Chapter 25:

Electromagnetic Induction

Electromagnetic Induction The discovery that an electric current in a

wire produced magnetic fields was a turning point in physics and the technology that followed.

The question now was: If an electric current could produce a magnetic field, then…

Could a magnetic field produce electric current?

In 1831, Michael Faraday in England and Joseph Henry in the U.S. independently discovered that the answer was…

Yes… actually a changing magnetic field produces a current.

Faraday and Henry both discovered that electric current could be produced simply by moving a magnet in or out of a wire coil.

No voltage (emf) or other battery source was needed—only the motionof a magnet in a coil (or a single wire loop).

They discovered that voltage was induced by the relative motion between a wire and a magnetic field!

emf = “Electromotive Force” Voltage is sometimes also referred to as the electromotive

force, or emf. Inside the battery, there is a chemical reaction which is

transferring electrons from one terminal to the other (the positive terminal is losing electrons and those electrons are moving through the battery to the negative terminal, gaining electrons). Because of the positive and negative charges on the battery terminals, an electric potential difference exists between them. This electric potential does work on each charge moving through the battery and ultimately, through the circuit. The Electromotive Force is doing the work.

The maximum potential difference is specifically called the “Electromotive Force” (emf) of the battery.

Specifically, if a battery is rated at 12 Volts, we say that it has a voltage of 12 volts, but does it always delivers 12 volts or 12 Joules/Coulomb?

Basically, you can think of emf as voltage.

Circuits with Batteries

In a circuit with a battery, the battery creates an electric field w/in and parallel to the wire, directed from the positive toward the negative terminal.

The electric field exerts a force (doing work) on the free electrons in the wire, and they respond by moving.

Thus, current is produced!

But, Faraday* discovered that a battery is notneeded to induce current or voltage!!

…voltage was induced by the relative motion between a wire and a magnetic field!

i.e., Voltage (emf) is induced (brought about) whether the magnetic field of a magnet moves past a stationary conductor (wire), or the conductor moves through a stationary magnetic field.

Phenomenal!

This phenomenon of inducing voltage by changing the magnetic field around a conductor is called electromagnetic induction.

*Faraday is given credit b/c he investigated “E-I” in more detail and published his findings first. His name goes with electromagnetic induction.

A changing magnetic field induces a current in a coil of wire. Since current cannot exist without emf (voltage), a changing magnetic field induces emf.

Fig 22.1

Another way emf and current can be induced is by changing the area of a coil in a fixed magnetic field. This is effectively changing magnetic field in the coil.

http://phet.colorado.edu/new/simulations/sims.php?sim=Faradays_Electromagnetic_Lab

As long as there is a complete (closed) circuit, both emf and current will be induced. If there is a break in the circuit, no current is induced but there still is an induced emf.

Changing a magnetic field and changing the area of a coil are methods that can be used to create an induced emf.

This phenomenon is called electromagnetic induction.

Motional EMF EMF is induced in a moving

conductor (the rod that the person is moving has an induced emf… if stop movingno more emf)

RHR#1: as conductor is moved with velocity, v, to the right (thumb), magnetic field, B, is into page (straight fingers), and the magnetic force (not shown) experienced by + charge is up to top of rod, and –charge is down to bottom of rod.

Charges separate due to magnetic force, inducing an emf, called motional emf because this emf is caused by the motion of charges through a magnetic field.

Fig 22.4

The positive and negative charges accumulate (keep separating to the ends of the rod) until the magnetic force (causing them to separate) is balanced by the electric force between the charges (since + and – attract, this is an attractive force).

Equilibrium is achieved when no further charge separation occurs because:

Magnetic Force separating (repelling) charges = Electric Force attracting charges, then…

Maximum emf is achieved!

Magnetic Force separating (repelling) charges = Electric Force attracting charges, then…

Maximum emf () is achieved! Electric Field causes attraction between + and –

charges, or F=Eq Magnetic Field causes separation between + and –

charges, or F=Bqv Electric Force = Magnetic Force when emf is

maximum: Eq = Bqv (What divides out?) Recall: Electric Field also is defined as V/d, E = V/d = /L (/L)q = Bqv

= B L v Motional emf when v, B, L are all ┴ to each other.

Suppose the rod is moving at a speed of 5.0 m/s in a direction ┴

to a 0.80 T magnetic field. The rod has a length of 1.6 m and a negligible electrical resistance. The rails also have a negligible electrical resistance. The light bulb, however, has a resistance of 96 Ω. Find…

(a) the emf produced by the rod,

(b) the induced current in the circuit,

(c) the electrical power delivered to the bulb, and

(d) the energy used by the bulb in 60.0 seconds.

Did you solve and find…?

(a) = B L v = (0.80T) (1.6m) (5.0 m/s) =

= 6.4V

(b) I = V/R = /R = 6.4V/96Ω =

I = 0.067A

(c) P = IV = I = (0.067A) (6.4V) =

P=0.43 W

(d) E = Pt = (0.43W) (60.0 s) =

E = 26J

Who or what provides this 26J of nrg to the bulb?

When emf causes a current, a second magnetic force enters the picture…

RHR#1: Current I is up (thumb), magnetic field is into page (straight fingers) and magnetic force F is to the left (palm of hand), or opposite velocity.

To keep the rod moving to the right with a constant velocity v, a counterbalancing force must be applied by an external agent (delivered by the person in this case) so that it acts opposite the magnetic force.

Who or what provides the 26 J of electrical energy to the light bulb?

The magnetic force caused by the induced current which must be counterbalanced by the hand for constant velocity is…

F=ILBsin=(0.067A)(1.6m)(0.80T)=0.086N Since the person is doing this work, W=Fd (where F and d must be parallel) And since the rod moves at constant velocity, v=d/t or

d=v t W=F(d) = F (v t) = (0.086N) (5.0m/s) (60.0s) = W=26J Energy is conserved!! The work done by the person is the

same as the energy used by the bulb!! Just like a battery converts chemical energy into electrical

energy to light a bulb in a circuit with a battery… The moving rod and magnetic force convert mechanical

energy into electrical energy to light the bulb!

Can the current flow in this direction and still cause the bulb to light?

The current cannot be directed clockwise in this circuit, because the magnetic force exerted on the rod would then be in the same direction as the velocity.

The rod would accelerate to the right and create energy on its own…

Impossible!

This violates the law of conservation of energy.

Magnetic Flux…Relating magnetic field and the surface through which it passes.

Recall… = B L v and starting at zero time…

In a time to, the moving rod goes from position zero to xo at a constant velocity, v, and v =x/t

= B L v = BL(x/t) = B(L x/t)

And, for a rectangle, Area=(length)(width), or A=Lx,

= B(A/t) = BA/t

And, BA is given the name magnetic flux and is represented by the symbol Φ (phi)

Φ = B ┴ A (Units: Tm2 = Wb = weber)

The magnitude of the emf is equal to the change in magnetic flux divided by the time interval during which the change occurs, or the rate at which the magnetic flux changes:

= BA/t = Φ/t

By convention, sometimes a minus sign is written to show that the direction of the current induced in the circuit is producing a magnetic force that opposes the motion of the rod (slowing the rod)…

= - Φ/t

What if the magnetic field B is at an angle (other than 90o) to the surface swept out by the rod?

= Φ/t when the magnetic field B is ┴ to the surface, such as if B were directed along the (red) Normal line above which we call the magnetic moment.

When B is not ┴ to the surface (as in the black line above), need to find the component of B that is ┴ to the surface using cosine…

Φ = BA = [B(cos)]A = BA (cos) In general: Φ = BA (cos) If B or A are not consistent, then use the average.

Magnetic flux is proportional to the number of field lines which pass through the area or loop.

This picture shows three orientations of a rectangular coil (drawn as an edge view), relative to the magnetic field lines.

The magnetic field lines that pass through the coil are those in the regions shaded in blue.

Notice, when oriented such as in (c), there is no magnetic flux. Φ = BA (cos) b/c angle is 90o… B is ┴ to the Normal to the surface (dashed line ┴ to B)

(a) (b) (c)

A rectangular coil of wire is situated in a constant magnetic field whose magnitude is 0.50T. The coil has an area of 2.0m2. Determine the magnetic

flux for the three orientations shown in the figure below.

Φ = BA (cos)

a) =0o, Φ = BA (cos) =(0.50T)(2.0m2)(cos 0o)=

1.0 Wb

b) =60o, Φ = BA (cos) =(0.50T)(2.0m2)(cos 60o)=

0.50 Wb

c) =90o, Φ = BA (cos) =(0.50T)(2.0m2)(cos 90o)=

0 Wb