law.

Example with numbers Suppose dB/dt = - 1300 Gauss per sec and R= 8.5 cm Find E at r = 5.2 cm Find E at 12.5 cm

Self Inductance Self induction When a current flows in a circuit, it creates a magnetic flux which links its own circuit. This is called self-induction. (‘Induction’ was the old word for the flux linkage ΦB). The strength of B is everywhere proportional to the I in the circuit so we can write L is called the self-inductance of the circuit L depends on shape and size of the circuit. It may also be thought as being equal to the flux linkage ΦB when I = 1 amp. The unit of inductance is the henry

Self Inductance Calculation of self inductance : A solenoid Accurate calculations of L are generally difficult. Often the answer depends even on the thickness of the wire, since B becomes strong close to a wire. In the important case of the solenoid, the first approximation result for L is quite easy to obtain: earlier we had Hence Then, So L is proportional to n2 and the volume of the solenoid

Self Inductance Calculation of self inductance: A solenoid (cont’d) Example: the L of a solenoid of length 10 cm, area 5 cm2, with a total of 100 turns is L = 6.28×10−5 H 0.5 mm diameter wire would achieve 100 turns in a single layer. Going to 10 layers would increase L by a factor of 100. Adding an iron or ferrite core would also increase L by about a factor of 100.  The expression for L shows that μ0 has units H/m, c.f, Tm/A obtained earlier

Self Inductance The magnetic flux inside the solenoid: Calculation of self inductance: A toroidal solenoid The magnetic flux inside the solenoid: Then the self-inductance of the solenoid: If N = 200 turns, A = 5.0 cm2 , and r = 0.10 m: Then when the current increases uniformly from 0.0 to 6.0 A in 3.0 ms, the self-induced emf E will be:

Self Inductance Stored energy in magnetic field Why is L an interesting and very important quantity? This stems from its relationship to the total energy stored in the B field of the circuit which we shall prove below. When I is first established, we have a finite (self-induced emf) The source of I does work against the self-induced emf in order to raise I to its final value. power = work done per unit time

Energy per unit volume in the field Self Inductance Stored energy in magnetic field: Example Returning to our expression for the energy stored in an inductance we can use it for the case of a solenoid. Using formulae we have already obtained for the solenoid and Hence: Energy per unit volume in the field

Self Inductance Inductor A circuit device that is designed to have a particular inductance is called an inductor or a choke. The usual symbol is: a I variable source of emf L b

M2 1—Mutual Inductance of the coils Transformer and mutual inductance The classic examples of mutual inductance are transformers for power conversion and for making high voltages as in gasoline engine ignition. A current I1 is flowing in the primary coil 1 of N1 turns and this creates flux B which then links coil 2 of N2 turns. The mutual inductance M2 1 is defined such that the induction Φ2 is given by M2 1—Mutual Inductance of the coils Also Generally, M 1 2 = M 2 1

Mutual Inductance Changing current and induced emf Consider two fixed coils with a varying current I1 in coil 1 producing magnetic field B1. The induced emf in coil 2 due to B1 is proportional to the magnetic flux through coil 2: f2 is the flux through a single loop in coil 2 and N2 is the number of loops in coil 2. But we know that B1 is proportional to I1 which means that F2 is proportional to I1. The mutual inductance M is defined to be the constant of proportionality between F2 and I1 and depends on the geometry of the situation. The induced emf is proportional to M and to the rate of change of the current .

Mutual Inductance Example Now consider a tightly wound concentric solenoids. Assume that the inner solenoid carries current I1 and the magnetic flux on the outer solenoid FB2 is created due to this current. Now the flux produced by the inner solenoid is: The flux through the outer solenoid due to this magnetic field is:

Mutual Inductance Example of inductor: Car ignition coil Two ignition coils, N1=16,000 turns, N2=400 turns wound over each other. l=10 cm, r=3 cm. A current through the primary coil I1=3 A is broken in 10-4 sec. What is the induced emf ? Spark jumps across gap in a spark plug and ignites a gasoline-air mixture

The R-L Circuit Current growth in an R-L circuit Consider the circuit shown. At t < 0 the switch is open and I = 0. The resistance R can include the resistance of the inductor coil. The switch closes at t = 0 and I begins to increase, Without the inductor the full current would be established in nanoseconds. Not so with the inductor. Kirchhoff’s Loop Rule: Power balance Multiply by I:

Power dissipated as heat in the resistor The R-L Circuit Current growth in an R-L circuit (cont’d) Rate at which energy is stored up in the inductor. Power supplied by the battery Power dissipated as heat in the resistor If energy in inductor is then: or Integrate from t = 0 (I = 0) to t =  (I = If) So, the energy stored in an inductor carrying current I is :

The R-L Circuit Current growth in an R-L circuit (cont’d) Kirchhoff’s Loop Rule: At t = 0+, I = 0 I then increases until finally dI/dt = 0 Compare with: Current in an LR circuit as function of time

The R-L Circuit Current growth in an R-L circuit (cont’d) Integrating between (I = 0, t = 0) and (I = I, t = t)

The R-L Circuit Current growth in an R-L circuit (cont’d) Now we raise e to the power of each side

The R-L Circuit Discharging an R-L circuit Add switch S2 to be able to remove the battery. And add R1 to protect the battery so that it is protected when both switches are closed. First S1 has been closed for a long enough time so that the current is steady at its final value I0. At t=0, close S2 and open S1 to effectively remove the battery. Now the circuit abcd carries the current I0. Kirchhoff’s loop rule:

The R-L Circuit Discharging an R-L circuit (cont’d) Now let’s calculate the total heat produced in resistance R when the current decreases from I0 to 0. Rate of heat production: Energy dissipated as heat in the resistor: The current as a function of time: The total energy: The total heat produced equals the energy originally stored in the inductor

The L-C Circuit Complex number and plane Complex number : z = x + iy real part Re(z)=x, imaginary part Im(z)=y

The L-C Circuit Simple harmonic oscillation

The L-C Circuit Simple harmonic oscillation (cont’d)

The L-C Circuit Simple harmonic oscillation (cont’d)

The L-C Circuit An L-C circuit and electrical oscillation S Consider a circuit with an inductor and a capacitor as shown in Fig. Initially the capacitor C carries charge Q0 S At t=0 the switch closes and charge flows through inductor producing self-induced emf. The current I is by definition: Acceleration equation for a mass on a spring Kirchhoff’s loop rule:

The L-C Circuit An L-C circuit and electrical oscillation (cont’d) The solution of this equation is simple harmonic motion. Now let’s figure out what A and f are. For that choose initial condition as: I(0)=0 and Q(0)=Q0. Then A=Q0 and f=0. The charge and current are 90o out of phase with the same angular frequency w I is at maximum when Q=0, and Q is at maximum when I=0.

The L-C Circuit An L-C circuit and electrical oscillation (cont’d) The charge and current are 90o out of phase with the same angular frequency w I is at maximum when Q=0, and Q is at maximum when I=0. -I(t)

The L-C Circuit An L-C circuit and electrical oscillation (cont’d) The electric energy in the capacitor: The electric energy oscillates between its maximum Q02 and 0. The magnetic energy in the inductor: The magnetic energy oscillates between its maximum Q02 /(2C) and 0. Utot=Ue+Um constant Ue(t) Um(t)

The L-R-C Circuit Another differential equation

The L-R-C Circuit Another differential equation (cont’d)

The L-R-C Circuit Another differential equation (cont’d)

The L-R-C Circuit An L-R-C circuit and electrical damped oscillation At t=0 the switch is closed and a capacitor with initial charge Q0 is connected in series across an inductor. Initial condition: A loop around the circuit in the direction of the current flow yields: Since the current is flowing out of the capacitor,

The L-R-C Circuit An L-R-C circuit and electrical damped oscillation (cont’d) If R2< 4LC, the solution is: Note that if R=0,no damping occurs.