Chapter 7 MAGNETIC FIELD

Recommended Problems:

1,3,5,7,9,11,13,14,15,17,20,23,33,55,57.

INTRODUCTION

It was found when a magnet suspended from its center, it tends to

line itself up in a north-south direction (the compass needle). The

north end is called the North Pole (N-pole), and the south end is

called the South Pole (S-pole). These two poles cannot be

isolated, that is, poles always appear in pairs. Opposite poles

attract each other, while like poles repel each other.

As for the e. field, we represent the m. field by field lines called the

magnetic field lines. These lines can be drawn such that

(i) they point out from the north pole toward the south pole of a

magnet,

(ii) the direction of the magnetic field at any point is tangent to

these lines at that point, and

(iii) the number of lines per unit area is proportional to the strength

of the magnetic field.

Notice that the magnetic field lines continue inside the magnet

forming closed loops, unlike electric filed lines that begin on

positive charges and terminate on negative charges.

The earth is considered as a magnet with its geographic south

pole is the N-pole. This explains the direction of a suspended

magnet.

(a) (b)

Figure 13.1 (a) Magnetic field lines for a bar magnet. (b) Magnetic filed pattern for a bar magnet as displayed by iron filings.

MAGNETIC FIELD AND FORCES

A moving charge, or a current, creates a magnetic field B in the

space around it.

Any moving charge q exists in the region of B will be affected by a

magnetic force according to

sinqvBBvqF

where v is the velocity of the charge q and is the angle between

v and B.

Note that F is zero if v and B are parallel and maximum if v and B

are perpendicular.

The SI unit of B is N.s/C.m, which is called Tesla (T). The cgs unit

of B is the Gauss (G), which is related to the Tesla through

1 T = 104 G

Let the outstretched fingers of your

right hand to point along the

direction of v and then bend them

toward the direction of B. The thumb

then gives the direction of the

magnetic force. This is true if q is

positive only, and if q is negative the

direction must be reversed.

In contrary to the electrostatic force which is parallel to E, the

direction of F is always normal to both v and B. It is given by the

right hand rule:

A dot is adopted for a direction out of page which represents

the tip of arrow coming toward you.

A cross is adopted for a direction into the page which

represents the feathered tails of arrow fired away from you.

B

jkiBvqF ˆˆˆ

jkiBvqF ˆˆˆ

Consider two opposite

charged particles

moving to the right and

entering a m.field into

the page.

The +ve charge will deflect upward while the –ve charge will

deflect downward.

Since F and v are perpendicular, then

0dt

dsFvF

This means that the magnetic force is always perpendicular to the

displacement ds, and hence it does not do any work in a

displaced particle.

The work-energy theorem states that the net work exerted on a

particle must equal the change in its kinetic energy, i.e.,

KWnet

the magnetic force cann’t change the speed of the particle.

Example 29.1 An electron moves with a

speed of 8.0106 m/s along the x-axis in a region

of uniform m. field of 0.025 T, directed at an angle

60o to the x-axis. Calculate the magnitude and the

direction of the magnetic force acting on the

electron as it enters the region of B. x

y

v

B

60o

Solution: Expressing v and B in vector notations we have

m/sˆ100.8 6 iv Tˆ022.0ˆ013.0ˆ60sin025.0ˆ60cos025.0 jijiB o

jiiBvqF ˆ022.0ˆ013.0ˆ100.8106.1 619

Nkji ˆ108.2ˆˆ022.0100.8106.1 14619

• Test Your Understanding (1)

A magnetic field exerts a force on a charged particle:

a) always

b) ) if the particle is at rest

c) if the particle is moving along the field lines.

d) if the particle is moving across the field lines.

MOTION OF A CHARGED PARTICLE IN

MAGNETIC FIELD

In the previous section we said that the static magnetic force

could not alter the speed of the particle.

Hence, the only thing the magnetic force can do is to change the

direction of the velocity.

The acceleration, therefore, is centripetal, and so the force. This

means that a moving particle in a magnetic field follows a circular

path.

If the magnetic is uniform and the initial velocity of the particle is

normal to the direction of B, the particle will move in a circle.

Since the magnetic force is always perpendicular to both v and B,

the circle in a plane normal to the direction of the magnetic field.

Now we have r

vmBvqF

2

If the velocity of the particle is perpendicular to the direction of B,

r

vmqvB

2

qB

mvr

The angular frequency and the period of the motion are

m

qB

r

v

qB

mT

22

v

F

+q

The sense of rotation can be determined

again from the right hand rule. The

thumb must now point toward the center,

the direction of the force. If B is directed

into the page. A positive charge will

rotate counterclockwise, while a negative

charge will rotate clockwise.

If the velocity of the charged particle and the magnetic field are

not perpendicular the path is a helix whose axis is parallel to B.

Example 29.7 An electron is accelerated from rest by a

potential difference of 350 V. It then enters a region of uniform

magnetic field and follows a circular path of radius 7.5 cm. If its

velocity is perpendicular to B. Calculate

a) The magnitude of B, b) the angular speed of the electron.

Solution a) To find the speed of the electron we use

UK eVmv221

m/s101.1

1011.9

350106.122 7

31

19

m

eVv

T104.8075.0106.1

1011.11011.9Now 4

19

731

qr

mvB

b) The angular speed is

sadr

v/r105.1

075.0

1011.1 87

• Test Your Understanding (2)

A magnetic field CAN’T:

a) exert a force on a charged particle

b) change the velocity of a charged particle

c) change the speed of a charged particle.

d) change the momentum of a charged particle.

IdtdqBut

BvIdtFd

lddtv

thatKnowing BlIdFd

For straight wires we can integrate the last equation to obtain

BlIF

If the wire is not straight we have to find the force on small

segment and then integrate, i.e.,

f

i

BLdIF

If B is uniform

BLiBLdIFf

i

is the straight line from i to f.

Lwith

Note that for closed loop

0L

The m. force on any closed current-loop in a uniform m. field

must be zero.

The direction of the magnetic force on a wire is determine by the

right hand rule described before with the four fingers now point

toward the direction of the current instead of the velocity.

L’

MAGNETIC FORCE ON CURRENT-CARRYING

WIRE

It is known that current is a collective of moving charges of the

same sign. This means that any current-carrying wire experiences

a magnetic force when placed in a magnetic field. This force is the

vector sum of the individual forces on the charged particles.

Let us consider a straight portion of

wire of length l and cross-sectional

area A. The wire is assumed to carry a

current I and to be placed in a uniform

magnetic field B as shown.

× × × × × × × × × × ×

× × × × × × × × × × ×

× × × × × × × × × × ×

Bin vd

L

A

The magnetic force on a single charge dq moving with drift

velocity vd is

BvdqFd

Example 29.2 A wire bent into a

semicircle of radius R and carries a

current I. The wire lies in the x-y plane,

and a uniform m.f. is directed along the

+ve y-axis. Find Fm on the straight portion

and on the curved portion of the wire.

R

I

B

Solution

For the straight portion we have

BLIF

jBiRI ˆˆ2 kIRBjiIRB ˆ2ˆˆ2

And for the curved portion we have

BLIF

kIRBjBiRI ˆ2ˆˆ2

Not that the net force for the whole wire is zero as expected.

• Test Your Understanding (3)

The four wires shown all carry the same current from point A to

point B through the same magnetic field. In all four parts of the

figure, the points A and B are 10 cm apart. Rank the wires

according to the magnitude of the magnetic force exerted on

them, from greatest to least.

a) (a), (b)= (c), (d)

TORQUE ON A CURRENT LOOP

a

b I

B

FR FL

Consider a rectangular loop of sides a

and b carrying a current I and immersed

in a uniform magnetic field B.

For the horizontal sides of the loop the

direction of current is parallel to B, so

the forces on the two sides is zero.

For the vertical sides we have.

BLIF LL

BLIF RR

kIbBijIbB ˆˆˆ

That is the two forces are opposite and equal in magnitude

If the loop is assumed to rotates about its center, these two forces

will produce a torque that tends to rotate the loop

counterclockwise. The magnitude of this torque is

kbBijIbB ˆˆˆ

22

aIbB

aIbB

Knowing that A=ab is the area of the loop

IAB

If B is not parallel to the plane of the loop the last equation can be

generalized to be

BBA

I A

Iwith

Is called the magnetic dipole moment.

The direction of is perpendicular to the plane of the loop and its

sense is determined by the right hand rule: Curling the four

fingers of the right hand in the direction of the current, the thumb

points in the direction of .

Example 29.3 A rectangular loop of 25 turns has dimensions

5.4 cm × 8.5 cm and carries a current of 15 mA. The loop exists in

a uniform magnetic field of 0.35 T and initially parallel to the plane

of the loop. What is the magnitude of the initial torque acting on

the loop?

Solution

Since the coil has 25 turns, we have

2334 A.m1072.11015105.84.525 NIA

Now noting that and B are perpendicular we have

N.m1002.635.01072.1 43 B