Physics 150 Electromagne5c induc5on Chapter 20 Mo5onal electromo5ve force Consider a conductor in a B-­‐field moving to the right. + + The electrons in the conductor will experience a downward magne5c force !
| FB |= evBsin θ = 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
Physics 150, Prof. M. Nikolic 2 Mo5onal EMF + + Charges will con5nue to separate un5l electric force become equal to the magne5c force è equilibrium evB = eE
L -­‐ -­‐ Electric field is also (Chapter 17) ΔV
E=
L
L – length of a conductor ΔV = vBL
If we connect the rod to a closed circuit, we could produce current in it. Physics 150, Prof. M. Nikolic ΔV vBL
I=
=
R
R
3 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.
Physics 150, Prof. M. Nikolic 4 Magne5c flux Similarly to electric flux Φ E = EA cosθ
we can define a magne*c flux Φ B = BA cosθ
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. Physics 150, Prof. M. Nikolic 5 Φ B = BA cosθ
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? Φ B = BA cosθ
What is given: θ = 320 l = 5.5 cm w = 7.2 cm Φ = 4.8 × 10-­‐5 T·∙m2 ΦB
B=
A cosθ
Area of a rectangle is A = lw l = 5.5 cm
w = 7.2 cm
ΦB
B=
lw cosθ
4.8 ×10 −5 Tm 2
B=
5.5 ×10 −2 × 7.2 ×10 −2 m 2 cos32 0
B = 14.2 ×10 −3 T
Physics 150, Prof. M. Nikolic 6 Faraday’s law The change in magne*c flux will induce electromo5ve force (EMF) in a conduc5ng loop. ΔΦ B
ε = −N
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: Φ Bf − Φ Bi
ΔΦ B
ε =N
=N
Δt
t f − ti
Φ Bf − Φ Bi
B f A f cosθ f − Bi Ai cosθi
ε =−
=−
t f − ti
t f − ti
Physics 150, Prof. M. Nikolic 7 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. B f − Bi
ε = − NA cosθ
t f − ti
2. The area of the loop changes. ε = vBw
3. The rela5ve orienta5on of the field and the loop changes. ε = NBAω sin ωt
Physics 150, Prof. M. Nikolic 8 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?
What is given: Bi = 0.25 T N = 50 r = 15 cm Δt = 0.12 s ΔΦ B
ε =N
Δt
Area of the circle is A = πr2 A = 3.14 ⋅ 0.152 m 2 = 0.071m 2
ε = 50 ⋅ 0.071m 2
Physics 150, Prof. M. Nikolic Only B field is changing B f − Bi
ε = NA cosθ
t f − ti
Magne5c field is perpendicular to the loop è θ = 0 è cosθ = 1 0 − 0.25T
= 7.36V
0.12s − 0
9 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 ε = NBAω sin ωt
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. Physics 150, Prof. M. Nikolic 10 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. ε = NBAω sin ωt
Physics 150, Prof. M. Nikolic 11 ε = NBAω sin ωt
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. What is given: N = 50 ω = 350 rpm r = 3 cm εmax = 17 V ε = NBAω sin ωt
Maximum (amplitude) EMF is when sinωt = 1 è εmax = NBAω Area of the circle is A = πr2 ε max
B=
NAω
A = 3.14 ⋅ 0.032 m 2 = 0.00283m 2
Angular speed is given in revolu5ons per minute è we have to convert it to rad/s rev
2π rad
rad
ω = 350
= 350
= 36.63
min
60s
s
Physics 150, Prof. M. Nikolic B=
17V
50 ⋅ 0.00283m 2 ⋅ 36.63rad / s
B = 3.28T
12 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 v S
N
Binduced
Physics 150, Prof. M. Nikolic • 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 13 Conceptual ques5on – Lenz’s law Q1
The downward magne5c field through the loop is decreasing as the magnet moves upward. Which direc5on does the current flow in the loop? A.
B.
C.
D.
Clockwise Counterclockwise There is no current Need more info 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 v S
N
Binduced
30 14 Conceptual ques5on – Induced current Q2
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.
B.
C.
D.
Clockwise Counterclockwise Zero Need more info Binduced
x 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 30 15 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. ΔV vBL
I=
=
R
R
⋅
Bind
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 Physics 150, Prof. M. Nikolic I=
5m / s ⋅ 0.75T ⋅ 0.45m
12.5Ω
I = 0.135A
What is given: v = 5 m/s L = 0.45 m R = 12.5 Ω B = 0.75 T 16 Motor: a generator run backwards Mechanical output AC Physics 150, Prof. M. Nikolic 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. 17 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) Physics 150, Prof. M. Nikolic 18 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. ε p = −N p
ΔΦ B
Δt
This causes a changing magne5c flux in the secondary coil. Physics 150, Prof. M. Nikolic ΔΦ B
ε s = −N s
Δt
19 Transformers Flux through the coils stays the same εp Np
=
εs N s
The “turns ra5o” gives the ra5o of the EMFs. ε p Is N p
= =
εs I p N s
• 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. Physics 150, Prof. M. Nikolic 20 Step down transformers 240,000 V in the power lines 2,400 V local substa5ons 120 V in houses Physics 150, Prof. M. Nikolic 21 Conceptual ques5on – Transformers Q2
A.
B.
C.
D.
240 V 120 V 60 V 0 V 20 turns 10 turns 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 ? εp Np
=
εs N s
Ns
εs = ε p
Np
ε s = 120V
20
= 240V
10
Double the turns è double the voltage Physics 150, Prof. M. Nikolic 30 22 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 Physics 150, Prof. M. Nikolic 23 Inductance – mutual inductance The induced EMF in coil 2: ε 2 = −N 2
ΔΦ 21
Δt
Flux through coil 2 is propor5onal to B1, which is propor5onal to I1 N2Φ21 = MI1
Where M is the mutual inductance. It depends only on constants and geometrical factors. ΔΦ 21
ΔI
= −M 1
Δt
Δt
ΔΦ12
ΔI 2
ε1 = − N1
= −M
Δt
Δt
ε 2 = −N2
Physics 150, Prof. M. Nikolic The unit of inductance is the Henry [1H = 1V·∙s/A]. 24 Inductance – self-­‐inductance Self-­‐inductance occurs when a current carrying coil induces an EMF in itself. The defini5on of self-­‐inductance (L) is NΦ = LI .
ΔΦ
ΔI
ε = −N
= −L
Δt
Δt
The unit of inductance (L) is the Henry [1H = 1V·∙s/A]. The most common inductor is solenoid ⎛ N 2 ⎞
⎟ A = µ 0 n 2 Aℓ
L = µ 0 ⎜
⎜ ℓ ⎟
⎝
⎠
Physics 150, Prof. M. Nikolic N = number of turns in solenoid l = length of solenoid A = cross sec5onal area of solenoid n = number turns per length 25 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? (a) (b) ⎛ N 2 ⎞
⎟ A = µ 0 n 2 Aℓ
L = µ 0 ⎜
⎜ ℓ ⎟
⎝
⎠
Ll
A=
µ0 N 2
ΔΦ
ΔI
ε = −N
= −L
Δt
Δt
I final − I initial
ε = −L
t final − tinital
7.3×10 −3 H ⋅ 0.24m
A=
1.26 ×10 −6 H / m ⋅ 450 2
ε = −7.3×10 −3 H
A = 68.6 cm 2
ε = 0.425 V
Physics 150, Prof. M. Nikolic 0A − 3.2A
0.055s − 0s
26 Energy storage in inductors An inductor stores energy in its magne5c field according to: 1 2
U = LI
2
We also want to consider the energy density (energy per unit volume) of a magne5c field: U
LI 2
1 2
uB =
=
=
B
Volume 2Volume 2µ 0
Physics 150, Prof. M. Nikolic 27 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. ! N2 $
1 2
L = µ0 # & A
U = LI
and 2
" ℓ %
! N2 $ 2
L = µ0 # & π r
" ℓ %
1.26 ×10 −6 H / m ⋅ 9000 2 ⋅ 3.14 ⋅ 0.02 2 m 2
L=
0.12m
L = 1.06H
uB =
U
U
=
Volume l ⋅ π r 2
Volume of a cylinder: V
uB =
= l ⋅ π r2
2.12J
0.12m ⋅ 3.14 ⋅ 0.02 2 m 2
uB = 14065J / m 3
1
U = 1.06H ⋅ (2A)2 = 2.12J
2
Physics 150, Prof. M. Nikolic 28 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 ε
−t/τ
I(t) = (1− e )
R
−t/τ
I(t) = I max (1− e )
Imax = ε/R maximum current in the circuit 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. Physics 150, Prof. M. Nikolic 29 LR circuits Voltage across the inductor vs. 5me: ΔV = ε e
Physics 150, Prof. M. Nikolic −t/τ
30 “Discharging ” inductor For a “discharging” inductor I (t ) = I 0 e
−t /τ
where I0 is the current in the inductor when t = 0. The LR circuit 5me constant τ plays the same role as in an RC circuit. Physics 150, Prof. M. Nikolic 31 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. What is given: a) Find the current in the coil axer 50 ms. L = 0.3 H b) What is the total energy stored in the coil? R =15.0 Ω c) What is the of energy dissipa5on rate in the coil axer 50 ms? ε = 12 V a) ε
I(t) = (1− e−t/τ )
R
I(t) =
t = 50 ms Time constant given by: τ = L/R = 0.30 H/15 Ω = 0.02 s 12V
−0.05s/0.02s
1−
e
(
)
15Ω
I(t) = 0.73A
Maximum current in the circuit Physics 150, Prof. M. Nikolic ε
I max = = 0.8A
R
32 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. What is given: a) Find the current in the coil axer 50 ms. L = 0.3 H b) What is the total energy stored in the coil? R =15.0 Ω c) What is the of energy dissipa5on rate in the coil axer 50 ms? ε = 12 V t = 50 ms b) Total energy stored in the coil U = 1 LI 2
max
depends on maximum current 2
1
U = 0.3H ⋅ (0.8A)2 = 0.096J
2
c) Energy dissipa5on rate is equal to the power dissipated P = I 2R
P = (0.73A)215Ω = 8W
Physics 150, Prof. M. Nikolic 33 Review Voltage and current for moving rod ΔV = vBL
ΔV vBL
I=
=
R
R
Electric generators ε = NBAω sin ωt
Transformers Magne5c flux ε p Is N p
= =
εs I p N s
Φ B = BA cosθ
Mutual inductance Faraday’s law ε 2 = −N2
ΔΦ B
ε = −N
Δt
Physics 150, Prof. M. Nikolic ΔΦ 21
ΔI
= −M 1
Δt
Δt
ΔΦ12
ΔI
ε1 = − N1
= −M 2
Δt
Δt
34 Review Faraday’s law for self inductance “Charging” an inductor ΔΦ
ΔI
ε = −N
= −L
Δt
Δt
ε
I(t) = (1− e−t/τ )
R
I(t) = I max (1− e−t/τ )
Inductance of a solenoid ⎛ N 2 ⎞
⎟ A = µ 0 n 2 Aℓ
L = µ 0 ⎜
⎜ ℓ ⎟
⎝
⎠
ΔV = ε e−t/τ
Energy stored in an inductor “Discharging” an inductor 1 2
U = LI
2
I (t ) = I 0 e −t /τ
Magne5c energy density U
LI 2
1 2
uB =
=
=
B
Volume 2Volume 2µ 0
Physics 150, Prof. M. Nikolic 35