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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

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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

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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.

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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= .

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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= .

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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.

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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.

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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

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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.