Physics’150 Electromagne5c’induc5oncsma31.csm.jmu.edu/.../Lectures/Phys150/20ElectromagneticInducti… · closed’circuitare’in’ relavemoon,’there’will’be’an’induced’EMF’

Physics 150

Electromagne5c induc5on

Chapter 20

Mo5onal electromo5ve force

Physics 150, Prof. M. Nikolic 2

Consider a conductor in a B-‐field moving to the right.

The electrons in the conductor will experience a downward magne5c force

|!FB |= evBsinθ = evB

-‐ -‐

+ +

The electrons in the bar will move toward the boJom of the bar.

This creates an electric field in the bar and results in a poten5al difference (EMF) between the top and boJom of the bar è electric force that tries to push nega5ve charges up

|!FE |= eE

Mo5onal EMF


-‐ -‐

+ + Charges will con5nue to separate un5l electric force become equal to the magne5c force è equilibrium

L evB = eE

Electric field is also (Chapter 17)

E = ΔVL

ΔV = vBL

If we connect the rod to a closed circuit, we could produce current in it. I =

ΔVR

=vBLR

L – length of a conductor

Faraday’s law


The English physicist Michael Faraday was the first to observe that moving a conductor through a magne5c field will generate an EMF.

It turned out that this is not the only case when magne5c fields can produce EMF and currents.

If the magne5c field changes, or if the magnet and conductor in closed circuit are in rela*ve mo*on, there will be an induced EMF

(and therefore current) in the conductor.

Faraday also discovered that: A static magnet will produce no current in a stationary coil.

Magne5c flux


Similarly to electric flux ΦE = EAcosθ

we can define a magne*c flux ΦB = BAcosθ

SI units: 1 T·∙m2 = 1 Weber = 1 Wb

We always look at the angle between the B field and the normal to the area.


Exercise: Flux A magne5c field is oriented at an angle of 320 to the normal of a rectangular area 5.5 cm by 7.2 cm. If the magne5c flux through this surface has a magnitude of 4.8 × 10-‐5 T·∙m2, what is the strength of the magne5c field?

ΦB = BAcosθ B = ΦB

Acosθ

Area of a rectangle is A = lw l = 5.5 cmw = 7.2 cm B = ΦB

lwcosθ

B = 4.8×10−5Tm2

5.5×10−2 × 7.2×10−2m2 cos320

T 3102.14 −×=B

ΦB = BAcosθ

What is given: θ = 320 l = 5.5 cm w = 7.2 cm Φ = 4.8 × 10-‐5 T·∙m2

Faraday’s law


The change in magne*c flux will induce electromo5ve force (EMF) in a conduc5ng loop.

ε = −N ΔΦB

Δt N – is number of turns in the loop

The minus sign just indicates the direc*on of the induced emf. To calculate the magnitude, we will use:

ε = N ΔΦB

Δt= N

ΦBf −ΦBi

t f − ti

ε = −ΦBf −ΦBi

t f − ti= −

Bf Af cosθ f −BiAi cosθit f − ti

Ways to induce EMF


The magne5c flux can change due to any combina5on of the following reasons: 1.  The size of the magne5c field changes.

2.  The area of the loop changes.

3.  The rela5ve orienta5on of the field and the loop changes.

ε = − NAcosθBf −Bit f − ti

tNBA ωωε sin=

ε = vBw


Exercise: EMF with changing B field A 0.25 T magne5c field into the page is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm. The magne5c field is reduced to zero in 0.12 s. What is the magnitude of the induced emf?

ε = N ΔΦB

ΔtOnly B field is changing

ε = NAcosθBf −Bit f − ti

Area of the circle is A = πr2

A = 3.14 ⋅0.152m2 = 0.071m2

Magne5c field is perpendicular to the loop è θ = 0 è cosθ = 1

ε = 50 ⋅0.071m2 0− 0.25T0.12s− 0

= 7.36V

What is given: Bi = 0.25 T N = 50 r = 15 cm Δt = 0.12 s

Electric generators


A generator is a device that converts mechanical energy to electrical energy.

Consider a current loop which rotates in a constant magne5c field è the angle between magne5c field and area of the loop is changing è EMF and current are produced

tNBA ωωε sin=A – area N – number of turns ω – angular speed

If this loop is part of a circuit, this EMF will induce an Alterna*ng Current (AC) in the circuit.

Electric generators


A coil of wire turns in a magne5c field. The flux in the coil is constantly changing, genera5ng an EMF in the coil.

tNBA ωωε sin=

Exercise: Generators


AC generator consists of a circular coil with 50 turns and radius of 3 cm. When the coil rotates at 350 rpm, the maximum EMF in the coil is 17 V. What is the strength of the magne5c field? Assume uniform magne5c field.

tNBA ωωε sin=

Maximum (amplitude) EMF is when sinωt = 1 è εmax = NBAω B = εmaxNAω

Area of the circle is A = πr2

A = 3.14 ⋅0.032m2 = 0.00283m2

Angular speed is given in revolu5ons per minute è we have to convert it to rad/s

ω = 350 revmin

= 350 2πrad60s

= 36.63 rads

B = 17V50 ⋅0.00283m2 ⋅36.63rad / s

B = 3.28T

tNBA ωωε sin=

What is given: N = 50 ω = 350 rpm r = 3 cm εmax = 17 V

Lenz’s law


The direc5on of induced EMFs and currents always oppose the change in flux that produced them.

•  The change in magne5c flux will induce EMF •  In closed circuit current will be induced too •  The induced current creates new magne*c field (Bind) such that will try to keep

the total magne5c flux constant

N

S v

inducedB

•  Consider a downward and increasing magne5c field

•  Generated current produces an upward magne5c field inside the loop to oppose the change.

•  Right hand rule gives the direc5on of the current – counterclockwise

The downward magne5c field through the loop is decreasing as the magnet moves upward. Which direc5on does the current flow in the loop?

A.  Clockwise B.  Counterclockwise C.  There is no current D.  Need more info

Conceptual ques5on – Lenz’s law Q1

N

S v

inducedB

B field is down and decreasing è flux is decreasing è induced field Bind has to be in the same

direc5on (down) to stop the flux decrease è by right hand rule current is clockwise

Physics 150, Prof. M. Nikolic 14 30

A square loop of wire lies in the plane of the page. A decreasing magne5c field is directed into the page. The induced current in the loop is:

A.  Clockwise B.  Counterclockwise C.  Zero D.  Need more info

Conceptual ques5on – Induced current Q2

x inducedB

x x

x x

B field is into the page and decreasing è flux is decreasing è induced field Bind has to be in the same direc5on (into

the page) to stop the flux decrease è by right hand rule current is clockwise

Physics 150, Prof. M. Nikolic 15 30


Exercise: Lenz’s law and moving loop Find the direc5on and magnitude of the current induced in the rod if the rod is moving 5 m/s to the right. The rod has length of L = 0.45 m and the en5re system with resistance of 12.5 Ω and is in uniform magne5c field coming out of the page with magnitude of 0.75 T.

indB

I = ΔVR

=vBLR

Direc5on: •  Rod is moving to the right è area is increasing è

flux is increasing •  Induced B field is in the opposite direc5on (into the

page) to stop the change •  by right hand rule current is clockwise

I = 5m / s ⋅0.75T ⋅0.45m12.5Ω

⋅

I = 0.135A

What is given: v = 5 m/s L = 0.45 m R = 12.5 Ω B = 0.75 T

Mechanical output

Motor: a generator run backwards


Runs an alterna5ng current through a coil of wire in a magne5c field. This generates a torque on the coil.

•  As the coil rotates in the external magne5c field, an addi5onal current is induced in it.

•  The direc5on of the induced current is, by Lenz’s Law, counter to the electric current we are running through the coil to make it rotate in the first place.

•  We some5mes say this counter-‐current is the result of Back EMF.

AC

Transformers


When sending power via power lines, we want to send as much as possible: P = IΔV

•  High voltage (500 kV) •  High current è big losses due to resistance in wires Ploss = IR2

House electronics, however, works at small voltages (120 V)

Transformers


A transformer is a device used to change the voltage in a circuit. AC currents must be used.

Apply a varying voltage to the primary coil.

This causes a changing magne5c flux in the secondary coil.

εp = −NpΔΦB

Δt

εs = −NsΔΦB

Δt

Transformers


Flux through the coils stays the same

εpεs=Np

Ns

The “turns ra5o” gives the ra5o of the EMFs.

εpεs=IsI p=Np

Ns

•  Ns>Np è εs>εp è higher output voltage è step up transformer

•  Ns<Np è εs<εp è lower output voltage è step down transformer

Power between coils stays the same (P=IΔV) è the current is the inverse of the EMF ra5o.

Step down transformers


240,000 V in the power lines

120 V in houses

2,400 V local substa5ons

Suppose we set up a transformer as shown, with a 120 V household outlet on the primary winding. What voltage is delivered at the secondary winding ?

A.  240 V B.  120 V C.  60 V D.  0 V

Conceptual ques5on – Transformers Q2

10 tu

rns

20 tu

rns

εpεs=Np

Ns

εs = εpNs

Np

εs =120V2010

= 240V

Double the turns è double the voltage Physics 150, Prof. M. Nikolic 22

30

Eddy currents


When a solid conductor is moved in a magne5c field, there is a force on the electrons (remember they are free to move in a conductor), which then move in the metal. This movement is called an eddy current.

The induced currents produce magne5c fields which tend to oppose the mo5on of the metal.

× × ×

×

× × ×

×

×

× ×

× × × ×

× × ×

× ×

Applica5on: The induc5on stove

Inductance – mutual inductance


The induced EMF in coil 2:

ε2 = −N2ΔΦ21

ΔtFlux through coil 2 is propor5onal to B1, which is propor5onal to I1

1212 MIN =Φ

Where M is the mutual inductance. It depends only on constants and geometrical factors.

tIM

tN

tIM

tN

Δ

Δ−=

Δ

ΔΦ−=

Δ

Δ−=

Δ

ΔΦ−=

21211

12122

ε

ε

The unit of inductance is the Henry [1H = 1V·∙s/A].

Inductance – self-‐inductance


Self-‐inductance occurs when a current carrying coil induces an EMF in itself.

.LIN =ΦThe defini5on of self-‐inductance (L) is

ε = −N ΔΦΔt

= −L ΔIΔt

The unit of inductance (L) is the Henry [1H = 1V·∙s/A].

The most common inductor is solenoid

ℓℓ

AnANL 20

20 µµ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

N = number of turns in solenoid l = length of solenoid A = cross sec5onal area of solenoid n = number turns per length


Exercise: Solenoid The inductance of a solenoid with a current of 3 A, 450 turns and a length of 24 cm is 7.3 mH. (a)  What is the cross-‐sec5onal area of the solenoid? (b)  What is the average induced emf in the solenoid if its current drops from 3.2 A

to 0 in 55 ms?

ℓℓ

AnANL 20

20 µµ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

(a)

A = Llµ0N

2

A = 7.3×10−3H ⋅0.24m1.26×10−6H /m ⋅ 4502

2cm 6.68=A

(b)

ε = −N ΔΦΔt

= −L ΔIΔt

ε = −LI final − Iinitialt final − tinital

ε = −7.3×10−3H 0A−3.2A0.055s− 0s

ε = 0.425 V

Energy storage in inductors


An inductor stores energy in its magne5c field according to:

2

21 LIU =

We also want to consider the energy density (energy per unit volume) of a magne5c field:

uB =U

Volume=

LI 2

2Volume=12µ0

B2


Exercise: Energy stored in solenoid A solenoid with radius of 2 cm, length 12 cm and 9000 turns carries a current of 2 A. a)  How much energy is stored in the solenoid? b)  Find the magne5c energy density.

2

21 LIU =

and L = µ0

N 2

ℓ

!

"#

$

%&A

L = 1.26×10−6H /m ⋅90002 ⋅3.14 ⋅0.022m2

0.12m

L = µ0N 2

ℓ

!

"#

$

%&πr2

L =1.06H

U =121.06H ⋅ (2A)2 = 2.12J

uB =U

Volume=

Ul ⋅πr2

Volume of a cylinder: V = l ⋅πr2

uB =2.12J

0.12m ⋅3.14 ⋅0.022m2

uB =14065J /m3

LR circuits


We can construct a circuit out of inductors and resistors. The circuit will behave similar to an RC circuit, with a 5me constant given by: τ = L/R

In this circuit, the inductor acts to keep the current in the circuit from instantly reaching the value of ε/R. Instead, some 5me must elapse before the circuit virtually reaches that current.

I(t) = εR1− e−t/τ( )

I(t) = Imax 1− e−t/τ( )

Imax = ε/R maximum current in the circuit

LR circuits


ΔV = εe−t/τVoltage across the inductor vs. 5me:

“Discharging ” inductor


For a “discharging” inductor τ/0)( teItI −=

The LR circuit 5me constant τ plays the same role as in an RC circuit.

where I0 is the current in the inductor when t = 0.


Exercise: LR circuit A coil has an inductance of 0.30 H and a resistance of 15.0 Ω. The coil is connected to a 12.0 V ideal baJery. a)  Find the current in the coil axer 50 ms. b)  What is the total energy stored in the coil? c)  What is the of energy dissipa5on rate in the coil axer 50 ms?

a) I(t) = εR1− e−t/τ( ) Time constant given by: τ = L/R = 0.30 H/15 Ω = 0.02 s

I(t) = 12V15Ω

1− e−0.05s/0.02s( )

I(t) = 0.73A

Maximum current in the circuit Imax =εR= 0.8A

What is given: L = 0.3 H R =15.0 Ω ε = 12 V t = 50 ms


Exercise: LR circuit A coil has an inductance of 0.30 H and a resistance of 15.0 Ω. The coil is connected to a 12.0 V ideal baJery. a)  Find the current in the coil axer 50 ms. b)  What is the total energy stored in the coil? c)  What is the of energy dissipa5on rate in the coil axer 50 ms?

b) Total energy stored in the coil depends on maximum current

U =12LImax

2

U =120.3H ⋅ (0.8A)2 = 0.096J

c) Energy dissipa5on rate is equal to the power dissipated

P = I 2R

P = (0.73A)215Ω = 8W

What is given: L = 0.3 H R =15.0 Ω ε = 12 V t = 50 ms

Review


ΔV = vBL

I = ΔVR

=vBLR

Voltage and current for moving rod

ΦB = BAcosθMagne5c flux

ε = −N ΔΦB

Δt

Faraday’s law

tNBA ωωε sin=Electric generators

εpεs=IsI p=Np

Ns

Transformers

tIM

tN

tIM

tN

Δ

Δ−=

Δ

ΔΦ−=

Δ

Δ−=

Δ

ΔΦ−=

21211

12122

ε

ε

Mutual inductance

Review


ε = −N ΔΦΔt

= −L ΔIΔt

Faraday’s law for self inductance

ℓℓ

AnANL 20

20 µµ =

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

Inductance of a solenoid

2

21 LIU =

Energy stored in an inductor

uB =U

Volume=

LI 2

2Volume=12µ0

B2

Magne5c energy density

I(t) = εR1− e−t/τ( )

I(t) = Imax 1− e−t/τ( )

ΔV = εe−t/τ

“Charging” an inductor

τ/0)( teItI −=

“Discharging” an inductor

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Physics’150 Electromagne5c’induc5oncsma31.csm.jmu.edu/.../Lectures/Phys150/20ElectromagneticInducti… · closed’circuitare’in’ relavemoon,’there’will’be’an’induced’EMF’