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Hong Kong Physics Olympiad Lesson 18
18.1 Permanent magnets
18.2 The magnetic force on moving charge
18.3 Electric versus magnetic forces
18.4 The magnetic force exerted on a current-carrying wire
18.5 Current loops and magnetic torque
18.6 Biot and Savart’s law
18.7 Ampere’s law
18.8 Forces between current wires
18.9 Magnetism
18.10 Induced EMF
18.11 Magnetic flux
18.12 Faraday’s law of induction
18.13 Lenz’s law
18.14 Motional EMF
18.1 Permanent magnets
A bar magnet can attract another magnet or repel it, depending on which ends of the
magnets are brought together. One end of a magnet is referred to as its north pole; the
other end is its south pole. The rule for whether two magnets attract or repel each
other: opposites attract; likes repel. Breaking a magnet in half results in the
appearance of two new poles on either side of the break. This behavior is
fundamentally different from that in electricity, where the two types of charge can
exist separately.
We saw a visual indication of the electric field E of a point charge using grass seed
suspended in oil. Similarly, the magnetic field B can be visualized using small iron
filings sprinkled onto a smooth surface. The filings are bunched together near the
poles of the magnets. This is where the magnetic field is most intense. The direction
2
of the magnetic field, B, at a given location is the direction in which the north pole of
a compass points when placed at that location. In general, magnetic field lines exit
from the north pole of a magnet and enter at the south pole.
18.2 The magnetic force on moving charge
The magnetic force depends on several factors:
• The charge of the particle, q;
• The speed of the particle, v;
• The magnitude of the magnetic field, B;
• The angle between the velocity vector and
the magnetic field vector, θ.
The mathematical relation of them in vector form is
)( BvqFrrr
×= .
One can rewrite it as a scalar expression, e.g. θsinqvBF = . The maximum force is
obtained when θ = 90o. The force vanishes when θ = 0
o. Now we define the magnetic
field B as θsinqv
FB = . The SI unit is 1 tesla = 1 T = 1 N/(A⋅m). The tesla is a fairly
large unit of magnetic strength, especially when compared with the magnetic field at
the surface of the Earth, which is roughly T5100.5 −× . Thus, another commonly used
3
unit of magnetism is the gauss (G), defined as follows: TG4101 −= . In terms of the
gauss, the Earth’s magnetic field on the surface of the Earth is approximately 0.5 G. A
bar magnet has a magnetic field of roughly 100 G.
Remark:
Magnetic field lines never cross one another. As the direction in which a compass
points at any given location is the direction of the magnetic field at that point. Since a
compass can point in one direction, there must be only one direction for the field B. If
field lines were to cross, however, there would be two directions for B at the crossing
point, and this is not allowed.
Example
Particle 1, with a charge q1 = 3.60 µC and a speed v1 = 862 m/s travels at right angles
to a uniform magnetic field. The magnetic force it experiences is N3
1025.4−× .
Particle 2, with a charge q2 = 53.0 µC and a speed
smv /1030.1 3
2 ×= moves at an angle of 55.0o
relative to the same magnetic field. Find (a) the
strength of the magnetic field and (b) the magnitude
of the magnetic force exerted on particle 2.
Answer:
Apply the formula: θsinqvBF = with θ = 90o. One obtains
TsmC
N
qv
FB
o37.1
90sin)/862)(1060.3(
1025.4
sin 6
3
=×
×==
−
−
θ.
Apply the formula: θsinqvBF = with θ = 55.0o. One obtains
NTsmF o 0773.00.55sin)37.1()/1030.1)(100.53( 36 =××= − .
The direction of the magnetic force is given by the magnetic force right-hand rule
(RHR), which states as follows.
To find the direction of magnetic force on a positive charge, start by pointing the
fingers of your right hand in the direction of the velocity, v. Now curl your fingers
forward the direction of B. Your thumb points in the direction of F.
4
If the charge is negative, the force points opposite to the
direction of your thumb. Note that the magnetic force F points
in a direction that is perpendicular to both B and the charge
velocity v.
As an example, the direction of a magnetic force F can be
indicated by the magnetic force RHR – extending our fingers to
the right (v) and then curling them into the page (B) – we see
that the magnetic force exerted on this particle is upward, as
indicated. If the charge is negative, the direction of F is
reversed.
Example
Three particles travel through a region of space where the magnetic field is out of the
page as shown in figure. For each of the three particles, state whether the particle’s
charge is positive, negative, or zero.
Answer:
Use the RHR to obtain the following. Particle 1: negative; particle 2: zero and particle
3: positive.
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Example
A particle with a charge of 7.70 µC and a speed of 435 m/s is acted
on by both an electric and a magnetic field. The particle moves along
the x axis in the positive direction, the magnetic field has a strength
of 3.20 T and points in the positive y direction, and the electric field
points in the positive z direction with a magnitude of
CN /1010.8 3× . Find the magnitude and direction of the net force
acting on the particle.
Answer:
Both forces point to the positive z direction.
Electric force: NCNCqEFE
236 1024.6)/1010.8)(1070.7( −− ×=××== .
Magnetic force: NTsmCqvBFB
26 1007.1)20.3)(/435)(1070.7( −− ×=×== .
Net force: NNNFFF BEnet
222 1031.7)1007.1()1024.6( −−− ×=×+×=+= .
18.3 Electric versus magnetic forces
Charges moving under electric field and magnetic field as shown in figure, undergo
circular motion. In an electromagnetic flowmeter, an artery passes between the poles
of a magnet. Charged ions in the blood will be deflected at right angles to the artery
by the magnetic field. The resulting charge separation produces an electric field that
opposes the magnetic deflection. When the electric field is strong enough the
deflection ceases and the blood flows normally through the artery. If the electric field
is measured, and the magnetic field is known, the speed of the blood flow is simply
B
Ev = (since qvBqE = ), as in a standard velocity selector.
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Remarks:
1. As the direction of velocity and the direction of magnetic force are
perpendicular to each other, the work done by the magnetic force is always
zero.
2. A charged particle with a velocity v that is
perpendicular to the magnetic field moves in a
circular path. The magnetic force acts as the
centripetal force, e.g. r
mvFcp
2
= and qvBFcp = .
Hence qvBr
mv=
2
gives the radius of circular path,
qB
mvr = . On the other hand,
qvBrmr
mvFcp === 2
2
ω , we can write qBm =ω ,
as ωrv = . Plugging in the relation T
πω
2= , we have
qBT
m =π2
, and the period T is given by qB
mT
π2= .
3. Helical motion is a combination of linear motion and circular motion.
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Example
In a device called a velocity selector, charged particles move through a region of
space with both an electric and a magnetic field. If the speed of the particle has a
particular value, the net force acting on it is zero. Assume that a positively charged
particle moves in the positive x direction, and the electric field is in the positive y
direction. Should the magnetic field be in (a) the positive z direction, (b) the negative
y direction, or (c) the negative z direction in order to give zero net force?
Answer:
The force exerted by the electric field is in the positive y
direction; hence, the magnetic force must be in the
negative y direction if it is to cancel the electric force. If
we simply try the three possible directions for B one at a
time, applying the magnetic force RHR in each case, we
find that only a magnetic field along the positive z axis
gives rise to a force in the negative y direction, as desired.
The answer is (a).
Example
An electron moving perpendicular to a magnetic field of
T31060.4 −× follows a circular path of radius 2.80 mm.
What is the electron’s speed?
Answer:
Form the equation qvBr
mv=
2
, we obtain
smkg
TCm
m
rqBv /1026.2
1011.9
)1060.4)(1060.1)(1080.2( 6
31
3193
×=×
×××==
−
−−−
.
Example
Two isotopes of uranium, 235
U ( kg251090.3 −× ) and 238
U ( kg251095.3 −× ), are sent
into a mass spectrometer with a speed of sm /1005.1 5× . Given that each isotope is
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singly ionized, and that the strength of the magnetic field is 0.750 T, what is the
distance d between the two isotopes after they complete half a circular orbit?
Answer:
The isotopes are singly ionized, which means that a single electron has been removed
from each atom. And the isotopes are now having charge of Ce19
1060.1−
×= .
cmTC
smkg
qB
mvr 1.34
)750.0)(1060.1(
)/1005.1)(1090.3(19
525
235 =×
××==
−
−
.
cmTC
smkg
qB
mvr 6.34
)750.0)(1060.1(
)/1005.1)(1095.3(19
525
238 =×
××==
−
−
.
The separation between the isotopes: cmcmcmrrd 1)1.346.34(222 235238 =−=−= .
18.4 The magnetic force exerted on a current-carrying wire
As a charged particle experiences a force when it moves across magnetic field lines.
The same thing happens when a wire carries current on it. Consider a straight segment
of length L of a wire with a current I flowing from left to right, presents in a magnetic
field B, as shown in figure. If the conducting charges move through the wire with an
average speed v, the time required for them to move
from one end of the wire segment to the other is
vLt /=∆ . The amount of charge that flows through the
wire in this time is vILtIq /=∆= . Therefore, the
force exerted on the wire is
θθ sinsin vBv
LIqvBF
== .
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Hence, we have θsinILBF = . Maximum force occurs when the current is
perpendicular to the magnetic field (θ = 90o) and is zero if the current is in the same
direction as B (θ = 0o). The direction of the magnetic force is given by the RHR,
where the direction of charge velocity v is now the direction of current I.
Example
When the switch is closed in the circuit, the wire between the poles of the horseshoe
magnet deflects downward. Is the left end of the magnet (a) a north magnet pole of (b)
a south magnetic pole?
Answer:
Once the switch is closed, the current in the wire is into the page, as shown in the
right figure. Applying the magnetic force RHR, we see that the magnetic field must
point from left to right in order for the force to be downward. Since magnetic field
lines leave from north poles and enter at south poles, it follows that the left end of the
magnet must be a north magnetic pole. The answer is (a).
18.5 Current loops and magnetic torque
A torque is experienced by a current loop as shown in
figure. If we imagine an axis of rotation through the
center of the loop, at the point O, it is clear that the forces
exert a torque that tends to rotate the loop clockwise.
IABhwIBw
IhBw
IhB ==
+
= )(
2)(
2)(τ ,
Plug in the area of the rectangular loop, A = hw. We have
IAB=τ .
If the plane of loop makes an angle with the magnetic field, we have θτ sinIAB= .
10
For the case that has n turns in a general loops, we have θτ sinnIAB= .
Applications of the magnetic torque are electric motors and galvanometer.
18.6 Biot and Savart’s law
Recall that in the Coulomb’s law, we have, say, a point charge q
gives an electric field at P and rr
qE ˆ
4
12
0πε=
r. For a charge
distribution, the electric field is at P due to dq is rr
dqEd ˆ
4
12
0πε=
r.
Now, for a magnetic field, we have the law of Biot and Savart
2
0 sin
4 r
dsIdB
θ
π
µ= ,
where
AmTAmT /1026.1/104 67
0 ⋅×≈⋅×= −−πµ and . θ
r
sdr
I
P
Magnetic field I
r
×
Current Current
Br
−field
Charge Charge Er
−Electric field
rr
dqdE
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0µ is called the permeability constant. In vector form, we have 2
0 ˆ
4 r
rsdIBd
×=
rr
π
µ.
Comparing with the Coulomb’s law, we realize that both are2
1
rlaws.
Example
Find the magnetic field at the center of a coil which carries a steady current I.
Answer:
The magnetic field through the center of coil is the
superposition of the magnetic field due to current segment.
Hence, ∑=2
0 sin
4 r
dsIB
θ
π
µ
As r is a constant and θ = 90o, we can write
r
Ir
r
Ids
r
IB
2)2(
44
0
2
0
2
0 µπ
π
µ
π
µ=== ∑ .
If the coil has N turns, the magnetic field is given by r
NIB
2
0µ= .
Remark:
If the solenoid has N turns, length l and carries
a current I, the magnetic field B at a point O on
the axis near the center of the solenoid is found
to be
nIl
NIB 0
0 µµ
== ,
where n is the number of turns per unit length of coil. If the coil is infinite long or
long enough that its length is ten times the diameter, the magnetic field at the end of
coil is half that at the center of coil, e.g. 2
0nIB
µ= .
Example
Find the magnetic field Bv
at a point P, a distance z from a long current wire, as
shown in the figure.
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Answer:
The magnetic field ∫×
=2
0ˆ
4 r
rldIB
vv
π
µ, where
θφ cossinˆ dldlrld ==×v
.
As θtanzl = , we have θθ dzdl )cos/(2
= .
We can write ∫−
= 2
2
2
0 cos
4
π
π
θ
π
µ
r
dlIB .
Now, θcos/ =rz ⇒ zr
θcos1= gives
z
I
z
Id
z
I
dz
zIB
π
µθ
π
µθθ
π
µ
θθθ
θπ
µ
π
π
π
π
π
π
2sin
4cos
4
coscos
cos4
02
2
02
2
0
2
2
2
2
2
0
===
=
−−
−
∫
∫
Hence, the magnetic field at P due to a long wire of current is given byz
IB
π
µ
2
0= .
Remark 1:
The magnitude of the magnetic field 1 m from a long, straight wire carrying a current
of 1 A is Tm
AAmT
z
IB
77
0 102)1(2
)1)(/104(
2
−−
×=⋅×
==π
π
π
µ.
This is a weak field, less than one hundredth the strength of the Earth’s magnetic field.
Remark 2
The magnetic field shown in the figure is due to the horizontal, current-carrying wire.
The current in the wire flows to the left. The following states the use of magnetic field
right-hand rule which points the direction of current:
θ
P
ldv
rv
O I
φ z
l
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If you point the thumb of your right hand along the wire to the left your
fingers curl into the page above the wire and out of the page below the
wire. Thus, the current flows to the left.
Example
A current loop of radius a lies on the yz-plane. Find the magnetic field at a distance x
far from the center of loop (P lies on the x-axis).
Answer:
From Biot-Savart’s law, we have
2
0ˆ
4 r
rldiBd
×=
rr
π
µ.
Therefore, the magnetic field due to a
small current segment is
( )22
0
4 ax
dlidB
+=
π
µ
The x-component of dB is
θcosdBdBx = .
Hence, we can write
( ) ( ) 21
2222
0
4 ax
a
ax
dlidBx
++=
π
µ
The y-component of dBy:
θsindBdBy =
( ) ( ) 2
122
22
0
4 ax
x
ax
dlidBy
++=
π
µ.
But the sum of these dBy gives zero contribution in y-direction.
( ) ( )∫ ∫
+
=
+
= dl
ax
ia
ax
adliBx
23
22
0
23
22
0
44 π
µ
π
µ, ∫ = adl π2
( ) 23
22
2
0
2 ax
iaBx
+
=µ
(circular loop)
θ
xdB
Bdr
ydBldv
θ
i z
y
x x
rr
a
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A coil consists of N closely spaced loops:
( ) 23
22
2
0
2 ax
NiaBx
+
=µ
Consider the limiting case for a single loop, where 0=x and 1=N , the magnetic
field at the center of the loop is given by a
iB
2
0µ= .
18.7 Ampere’s law
Electric currents can create magnetic fields. The direction of
the magnetic field is given by the magnetic field right-hand
rule. The magnetic field right-hand rule states the following:
To find the direction of the magnetic field due to a current-
carrying wire, point the thumb of your right hand along the
wire in the direction of the current I. Your fingers are now
curling around the wire in the direction of the magnetic field.
Experiments show that the field produced by a current-carrying
wire doubles if the current I is doubled. In addition, the field
decreases by a factor of 2 if the distance from the wire, r, is
doubled. Hence, we conclude that the magnetic field B must be
proportional to rI / ; that is r
IB ∝ .
Ampere’s law states that the sum of ldBrr
⋅ along the closed path is proportional to the
current enclosed by the path. Mathematically, we have
enclosedIldB 0µ=⋅∫rr
.
The proportional constant, µ0 is the permeability of free
space. Its value is
AmT /104 7
0 ⋅×= −πµ .
For example, the magnetic field at a distance r due to a long
wire of current can by obtained by the Ampere’s law.
enclosedIrBldB 0)2( µπ ==⋅∫rr
,
which gives r
IB
π
µ
2
0= .
15
Remark:
Recall that in the long solenoid, which has n turns per unit
length of solenoid and carries a current I, the magnetic
field B at a point O on the axis of solenoid is found to be
nIB 0µ= .
In fact, this expression can be obtained simply by using
the Ampere’s law. Consider the amperian loop as shown
in figure. The length, side 1 of the amperian loop is L,
which has N turns of coils. The magnetic field is nearly
uniform and tightly packed inside the loops. In the ideal
case of a very long, tightly packed solenoid, the
magnetic field outside is practically zero – especially
when compared with the intense field inside the solenoid.
We can use this idealization, in combination with
Ampere’s law, to calculate the magnitude of the field inside the solenoid.
enclosed
sidesidesideside
IBLLBLBLBLBldB 0
4
//
3
//
2
//
1
// 000 µ=+++=∆+∆+∆+∆=⋅ ∑∑∑∑∫rr
The answer is simply )(00 nLIINBL µµ == , which gives nIB 0µ= .
Example
If you want to increase the strength of the magnetic field inside a solenoid is it better
to (a) double the number of loops, keeping the length the same, or (b) double the
length, keeping the number of loops the same?
Answer:
Since nIB 0µ= , we know that the number of coil per unit length of solenoid governs
the magnitude of magnetic field. Doubling the number of loops and keeping the
length the same, results in doubling the variable n. So, the answer is (a).
Example
A solenoid is 20.0 cm long, has 200 loops, and carries a current of 3.25 A. Find the
force exerted on a 15.0-µC charged particle moving at 1050 m/s through the interior
of the solenoid, at an angle of 11.5o relative to the solenoid’s axis.
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Answer:
The magnetic field inside the solenoid:
⋅×=
⋅×=
== −−
TAm
AmTIL
NnIB
37
00 1008.4)25.3(200.0
200)/104( πµµ
The magnetic force on the charged particle:
NTsmCqvBF o 536 1028.15.11sin)1008.4)(/1050)(100.15(sin −−− ×=××== θ .
Example
A 52-µC charged particle moves parallel to a long wire
with a speed of 720 m/s. The separation between particle
and the wire is 13 cm, and the magnitude of the force
exerted on the particle is N7104.1 −× . Find (a) the
magnitude of the magnetic field at the location of the
particle and (b) the current in the wire.
Answer:
As the magnetic force is given by qvBF = , which gives the magnetic field
TsmC
N
qv
FB 6
5
7
107.3)/720)(102.5(
104.1 −
−
−
×=×
×== .
The magnetic field B at a distance r from the long current wire is given by r
IB
π
µ
2
0= .
Now, the current can be obtained as
AAmT
TmrBI 4.2
/104
)107.3)(13.0(227
6
0
=⋅×
×==
−
−
π
π
µ
π.
17
Example
Two wires separated by a distance of 22 cm carry currents in the same direction. The
current in one wire is 1.5 A, and the current in the other wire is 4.5 A. Find the
magnitude of the magnetic field halfway between the wires.
Answer:
The magnetic field due to I1:
Tm
A
r
IB 6
2
010
1 107.2)100.11(2
)5.1(
)2/(2
−
−×=
×==
π
µ
π
µ, into page.
The magnetic field due to I2:
Tm
A
r
IB 6
2
020
2 102.8)100.11(2
)5.4(
)2/(2
−
−×=
×==
π
µ
π
µ, out of page.
The net magnitude of magnetic field: 6
12 105.5 −×=−= BBB , out of page.
18.8 Forces between current wires
We know that current-carrying wire in a magnetic field
experiences a force. We also know that a current-
carrying wire produces a magnetic field. It follows, then,
that one current-carrying wire will exert a force on
another.
The force experienced by wire 2 in the figure, has a
magnitude
Ld
II
d
ILILBIF
π
µ
π
µ
22
21010
22 =
== .
Similarly, we notice that an equal magnitude but opposite
direction force is experienced by wire 1. To conclude,
Wires with parallel currents attract one another.
Wires with opposite currents repel one another.
18
18.9 Magnetism
There are three types of magnetism.
(a) Ferromagnetism – A ferromagnetic material produces a magnetic field
even in the absence of an external magnetic field. Permanent magnets are
constructed of ferromagnetic materials, e.g. bar magnets.
(b) Paramagnetism – A paramagnetic material has no magnetic field unless an
external magnetic field is applied to it. Is this case it develops a
magnetization in the direction of the external field.
(c) Diamagnetism – Diamagnetism is the effect of the production by a
material of a magnetic field in the opposite direction to an external
magnetic field that is applied to it. All materials show at least a small
diamagnetic effect.
18.10 Induced EMF
A simple experiment was demonstrated by Faraday to observe the induced emf. In the
experiment, two electrical circuit are involve. The first, called the primary circuit,
consist of a battery, a switch, a resistor to control the current, and a coil of several
turns around an iron bar. When the switch is closed on the primary circuit a current
flows through the coil, producing a magnetic field that is particularly intense within
the iron bar. The second circuit also has a coil wrapped around the iron bar, and this
coil is connected to an ammeter to detect any current in the circuit. Note, however,
that there is no direct physical contact between the two circuits.
When the switch is closed on the primary circuit the magnetic field in the iron bar
rises from zero to some finite amount, and the ammeter in the secondary circuit gives
zero reading. If the switch on the primary circuit is now opened, so that the magnetic
field decreases again to zero, the ammeter in the secondary circuit deflects briefly in
the opposite direction, and then returns to zero. The current in the second circuit is
19
referred to as the induced current, because the two circuit has no direct contact. As we
know that there should be an induced emf to produce such an induced current.
However, the changing magnetic field is caused by a changing current in the primary
circuit. The following demonstration shows the same idea.
18.11 Magnetic flux
The magnetic flux Φ is defined as the dot product of the magnetic field and the area of
the current loop, e.g. ABrr
⋅=Φ , in scalar representation, we have θcosBA=Φ ,
where θ is the angle between the magnetic field B and the unit vector of the area.
The SI unit of Φ is weber, where 1 weber = 1 Wb = 1 T⋅m2.
Example
The three loops of wire shown in the figure are all in a
region of space with a uniform, constant magnetic field.
Loop 1 swings back and forth as the bob on a pendulum;
loop 2 rotates about a vertical axis; and loop 3 oscillates
vertically on the end of a spring. Which loop or loops
have a magnetic flux that changes with time?
Answer:
20
Loop1 moves back and forth, and loop 3 moves up and down, but since the magnetic
field is uniform, the flux does not depend on the loop’s position. Loop 2, on the other
hand, changes its orientation relative to the field as it rotates; hence, its flux does
change with time. The answer is loop 2.
18.12 Faraday’s law of induction
Faraday found that the second coil in the experiment described above experiences an
induced emf which is given by the following relation:
dt
dNE
Φ−=
This is known as faraday’s law of induction, the minus sign in the right of the
expression indicates that the induced emf opposes the change in magnetic flux. The
variable N is the number of loops in a coil. The above relation can be written simply
as t
NE∆
∆Φ−= , if the change in magnetic flux is uniform with time.
Example
A bar magnet is moved rapidly toward a 40-turn, circular coil of wire. As the magnet
moves, the average value of θcosB over the area of the coil increases from 0.0125 T
to 0.450 T in 0.250 s. If the radius of the coil is 3.05 cm, and the resistance of its wire
is 3.55 Ω, find the magnitude of (a) the induced emf and (b) the induced current.
Answer:
The cross sectional area of the coil: 22 )0305.0( mrA ππ ==
The initial magnetic flux is given by
252 1065.3)0305.0()0125.0( mTmTABii ⋅×===Φ −π .
21
The final magnetic flux is given by
232 1032.1)0305.0()450.0( mTmTAB ff ⋅×===Φ −π .
Applying the Faraday’s law, we have
Vs
mTmT
tNE 205.0
250.0
1065.31032.1)40(||
2523
=⋅×−⋅×
=∆
∆Φ=
−−
.
Use the Ohm’s law to calculate the induced current: AV
R
VI 0577.0
55.3
205.0=
Ω== .
Remark:
If the magnet is now pulled back to its original position in the same amount of time,
the induced emf and current will have the same magnitudes; their directions will be
reversed.
18.13 Lenz’s law
Lenz’s law states that an induced current always flows in a direction that opposes the
change that caused it. As an illustration, consider the magnetic field which decreases
with time. The induced current flows through the ring so as to oppose this change,
producing a magnetic field within the ring in the same direction as the decreasing
magnetic field.
Example
If the north pole of a magnet is moved toward a conducting loop, the induced current
produces a north pole pointing toward the magnet’s north pole. This creates a
repulsive force opposing the change that caused the current. On the other hand, if the
north pole of a magnet is pulled away from a conducting loop, the induced current
produces a south magnetic pole near the magnet’s north pole. The result is an
attractive force opposing the motion of the magnet.
22
Example
The magnets shown in the figure are dropped from rest
through the middle of conducting rings. Notice that the
ring on the right has a small break in it, whereas the ring
on the left forms a closed loop. As the magnets drop
toward the rings does the magnet on the left have an
acceleration that is (a) more than, (b) less than, or (c) the
same as that of the magnet on the right?
Answer:
According to Lenz’s law, the induced current in the ring produces a magnetic field
that exerts a repulsive force on the magnet – to oppose its motion. In contrast, the ring
on the right has a break, so it cannot have a circulating current. As a result, it exerts
no force on its magnet. Therefore, the magnet in the left falls down with an
acceleration smaller than the gravitational acceleration. But, the right magnet falls
with the gravitational acceleration.
Consider a system in which a metal ring is falling out of
a region with a magnetic field and into a field-free
region, as shown in the figure. According to Lenz’s law,
the induced current in the ring is counterclockwise. The
reasons are as follows. The induced current must be in a
direction that opposes the change in the system. In this
case, the change is that fewer magnetic field lines are
piercing the area of the loop and pointing out of the
page. The induced current can oppose this change by generating more field lines out
of the page within the loop. This can be accomplished by an induced current
circulating in a counterclockwise direction.
23
Note also that an upward force is experienced at the top of the ring. No force is
experienced at the bottom of ring.
The retarding effect on a ring leaving a magnetic field, allows
us to understand the behavior of eddy currents. This is also
known as the magnetic braking. There are several points worth
notice.
• There is no direct physical contact, thus eliminating
frictional wear.
• The magnetic braking force is stronger if the speed of
the metal is greater.
The same idea applies to a metal sheet which is leaving the
magnetic field. Eddy current flows within the metal sheet and its direction is such that
the change is opposed.
18.14 Motional EMF
When the conducting rod of length l is moving with constant speed v, as shown in
figure. The change in magnetic flux ∆Φ in a time interval ∆t is given by
)( tvlBAB ∆=∆=∆Φ .
Applying Lenz’s law, we have
Bvlt
tBvl
tNE =
∆
∆=
∆
∆Φ= )1(
The current in the conducting rod is R
Bvl
R
EI == . The magnetic force experienced by
the conducting rod is R
vlBl
R
BvlBBIlFmagnetic
22
=
== . So, an equal magnitude but
opposite direction force, that is Fexternal, must be applied to maintain a movement of it
with a constant speed.