Electromagnetic Induction and Faraday’s Law

Induced EMF

Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus:

He found no evidence when the current was steady, but did see a current induced when the switch was turned on or off.

Therefore, a changing magnetic field induces an emf (electromotive force, or voltage). This emf will also be associated with an induced current if the charges are free to flow (complete circuit).

Faraday’s experiment used a magnetic field that was changing because the current producing it was changing; the previous graphic shows a magnetic field that is changing because the magnet is moving.

Induced emf in a conductor

If a conductor (as shown on next slide) moves in a magnetic field there is a force on the charges in the conductor. The electrons are the "mobile" charge carriers. They will be forced to move by the effects of the magnetic field. This will cause a potential difference to be created (induced) between the two ends of the conductor (piece of wire).

We can derive a relationship for the induced emf between two points in conductor (i.e. between the ends of a wire) by looking at the magnetic force on a positive charge in a conductor. In this case the B field is perpendicular to the length of the wire

!

The work done in moving the charge along the length(l) of wire is: W=Fd=qvBl Voltage = W/q = qvBl/q= Bvl

∴ ε = Bυl

The more general equations becomes:

ε = Bυl sinθ= induced emf (V) ε

θ = angle between B and v vectors.

Example: A solid metal rectangle (dimensions L= 25 cm, W=15 cm) moves to the right through a magnetic field directed into the page. If is has a speed of 280 m/s, and the field strength is 1.25 T determine the induced emf across its length (L) and width(W).

v

L

W

Solution: The force on a (+) charged particle would be upwards therefore there would be no charge separation along its length, only across its width. The induced emf (voltage) across its width is:

ε = BLυ sinθ = (1.25)(0.15)(280)sin900= 52.5 V

*It is important to note that it is the electrons that actually move downwards in the conductor, creating this induced voltage

Example: A 0.75 m conducting rod is moved at 8.0 m/s across a 0.25 T magnetic field along metal rails. The electrical resistance of the system is 5.0 Ω. What are the magnitude and direction of the current through point X?

An equivalent statement of Faraday’s law (to be sated later) related to a magnetic dipole entering or leaving a wire loop can be stated as follows. Lenz’s Law: A current will be induced in a wire loop (or solenoid) that will create a magnetic field to oppose the relative motion of the magnet and coil. Relative motion is mentioned because one could move the coil instead of the magnet, or move both. On could also consider this as conservation of energy. Where did the energy come from to induce the movement of electrons? Answer: the kinetic energies of the moving magnet and/or coil.

Lenz’s Law

Example: A bar magnet is being moved to the left into a solenoid. A current will be induced to oppose its motion.

N Sv

In what direction will the induced conventional current move through the resistor ? Left or right?

Solution: The answer is left.

To oppose the motion of the magnet the current induced should create a N pole on the right side of the solenoid. (Lenz’s Law) Using a RHR it means the conventional current would move left through the resistor as shown below.

N Sv

NS

Magnetic Flux

The induced emf in a wire loop is proportional to the rate of change of magnetic flux through the loop. The wire loop defines and area within the loop.

Magnetic flux (vector dot product) is defined as:

Aside note: Unit of magnetic flux: weber, Wb.

1 Wb = 1 T·m2 , Also, magnetic field (B) can be stated in Wb/m2.

Φ =B ⋅A = BAcosθ

Φ = BAcosθ

Φ = magnetic flux (Wb)

θ = angle between B and A vectors.

The area A can be represented by a vector perpendicular (normal) to the area. The magnitude of the vector is proportional to the area, and its direction give the orientation of the area. The angle is the angle made between the area vector and the magnetic field Example: If there is a magnetic field coming out of the page, what is the magnetic flux on the 3.00 cm x 5.00 cm rectangular loop if field has a strength of 0.65 T?

Solution:

Φ = BAcosθ

The area is: 0.0300 m x 0.0500 m= 1.50x10-3 m2

There are two possible choices for the direction of the area vector, into or out of the page (both are correct). Therefore could be 00 or 1800.

Use the formula above to calculate the magnetic flux:

= (0.65)(1.50x10-3 )(±1)= ±9.75x10-4 Wb

Φ = BAcosθ

*In this case one would normally state the magnitude 9.75x10-4 Wb .

This drawing shows the variables in the flux equation:

One can think of the magnetic flux as the product of the area (A) with the component of magnetic field perpendicular (B⊥) to the area.

Φ = B⊥A

The magnetic flux is proportional to the total number of lines passing through the loop.

Magnetic flux can be changed in different ways. Just by examining the formula for magnetic flux it can be deduced how to change flux.

Changing Magnetic Flux

Φ = BAcosθ

One obvious way is to change the intensity (B) of the magnetic field

Magnetic flux will change if the area of the loop changes:

Magnetic flux will change if the relative angle between the loop and the field changes:

Lenz’s Law & Faraday’s law of induction:

ε = −N ΔφΔt

An induced emf always gives rise to a current whose magnetic field opposes the original change in magnetic flux. Another way of stating this is that the induced emf gives rise to a current which attempts to restore the coil to its original state of flux.

N= number of coils = change in flux (Wb) ΔφΔt =change in time(s) ε = emf (V)

*The minus sign indicates the induced flux opposes the original change in flux

Example: A coil was orginally in the presence of no magnetic field. It was then subjected to a magnetic field into the page. What would have been the direction of induced current in the wire coil?

initial final

Solution: The induced current would create a field in an attempt to bring the coil back to its original state of flux( ). This means field lines coming out of the page, to counter act the field lines into the page, would be produced by an induced current travelling counter clockwise.

φinitial = 0

Example: A single circular loop of wire in a magnetic field was deformed into a rectangle. What was the direction of the induced current in the coil?

initial final

Solution: Lets assume that the initial magnetic flux abd final flux are:

φinitial = BA1 φ final = BA2

A2 is smaller than A1.

Therefore to bring the coil back to its original state of flux field lines must be added into the page to increase the flux. The current must be clockwise around the rectangular loop.

final

Example: A rectangular coil ( 4 cm x 5 cm) with 20 turns, lies with the plane of its area perpendicular to a magnetic field of strength 0.92 T. If the coil were turned through an angle of 1800 in a time of 25 ms, what average emf was induced?

B

Solution: The negative sign can be dropped in front of the N, as one is generally interested in the magnitude of

ε = N ΔφΔt

= Nφ final −φinitial

Δt= N

BAcosθ final −BAcosθinitialΔt

ε

I will arbitrarily choose the Area vector to the right which would mean θinital= 00.

ε = N ΔφΔt

= (20)[(0.92)(2x10−3)(cos180)− (0.92)(2x10−3)(cos0)]

25x10−3

= -2.94 V

= +2.94 V, We are generally interested in the magnitude.

*If the Area vector was initially chosen to be to the right (θinital= 1800), would have been calculated as +2.94 V ε

Eddy Currents

Induced currents can flow in bulk material as well as through wires. These are called eddy currents, and can dramatically slow a conductor moving into or out of a magnetic field.

Question: A magnet is dropped through a hollow metal pipe. What possible kinds of induced currents in the pipe could occur as it is dropping?

N

S

V

Solution: The moving magnet, according to Lenz’s law, will induce currents in the pipe to impeded its motion.

N

S

The induced circular currents shown would impede the magnet’s motion down the pipe.

Electric Generators

A generator is the opposite of a motor – it transforms mechanical energy into electrical energy. This is an ac generator:

The axle is rotated by an external force such as falling water or steam. The brushes are in constant electrical contact with the slip rings.

A dc generator is similar, except that it has a split-ring commutator instead of slip rings. It is essentially the same in design as the dc motor.

A sinusoidal emf is induced in the rotating loop (N is the number of turns, and A the area of the loop, θ=ωt):

See the derivation of this relationship on a separate pdf file.

Back EMF and Counter Torque An electric motor turns because there is a torque on it due to the current. We would expect the motor to accelerate unless there is some sort of drag torque. That drag torque exists, and is due to the induced emf , called a back emf. The diagram on the right shows an electric motor connected to a voltage source. This electrical diagram represents how the current and voltage behave in a rotating electric motor.

Vapplied

Motor

I

R

!back

R= resistance of motor (armature)

εback

A similar effect occurs in a generator – if it is connected to a circuit, current will flow in it, and will produce a counter torque. This means the external applied torque must increase to keep the generator turning.

Analyzing the diagram on the previous slide by using Kirchhoff’s voltage law, one obtains the following relationship:

Vapplied −εback − IR = 0 εback =Vapplied − IR

Be aware that on the physics 12 formula sheet:

or

Vback = ε − Ir

When a dc motor is first connected there is no back emf since it is initially not rotating. The current through the motor is at a maximum when the motor is not rotating. It immediately starts to rotate. The magnitude of the back emf is proportional to the speed of rotation. As the rotational speed increases so does the back emf. As the back emf increases the current decreases. The current decreases to a point where the motor just has enough force on it to keep it rotating. At this point the terminal rotational speed has been reached. All this happens almost instantly.

DC Motor on Startup

When it jams the max current is: Imax =VappliedR

Question: When would a dc motor be most likely to burn out?

Answer: It would be most likely to burn out if the motor jams. When it is jammed the coils are not rotating and the induced emf is zero. If the induced emf is zero there is a maximum current going through the coils. This would create the maximum amount of heat and possibly burn the insulation of the wire that make up the coils in the motor.

Question: A rotating motor has an applied voltage of 20 V across it, and a back emf is 5 V with a current of 7.5 A. If the speed of the motor doubles, what is now the current through the motor?

Solution: The motor behaves according to: εback =Vapplied − IR

With the initial given information one can determine the resistance of the motor.

R =Vapplied −εback

I= 2.00Ω

If the rotational speed of the motor doubles so does the back emf ( 10 V). Using the formula once again one can calculate the new current.

I =Vapplied −εback

R= 5.00 A

Applications of Induction

This microphone works by induction; the vibrating membrane induces an emf in the coil

Differently magnetized areas on an audio tape or disk induce signals in the read/write heads.

A seismograph has a fixed coil and a magnet hung on a spring (or vice versa), and records the current induced when the earth shakes.

A ground fault circuit interrupter (GFCI) will interrupt the current to a circuit that has shorted out in a very short time, preventing electrocution.