motional emf

Motional EMF

AP Physics C

Montwood High School

R. Casao

• Motion EMF is the EMF produced in a conductor moving through a magnetic field.

• Consider a straight conductor of length l moving with constant velocity through a uniform magnetic field directed into the paper.

• The conductor is moving perpendicular to the field.• Electrons in the conductor experience a force along the conductor given by F = q·(v x B).• Under the influence of this force, the electrons move to the lower end.

• The lower end of the rod becomes negatively charged; the upper end of the rod becomes positively charged.

• An electric field is produced in the conductor as a result of the charge separation.

• The charge at the ends builds up until the magnetic force q·v·B is

balanced by the electric force q·E.• At this point, the charge stops flowing and the condition for equilibrium requires that: q·E = q·v·B or E = v·B.

• Since the electric field is constant, the electric field produced in the conductor is related to the

potential difference across the ends by V = E·l.• Thus: V = E·l = B·l·v, where the upper end of the

conductor is at a higher potential than the lower end.

• A potential difference is maintained as long as there is motion through the field.• If the motion is reversed, the polarity of the voltage is also reversed.

• If the moving conductor is part of a closed conducting path, the changing magnetic flux can cause an induced current in the closed circuit.

• Consider a circuit consisting of a conducting bar of

length l sliding along two fixed parallel conducting rails as shown in the figure.

• Assume that the moving bar has zero resistance and that the stationary part of the circuit has a resistance R.• A uniform and constant magnetic field B is applied perpendicular to the plane of the circuit.

• As the bar is pulled to the right with a velocity v, under the influence of an applied force Fapp, free charges in the bar experience a magnetic force along the length of the bar.

• This magnetic force sets up an induced current since the charges are free to move in a closed

conducting path.• The rate of change of magnetic flux thru the loop and the corresponding induced EMF across the moving bar are proportional to the change in the area of the loop as the bar moves thru the magnetic field.

• If the bar is pulled to the right with a constant velocity, the work done by the applied force is dissipated in the form of joule heating in the circuit’s resistive element.

• The area of the circuit at any instant is l·x, the external magnetic flux through

the circuit is Φm = B·l·x, where x is the width of the circuit, which changes with time.• Using Faraday’s law:

vBdtdx

BEMF

dtxBd

dtdΦ

EMF m

ll

l

• If the resistance of the circuit is R, the magnitude of the induced current is:

• The equivalent circuit diagram is shown below. • Since there is no real battery in the circuit, the external force does work on the conductor, thereby moving charges thru the magnetic field.• This causes the charges to move along the conductor with an average drift velocity, and a current is set

up.

RvB

R

EMFI

l

• The total work done by the applied force during some time interval should equal the electrical energy that the induced EMF supplied in the same period.

• If the bar moves with constant speed, the work done must equal the energy dissipated as heat in the resistor in this time interval.

• As the conductor of length l moves through the uniform magnetic field B, it experiences a

magnetic force Fm of magnitude I·l·B.

• The direction of the force Fm is opposite the motion of the bar, or to the left in the figure.

• If the bar is to move with a constant velocity, the applied force must be equal to and opposite the magnetic force, or to the right in the figure.

• If the magnetic force acted in the direction of motion, it would cause the

bar to accelerate once it was in motion, thereby increasing its velocity.• An acceleration would violate the law of conservation of energy.

app

22 2 2

2

P F v I B v I B v

B vI P=I V EMF B v

R

B vB v B vP B v

R R REMF B v

EMFP

R

l l

ll

ll ll

l

• The power delivered by the applied force is:

• This power is equal to the rate at which energy is dissipated in the resistor, I2·R.

• This power is also equal to the power I·EMF supplied by the induced EMF.

• This is a clear demonstration of the conversion of mechanical energy into electrical energy and finally into thermal energy (joule heating).

Example: Induced EMF in a Rotating Bar

• A conducting bar of length l rotates with a constant angular velocity about a pivot at one end. A uniform magnetic field B is directed perpendicular to the plane of rotation. Find the induced EMF between the ends of the bar.

• Divide the length of the bar into small elements of length labeled dr, whose tangential velocity is v.

• Tangential velocity v is constant for each element of length dr; however, the tangential velocity of each length of dr

increases from the point O (where it is 0 m/s) to the end of the bar (where it is a maximum.

• The EMF induced in each element dr is a component of the total EMF in the conductor.

• The component of the EMF of each element of length dr moving perpendicular to a field B is:

dEMF = B·v·dr• Each element dr of the bar is moving perpendicular to B, so there is an EMF generated across each element dr.

• Summing up the EMFs induced across all the elements, which are in series, gives the total EMF between the ends of the bar.

• Mathematically: EMF = B·v·dr• To integrate the expression, linear speed v is related to angular speed by v = r·.• Since B and are constant:

l

0drrBEMF

drrBEMF

drvBEMF

• Completing the integration from the pivot point of

the rod at 0 to the end of the rod at l:

2B

EMF

2B

20

2BEMF

1 1r

B1 1

rBEMF

2

222

0

2

0

11

l

ll

ll

Magnetic Force on a Sliding Bar

• A bar of mass m and length l moves on two frictionless parallel rails in the presence of a uniform magnetic field directed into the page. The bar is given an initial velocity vo to the right and released. Find the velocity of the bar as a function of time.– The induced current is counterclockwise and the

force Fm = -I·l·B, where the negative sign denotes that the force is to the left and retards the motion.

Magnetic Force on a Sliding Bar– Fm is the only horizontal force acting on the bar and

applying Newton’s second law:

– The induced current is equal to:

BIdtdv

mamFx l

RvB

I

l

• Substituting and rearranging:

• Write the expression as a differential equation:

mRB-

dtdv

RB-

BR

vBdtdv

m

BIdtdv

m

22

22

v

v

l

ll

l

l

dtmR

Bv

dv 22

l

• Integrate the equation from v = vo to v = v and from t = 0 s to t = t.

o

o

2 2v t

v 0

2 2tv

v 0

2 2t

o 0

2 2

o

2 2

o

1 Bdv dt

v R m

Bln v dt

R m

Bln v-ln v t

R m

v Bln t 0

v R m

v Bln t

v R m

l

l

l

l

l

• Exponentiate both sides of the equation:

• The velocity of the bar decreases exponentially with time under the action of the magnetic retarding force.

mRtB

o

mRtB

mRtBv

vln

22

22

22

o

vtv

v

l

l

l

e

ev

ee

o