(28 – 1)

Chapter 28 Magnetic Fields

In this chapter we will cover the following topics:

Magnetic field vector Magnetic force on a moving charge Magnetic field lines Motion of a moving charge particle in a uniform magnetic field Magnetic force on a current carrying wire Magnetic torque on a wire loop Magnetic dipole, magnetic dipole moment Hall effect Cyclotron particle accelerator

B

BF

What is the equivalent resistance between points F and G? Each Resistor is 8 Ω.(A) 2.5 Ω(B) 8.0 Ω(C) 2.0 Ω(D) 4.0 Ω(E) none of these

One can generate a magnetic field using one of the

following methods:

Pass a current through a wire and thus form what is knows

as an "electromagnet".

Use a "permanent" ma

What produces a magnetic field

gnet

Empirically we know that both types of magnets attract

small pieces of iron. Also if supended so that they can

rotate freely they align themselves along the north-south

direction. We can thus say that these magnets create in

the surrounding space a " " which

manifests itself by exerting a magnetic force .

We will use the magnetic force to define precicely

the magnetic field

B

B

F

magnetic field

vector . B

(28 – 2)

The magnetic field vector is defined in terms of the force it

exerts on a charge which moves with velocity . We inject

the charge in a region where we wish to determine

BF

q v

q B

Definition of B

at

random directions, trying to scan all the possible directions.

There is one direction for which the force on is zero. This

direction is parallel with . For all other directions is n

B

B

F q

B F

ot zero

and its magnitude where is the angle between

and . In addition is perpendiculat to the plave defined

by and . The magnetic force vector is given by the e

sin

quation:

B

B

B

v B F

v B

F q v

F

The defining equation is sin

If we shoot a particle with charge = 1 C at right

angles ( 90 ) to with speed = 1m/s and the

magnetic force 1 N, then = 1 tesla

B

B

q

F q v

q

B v

F

v B

B

SI unit of B :

BF qv B

sinBF q vB

(28 – 3)

The vector product of the vectors and

is a vector

The magnitude of is given by the equation:

The direction of is perpendicu

si

lar

n

c a b a b

c

c

c ab

c

The Vector Product of two Vectors

to the plane P defined

by the vectors and

The sense of the vector is given by the :

Place the vectors and tail to tail

Rotate in the plane P along the shortest an

a b

c

a b

a

right hand rule

a.

b.

gle

so that it coincides with

Rotate the fingers of the right hand in the same direction

The thumb of the right hand gives the sense of

The vector product of two vectors is also known as

b

c

c.

d.

the " " productcross

(28 – 4)

The vector components of vector are given by the equations:

,

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ , ,

x y z z y

x y z x y z x y z

c a b a b

a a i a j a k b b i b j b k c c i c j c k

c

c = a×bThe Vector Product in terms of Vector Components

Those familiar with the use of determinants can use the expres

,

The order of the two vectors in the cross product is impor

si

c c

t

nt

on

a

x y z

x y

y z x x z y y x

z

z x

i j k

a b a a a

b b b

a b a b a b a b

b

Note

N

:

ote :

a a b

(28 – 5)

In analogy with the electric field lines we

introduce the concept of magnetic field lines which help visualize

the magnetic field vector without using equations.

The relation

b

B

Magnetic Field Lines :

etween the magnetic field lines and are: B

1. At any point P the magnetic field vector B is tangent to the magnetic field lines

P

PB

magnetic field line

2. The magnitude of the magnetic field vector B is proportional

to the density of the magnetic field lines

magnetic field linesPQ

PB

QB

P QB B

(28 – 6)

The magnetic field lines of a permanent magnet are shown

in the figure. The lines pass through the body of the magnet

and form loops. This is in contr

Magnetic field lines of a permanent magn

d

et

close ast to the electric field

lines that originate and terminate on elecric charges.

The closed magnetic field lines enter one point of the magnet

and exit at the other end. The end of the magnet from which

the lines emerge is known as the of the magnet.

The other end where the lines enter is called the

of the magnet. The two poles of the magnet cannot be

separated. Together they

north pole

south pole

form what is known as a

" "magnetic dipole

(28 – 7)

hitt

The current in the 5.0 Ω resistor in the circuit shown is:

(1)0.42A (2) 0.67A (3) 1.5A (4) 2.4A (5) 3.0A

EF

BF

Cathode

AnodeEF qE

BF qv B

A cathode ray tube is shwon in the figure. Electrons are

emitted from a hot filament known as the "cathode". They are accelerated by a

voltage applied between the cathode V

Discovery of the electron :

and a second electrode known as the "anode".

The electrons pass through a hole in the anode and they form a narrow beam. They

hit the fluorescent coating of the right wall of the cathode ray tube where they produce

a spot of light. J.J. Thomson in 1897 used a version of this tube to investigate the nature

of the particle beam that caused the fluorescent spot. He applied constant electric and

magnetic fields in the tube region to the right of the anode. With the fields oriented as

shown in the figure the electric force and the magnetic force have opposite

directions. By adjusting aE BF F

B

nd Thomson was able to have a zero net forceE

(28 – 8)

.

.electron

B

v

F

C

r

(also known as )

A particle of mass and charge when injected with a speed

at right angles to a uniform magnetic field , fo

m q

v B

Motion of a charged particle in a uniform magnetic field

cyclotron motion

llows a

circular orbit, with uniform speed. The centripetal force

required for such motion is provided by the magnetic force

BF qv B

2

The circular orbit of radius for an electron is shown in the figure. The magnetic force

2 2 2. The period

1The corresponding frequency . The angu

2

B

r

v r mv mF q vB ma m T

r v q Bv q B

q Bf

m

r

T

mv

q B

lar frequency 2

The cyclotron period does not depend on the speed . All particles of the

same mass complete their circular orbit during the same time regardless of speed

Fa

q Bf

mv

T

Note 1 :

Note 2 : st particles move on larger radius circular orbits, while slower particles move

on smaller radius orbits. All orbits have the same period T (28 – 9)

mvr

q B q B

m

We now consider the motion of

a charged in a uniform magnetic

field when its initial velocity

forms and angle with .

We decompose into two

components.

B

v B

v

Helical paths

One component parallel to and the other perpendicular to (see fig.a)

cos sin The particle executes two independent motions.

One is the cyclotron motion is in the plane

v B v B

v v v v

perpendicular to we have

2analyzed in the previous page. Its radius . Its period

The second motion is along the direction of and it is linear motion with constant

speed .

B

mv mr T

q B q B

B

v

The combination of the two motions results in a helical path (see fig.b)

2 cosThe pitch of the helix is given by:

mvp p Tv

q B

mvr

q B

2 mT

q B

(28 – 10)

Two isolated conducting spheres are separated by a large distance. Sphere 1 has a radius of R and an initial charge 2Qwhile sphere 2 has a radius of 3R and an initial charge 8Q. A very thin copper wire is now connected to the spheres.Charge flows between the spheres until they reach the same electrical potential. How much charge will be transferredfrom sphere 2 to sphere 1? (Note that the charge transferred can be positive, negative or zero .)

(1) Q/2 (2) 2Q (3) –Q/3 (4) -3Q (5) None of these

BF

Consider a wire of length which carries a current as shown in

the figure. A uniform magnetic field is present in the vicinity

of the wire. Experimentaly

L i

B

Magnetic force on a current carrying wire.

it was found that a force is

exerted by on the wire, and that is perpendicular

to the wire. The magnetic force on the wire is the vector sum

of all the magnetic forces exerted by on the

B

B

F

B F

B

electrons that

constitute . The total charge that flows through the wire

in time is given by:

Here is the drift velocity of the electrons

in the wire.

The magnetic force sin 9

dd

B d

i q

t

Lq it i v

v

F qv B

0 dd

Li v B iLBv

BF iLB

(28 – 11)

. dL

B

dF

i

If we assume the more general case for which the

magnetic field froms and angle with the wire

the magnetic force equation can be writ

B

Magnetic force on a straight wire in a uniform

magnetic field.

ten in vector

form as: Here is a vector whose

magnitude is equal to the wire length and

has a direction that coincides with that of the current.

The magnetic force magnitude

B

B

F iL B L

L

F

sin

In this case we divide the wire into elements of

length which can be considered as straight.

The magnetic f

iLB

dL

Magnetic force on a wire of arbitrary shape

placed in a non - uniform magnetic field.

orce on each element is:

The net magnetic force on the

wire is given by the integral:

B

B

dF idL B

F i dL B

=

BF iL B

BdF idL B=

BF i dL B

(28 – 12)

C

C

Top viewSide view

Consider the rectangular loop in fig.a with sides of lengths and which carries

ˆa current . The loop is placed in a magnetic field so that the normal to the lo

a b

i n

Magnetic torque on a current loop

1 3

2 4

op

forms an angle with . The magnitude of the magnetic force on sides 1 and 3 is:

sin 90 . The magnetic force on sides 2 and 4 is:

sin(90 ) cos . These forces cancel i

B

F F iaB iaB

F F ibB ibB

2 4

1 3

n pairs and thus 0

The torque about the loop center C of and is zero because both forces pass

through point C. The moment arm for and is equal to ( / 2)sin . The two

torques tend to ro

netF

F F

F F b

1 3

tare the loop in the same (clockwise) direction and thus add up.

The net torque + =( / 2)sin ( / 2)sin sin siniabB iabB iabB iAB

sinnet iAB

0netF

(28 – 13)

U B U B

The torque of a coil that has loops exerted

by a uniform magnetic field and carrries a

current is given by the equation:

We define a new vector associated wi

N

B

i NiAB

Magnetic dipole moment :

th the coil

which is known as the magnetic dipole moment of

the coil.

The magnitude of the magnetic dipole moment

Its direction is perpendicular to the plane of the coil

The sense of is defined by the right hand rule. We curl the fingers of the right hand

in the

NiA

direction of the current. The thumb gives us the sense. The torque can

expressed in the form: sin where is the angle between and .

In vector form:

The potential energy of the

B B

B

coil is:

has a minimum value of for 0 (position of equilibrium)

has a maximum value of for 180 (position of equilibrium)

For both pos

cos

itions the n

U

U B

B

B

U B

stable

unstable

Note :

et torque 0

B

(28 – 14)

U B

L L

L

R R

R

In 1879 Edwin Hall carried out an experiment in which

he was able to determine that conduction in metals is due

to the motion of charges (electrons). He was also

able to determin

ne

The Hall

gat

e t

ive

ffec

e the concentration of the electrons.

He used a strip of copper of width and thickness . He passed

a current along the length of the strip and applied a magnetic

field perpendicular to the stri

n

d

i

B

p as shown in the figure. In the

presence of the electrons experience a magnetic force that

pushes them to the right (labeled "R") side of the strip. This

accumulates negative charge on the R-sid

BB F

e and leaves the left

side (labeled "L") of the strip positively charged. As a result

of the accumulated charge, an electric field is generated as

shown in the figure so that the electric force bala

E

nces the magnetic

force on the moving charges.

( ). From chapter 26 we have:

( )

E B d

d d

d

F F eE ev B

E v B J nev

J i iv

ne Ane dne

eqs.1

eqs.2 (28 – 15)

L L

L

R R

R

( ). / ( )

Hall measured the potential difference between the left and

the right side of the metal strip. (eqs.3)

We substitute from eqs.3 and

d dE v B v i dne

V

V Ed

E

eqs.1 eqs.2

from eqs.(2) into eqs.1 and get:

(eqs.4)

Fig. a and b were drawn assuming that the carriers are electrons.

In this case if we define we get a value.

f

I

d

L R

v

V iB

d dne

V V V

Bin

V e

positive

we assume that the current is due to the motion

of positive charges (see fig.c) then positive charges accumulate on

the R-side and negative charges on the L-side and thus

is now a numbeL RV V V

negative r.

By determining the polarity of the voltage develops between the

left and right hand side of the strip Hall was able to prove that

current was composed of moving electrons. From the value of

using e

V

quation 4 he was able to determine the concentration of

the negative charge carriers.(28 – 16)

The cyclotron accelerator consissts of two hollow

conductors in the shape of the letter dee (these are

known as the "dees" of the cyclotron. Between the

two dees an o

The cyclotron particle accelerator

scillator of frequency creates an

oscillating electric field that exists only in the gap

between the two dees. At the same time a constant

magnetic field is applied perpendicular to the

plane

oscf

E

B

of the dees.

mvr

eB

2

eBf

m

In the figure we show a cyclotron accelerator for protons. The protons follow circular

orbits of radius and rotate with the same frequency . If the cyclotron2

frequency matches the oscil

mv eBr f

eB m

lator frequency thne the protons during their trip through

the gap between the dees are accelerated by the electric field that exists in the gap.

The faster protons travel on increasingly larger radius orbits. Thus the electric

field changes the speed of the protons while the magnetic field changes only

the direction of their velocity and forces them to move on circular (cyclotron) orbits.

(28 – 17)

Four identical light bulbs A, B, C and D are connected in series to a constant voltage source. A wire is then connectedacross B. What is the brightness of A relative to its former brightness?

(1) 0.44 (2) 0.56 (3) 1.78 (4) 2.25 (5) none of these

Hitt