6/10/2011
hitt
Phy2049: Magnetism
• Last lecture: Biot-Savart’s and Ampere’s
law:
– Magnetic Field due to a straight wire
– Current loops (whole or bits)and solenoids
• Today: reminder and Faraday’s law.
Two long straight wires pierce the plane of the paper at vertices of an
equilateral triangle as shown. They each carry 3A but in the opposite
direction. The wire on the left has the current coming out of the paper
while the wire on the right carries the current going into the paper. The
magnetic field at the third vertex (P) has the magnitude and direction
(North is up):
(1) 20 µT, east (2) 17 µT, west (3) 15 µT, north (4)
26 µT, south (5) none of these
4 cm
X
Magnetic Field Units
Force Between Two Parallel Currents
• From the expression for force on a current-carrying wire:
– B = Fmax / I L
– Units: Newtons/A⋅m ≡ Tesla (SI unit)
– Another unit: 1 gauss = 10-4 Tesla
• Some sample magnetic field strengths:
– Earth: B = 0.5 gauss = 0.5 x 10-4 T = 5 x 10-5 T
– Galaxy: B ∼ 10-6 gauss = 10-10 T
– Bar magnet: B ∼ 100 – 200 gauss
– Strong electromagnet: B = 2 T
– Superconducting magnet: B = 20 – 40 T
– Pulse magnet: B ∼ 100 T
– Neutron star: B ∼ 108 – 109 T
– Magnetar: B ∼ 1011 T
Force Between Two Anti-Parallel
Currents
• Force on I2 from I1
µ II
µ I 
F2 = I 2 B1L = I 2  0 1  L = 0 1 2 L
2π r
 2π r 
– RHR ⇒ Force towards I1
• Force on I1 from I2
µ II
µ I 
F1 = I1B2 L = I1  0 2  L = 0 1 2 L
2π r
 2π r 
– RHR ⇒ Force towards I2
I2
Parallel Currents (cont.)
• Look at them edge on to see B fields more clearly
I2
I2
I1
• Magnetic forces repel two antiparallel currents
B
B
2
Antiparallel: repel
1
F
µ II
µ I 
F1 = I1B2 L = I1  0 2  L = 0 1 2 L
2π r
 2π r 
– RHR ⇒ Force away from I2
I1
• Magnetic forces cause attraction between two parallel currents
• Force on I2 from I1
µ II
µ I 
F2 = I 2 B1L = I 2  0 1  L = 0 1 2 L
2π r
 2π r 
– RHR ⇒ Force away from I1
• Force on I1 from I2
I2
I1
2
1
2
1
F
B
B
2
1
I1
F
Parallel: attract
F
1
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B Field @ Center of Circular Current
Loop
• Radius R and current i: find B field at center of loop
µi
B= 0
2R
– Direction: RHR #3 (see picture)
Current Loop Example
• i = 500 A, r = 5 cm, N=20
µ 0i
B=N
2r
=
( 20 ) ( 4π ×10−7 ) 500
2 × 0.05
= 1.26T
• If N turns close together
B=
N µ 0i
2R
Challenge: How much heat is produced?
B Field of Solenoid
• Formula found from Ampere’s
law
– i = current
– n = turns / meter
B = µ0in
– B ~ constant inside solenoid
– B ~ zero outside solenoid
– Most accurate when L>>R
• Example: i = 100A, n = 10
turns/cm
– n = 1000 turns / m
B = ( 4 π × 1 0 − 7 ) (1 0 0 ) (1 0 3 ) = 0 .1 3 T
Field at Center of Partial Loop
• Suppose loop covers angle φ
µ 0i  φ 


2 R  2π 
B=
• Use example where φ = π (half circle)
– Define direction into page as positive
µ i  π  µ 0i  π 
B= 0 
−


2 R1  2π  2 R2  2π 
B=
µ 0i  1
The magnetic field of a magnetic dipole.
Partial Loops (cont.)
• Note on problems when you have to evaluate a B field at a
point from several partial loops
– Only loop parts contribute, proportional to angle
(previous slide)
– Straight sections aimed at point contribute nothing
– Be careful about signs, e.g.in (b) fields partially cancel,
whereas in (a) and (c) they add
1 
 − 
4  R1 R2 
Consider the magnetic field generated by a wire coil of
radius R which carries a current i. The magnetic field
at a point P on the z-axis is given by:
B=
r
r
µ µ
B( z ) = o 3
2π z
µ oiR 2
2 ( R2 + z 2 )
3/ 2
Here z is the distance between
P and the coil center. For points far from the loop
(z
R ) we can use the approximation: B ≈
µo iπ R 2
µo iR 2
2 z3
µ iA µ µ
= o 3= o 3
2π z
2π z
B=
Here µ is the magnetic
2 pz 3
dipole moment of the loop. In vector form:
r
r
µ µ
B( z ) = o 3
2π z
The loop generates a magnetic field that has the same
form as the field generated by a bar magnet.
(29 – 14)
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Chapter 30
Induction and Inductance
Faraday's experiments
In a series of experiments Michael Faraday in England
(30 – 2)
and Joseph Henry in the US were able to generate
electric currents without the use of batteries
Below we describe some of these experiments that
In this chapter we will study the following topics:
helped formulate whats is known as "Faraday's law
of induction"
-Faraday’s law of induction
-Lenz’s rule
-Electric field induced by a changing magnetic field
The circuit shown in the figure consists of a wire loop connected to a sensitive
ammeter (known as a "galvanometer"). If we approach the loop with a permanent
magnet we see a current being registered by the galvanometer. The results can be
summarized as follows:
1. A current appears only if there is relative motion between the magnet and the loop
2. Faster motion results in a larger current
Electric current without a battery
-Inductance and mutual inductance
- RL circuits
-Energy stored in a magnetic field
(30 – 1)
In the figure we show a second type of experiment
in which current is induced in loop 2 when the
(30 – 3)
loop 1
switch S in loop 1 is either closed or opened. When
the current in loop 1 is constant no induced current
is observed in loop 2. The conclusion is that the
magnetic field in an induction experiment can be
loop 2
generated either by a permanent magnet or by an
electric current in a coil.
Faraday summarized the results of his experiments in what is known as
"Faraday's law of induction"
An emf is induced in a loop when the number of magnetic field lines that
pass through the loop is changing
3. If we reverse the direction of motion or the polarity of the magnet, the current
reverses sign and flows in the opposite direction.
The current generated is known as "induced current"; the emf that appears
is known as "induced emf"; the whole effect is called "induction"
r r
Φ B = ∫ B ⋅ dA
n̂
r
B
φ
dA
Magnetic Flux Φ B
The magnetic flux through a surface that borders
a loop is determined as follows:
1. we divide the surface that has the loop as its border
into area elements of area dA.
2. For each element we calculate the magnetic flux through it: d Φ B = BdA cos φ
r
Here φ is the angle between the normal nˆ and the magnetic field B vectors
at the position of the element.
r r
3. We integrate all the terms. Φ B = ∫ BdA cos φ = ∫ B ⋅ dA
SI magnetic flux unit: T ⋅ m 2 known as the Weber (symbol Wb)
We can express Faraday's law of induction in the folowing form:
The magnitude of the emf E induced in a conductive loop is equal to rate
Faraday's law is not an explanation of induction but merely a description of
of what induction is. It is one of the four "Maxwell's equations of electromagnetism"
all of which are statements of experimental results. We have already encountered
Gauss' law for the electric field, and Ampere's law (in its incomplete form)
r r
Φ B = ∫ BdA cos φ = ∫ B ⋅ dA
n̂
φ
dA
r
B
Methods for changing ΦB through a loop
1. Change the magnitude of B within the loop
2. Change either the total area of the coil or
the portion of the area within the magnetic field
r
3. Change the angle φ between B and nˆ
dΦB
= ω NabB sin (ωt )
dt
ω = 2π f →
n̂
E=−
dΦB
dt
dΦB
dt
(30 – 4)
Lenz's Rule
We now concentrate on the negative sign
in the equation that expresses Faraday's law.
The direction of the flow of induced current
in a loop is acurately predicted by what is
known as Lenz's rule.
induced current opposes the change in the magnetic flux that induces the current
Lenz's rule can be implemented using one of two methods:
E=−
φ
E=−
An induced current has a direction such that the magnetic field due to the
An Example. Problem 30-11
Φ B = NAB cos φ = NabB cos (ωt )
r
B
at which the magnetic flux ΦB through the loop changes with time
1. Opposition to pole movement
In the figure we show a bar magnet approaching a loop. The induced current flows
in the direction indicated because this current generates an induced magnetic field
E = 2π fNabB sin ( 2π t )
that has the field lines pointing from left to right. The loop is equivalent to a
magnet whose north pole haces the corresponding north pole of the bar magnet
approaching the loop. The loop repels the approaching magnet and thus opposes
loop
(30 – 5)
the change in Φ B which generated the induced current.
(30 – 6)
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2. Opposition to flux change
N
S
2. Opposition to flux change
Example a : Bar magnet approaches the loop
with the north pole facing the loop.
N
magnet
motion
S
Example b : Bar magnet moves away from the loop
with north pole facing the loop.
magnet
motion
r
As the bar magnet approaches the loop the magnet field B points towards the left
and its magnitude increases with time at the location of the loop. Thus the magnitude
of the loop magnetic flux Φ B also increases. The induced current flows in the
r
counterclockwise (CCW) direction so that the induced magnetic field Bi opposes
r
r
r r
the magnet field B. The net field Bnet = B − Bi . The induced current is thus trying
to prevent Φ B from increasing. Remember it was the increase in Φ B that generated
the induced current in the first place.
r
As the bar magnet moves away from the loop the magnet field B points towards the left
and its magnitude decreases with time at the location of the loop. Thus the magnitude
of the loop magnetic flux Φ B also decreases. The induced current flows in the
r
clockwise (CW) direction so that the induced magnetic field Bi adds to
r
r
r r
the magnet field B. The net field Bnet = B + Bi . The induced current is thus trying
to prevent Φ B from decreasing. Remember it was the decrease in Φ B that generated
the induced current in the first place.
(30 – 7)
S
(30 – 8)
2. Opposition to flux change
2. Opposition to flux change
Example c : Bar magnet approaches the loop
with south pole facing the loop.
Example d : Bar magnet moves away from the loop
with south pole facing the loop.
S
N
N
magnet
motion
magnet
motion
r
As the bar magnet approaches the loop the magnet field B points towards the right
and its magnitude increases with time at the location of the loop. Thus the magnitude
of the loop magnetic flux Φ B also increases. The induced current flows in the
r
clockwise (CW) direction so that the induced magnetic field Bi opposes
r
r
r r
the magnet field B. The net field Bnet = B − Bi . The induced current is thus trying
to prevent Φ B from increasing. Remember it was the increase in Φ B that generated
the induced current in the first place.
r
As the bar magnet moves away from the loop the magnet field B points towards the
right and its magnitude decreases with time at the location of the loop. Thus
the magnitude of the loop magnetic flux Φ B also decreases. The induced current
r
flows in the counterclockwise (CCW) direction so that the induced magnetic field Bi
r
r
r r
adds to the magnet field B. The net field Bnet = B + Bi . The induced current is thus
trying to prevent Φ B from decreasing. Remember it was the decrease in Φ B that
generated the induced current in the first place.
(30 – 9)
(30 –10)
Induction and energy transfers
The rate at which thermal energy is dissipated on R
By Lenz's rule, the induced current always opposes
the external agent that produced the induced current.
Thus the external agent must always do work on the
2
B 2 L2v 2
 BLv 
Pth = i 2 R = 
(eqs.1)
 R=
R
R


The magnetic forces on the wire sides are shown
r
r
in the figure. Forces F2 and F3 cancel each other.
r
r r
BLv
Force F1 = iL × B
F1 = iLB sin 90° = iLB =
LB
R
2 2
B Lv
F1 =
The rate at which the external agent is
R
B 2 L2v 2
producing mechanical work Pext = F1v =
(eqs.2)
R
If we compare equations 1 and 2 we see that indeed the
loop-magnetic field system. This work appears as
thermal energy that gets dissipated on the resistance
R of the loop wire.
Lenz's rule is actually a different formulation of
the principle of energy conservation
Consider the loop of width L shown in the figure.
Part of the loop is located in a region where a
uniform magnetic field B exists. The loop is being
pulled outside the magnetic field region with constant
speed v. The magnetic flux through the loop
Φ B = BA = BLx The flux decreases with time
E=
(30 –11)
d ΦB
dx
= BL = BLv
dt
dt
i=
E BLv
=
R
R
mechanical work done by the external agent that moves
the loop is converted into thermal energy that appears
on the loop wires.
(30 –12)
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Induced electric fields
Consider the copper ring of radius r shown in the
Eddy currents
We replace the wire loop in the previous example
with a solid conducting plate and move the plate
figure. It is paced in a uniform magnetic field B
pointing into the page, that icreases as function
of time. The resulting change in magnetic flux
out of the magnetic field as shown in the figure.
r
The motion between the plate and B induces a
current in the conductor and we encounter an opposing
induces a current i in the counterclock-wise
(CCW) direction.
force. With the plate the free electrons do not follow
one path as in the case of the loop. Instead the electrons
swirl around the plate. These currents are known as
"eddy currents". As in the case of the wire loop the net
result is that mechanical energy that moves the plate is
The presence of the current i in the conducting ring implies that an induced
r
electric field E must be present in order to set the electrons on motion.
Using the argument above we can reformulate Faraday's law as follows:
A changing magnetic field produces an electric field
transformed into thermal energy that heats up the plate.
Note : The induced electric field is generated even in the absense of the
copper ring.
(30 –14)
(30 –13)
Consider the circular closed path of radius r shown in
the figure to the left. The picture is the same as that
in the previous page except that the copper ring has
Inductance
Consider a solenoid of length l that has N loops of
r
B
The emf is also given by Faraday's law:
dΦB
E=−
(eqs.2) If we compare eqs.1 with eqs.2
dt
r r
dΦB
we get: ∫ E ⋅ ds = −
dt
r r
∫ E ⋅ ds = ∫ Eds cos 0 = E ∫ ds = 2π rE
dΦB
dB
ΦB = π r B →
= π r2
dt
dt
r dB
dB
→ 2π rE = π r 2
→E=
dt
2 dt
N
windings per unit length. A current
l
i flows through the solenoid and generates a unifrom
magnetic field B = µ o ni inside the solenoid.
area A each, and n =
been removed. The path is now an abstract line.
The emf along the path is given by the equation:
r r
E = ∫ E ⋅ ds
(eqs.1)
l
The solenoid magnetic flux Φ B = NBA
L = µo n 2 lA
The total number of turns N = nl → Φ B = ( µ o n 2lA ) i
The result we got for the
special case of the solenoid is true for any inductor. Φ B = Li.
Here L is a
constant known as the inductance of the solenoid. The inductance depends
on the geometry of the particular inductor.
2
Inductance of the solenoid
For the solenoid L =
Φ B µo n 2 lAi
=
= µ o n 2lA
i
i
(30 –15)
E = −L
di
dt
Self Induction
In the picture to the right we
already have seen how a change
in the current of loop 1 results
in a change in the flux through
loop 2, and thus creates an
induced emf in loop 2
loop 1
loop 2
(30 –16)
i(t ) =
E
(1 − e−t /τ ) τ = RL
R
(30 –17)
(thus the name "RL circuit"). Our objective is to
calculate the current i as function of time t. We
write Kirchhoff's loop rule starting at point x and
moving around the loop in the clockwise direction.
di
di
−iR − L + E = 0 → L + iR = E
dt
dt
If we change the current through an inductor this causes
a change in the magnetic flux Φ B = Li through the inductor
d ΦB
di
=L
Using Faraday's
dt
dt
law we can determined the resulting emf known as
d ΦB
di
self induced emf. E = −
= −L
dt
dt
SI unit for L : the Henry (symbol: H)
according to the equation:
An inductor has inductance L = 1 H if a current
change of 1 A/s results in a self-induced emf of 1 V.
RL circuits
Consider the circuit in the upper figure with the switch
S in the middle position. At t = 0 the switch is thrown
in position a and the equivelent circuit is shown in
the lower figure. It contains a battery with emf E,
connected in series to a resistor R and an inductor L
The initial condition for this problem is: i(0) = 0. The solution of the differential
equation that satisfies the initial condition is:
E
(1 − e−t /τ )
R
the RL circuit.
i (t ) =
The constant τ =
L
is known as the "time constant" of
R
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E
L
Here τ =
(1 − e−t /τ )
R
R
The voltage across the resistor VR = iR = E (1 − e −t /τ ) .
i (t ) =
UB =
Li 2
2
Energy stored in a magnetic field
We have seen that energy can be stored in the electric field
of a capacitor. In a similar fashion energy can be stored in
the magnetic field of an inductor. Consider the circuit
shown in the figure. Kirchhoff's loop rules gives:
di
= Ee −t /τ
dt
The solution gives i = 0 at t = 0 as required by the
The voltage across the inductor VL = L
initial condition. The solution gives i (∞) = E / R
The circuit time constant τ = L / R tells us how fast
the current approaches its terminal value.
i (t = τ ) = ( 0.632 )( E / R )
i (t = 3τ ) = ( 0.950 )( E / R )
i (t = 5τ ) = ( 0.993)( E / R )
If we wait only a few time constants the current, for all
practical purposes has reached its terminal value ( E / R ) .
di
di
If we multiply both sides of the equation we get: Ei = Li + i 2 R
+ iR
dt
dt
The term Ei describes the rate at which the batter delivers energy to the circuit
E=L
The term i 2 R is the rate at which thermal energy is produced on the resistor
di
Using energy conservation we conclude that the term Li
is the rate at which
dt
dU B
di
energy is stored in the inductor.
= Li → dU B = Lidi We integrate
dt
dt
i
(30 –19)
both sides of this equation: U B = ∫ Li′di′ =
o
L i′2 
2
i
o
=
Li 2
2
(30 –20)
Energy density of a magnetic field
r
B
Mutual Induction
Consider the soleniod of length l and loop area A
that has n windings per unit length. The solenoid
carries a current i that generates a uniform magnetic
field B = µo ni inside the solenoid. The magnetic field
l
N2
N1
Consider two inductors
which are placed close enough
so that the magnetic field of one
can influence the other.
outside the solenoid is approximately zero.
B2
uB =
2µo
E2 = − M 21
1 2 µ o n 2 Ali 2
Li =
2
2
This energy is stored in the empty space where the magnetic field is present
U
We define as energy density u B = B where V is the volume inside
V
2
µ n 2 Ali 2 µ o n 2i 2 µ o n 2i 2 B 2
the solenoid. The density u B = o
=
=
=
2 Al
2
2 µo
2µo
The energy U B stored by the inductor is equal to
In fig.a we have a current i1 in inductor 1. That creates a magnetic field B1
in the vicinity of inductor 2. As a result, we have a magnetic flux Φ 2 = M 21i1
through inductor 2. If curent i1 varies with time, then we have a time varying
This result, even though it was derived for the special case of a uniform
magnetic field, holds true in general.
(30 –21)
N2
flux through inductor 2 and therefore an induced emf across it.
dΦ2
di
E2 = −
= − M 21 1
M 21 is a constant that depends on the geometry
dt
dt
of the two inductors as well as their relative position.
(30 –22)
(30 –23)
N1
di1
dt
N2
N1
di
E1 = − M 12 2
dt
E1 = − M 12
di2
dt
E2 = − M 21
di1
dt
(30 –24)
It can be shown that the constants M 12 and M 21 are equal. M 12 = M 21 = M
In fig.b we have a current i2 in inductor 2. That creates a magnetic field B2
in the vicinity of inductor 1. As a result, we have a magnetic flux Φ1 = M 12i2
through inductor 1. If curent i2 varies with time, then we have a time varying
flux through inductor 1 and therefore an induced emf across it.
d Φ1
di
= − M 12 2
M 12 is a constant that depends on the geometry
dt
dt
of the two inductors as well as their relative position.
E1 = −
The constant M is known as the "mutual inductance" between the two coils.
Mutual inductance is a constant that depends on the geometry of the two
inductors as well as their relative position. The SI unit for M : the Henry (H)
The expressions for the induced emfs across the two inductors become:
E1 = − M
di2
dt
E2 = − M
di1
dt
Φ1 = Mi2
Φ 2 = Mi1
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Electrons are going around a circle in a counterclockwise
direction as shown. At the center of the circle they produce a
magnetic field that is:
Long parallel wires carry equal currents into or out of the page.
Rank according to the magnitude of the net magnetic field at the
center of the square, highest first. (Parenthesis means the same
magnitude).
e
A. into the page
B. out of the page
C. to the left
D. to the right
E. zero
1. C,D, (A,B)
4. D, (A, B), C
2. A, B, (C,D)
3. B, A, C, D
Long, straight, parallel wires carry equal currents into or out of
page. Rank according to the magnitude of the force on the central
wire.
1. d, c, a, b
5. b, d, c, a
2. a, b, c, d
3. b, c, d, a
4. c, a, b, d
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