Written by Dr. John K. Dayton MAGNETIC FIELDS THE MAGNETIC FIELD FORCES ON MOVING CHARGES THE...

Written by Dr. John K. Dayton

MAGNETIC FIELDS

• THE MAGNETIC FIELD• FORCES ON MOVING CHARGES

• THE MAGNETIC FIELD OF ALONG, STRAIGHT WIRE• THE MAGNETIC FIELD OF A

LOOP OR COIL• THE MAGNETIC FIELD OF A

SOLENOID

THE MAGNETIC FIELD:

N

S

The magnetic field is designated by the symbol B. The source of all magnetic fields are moving electric charges. Even the magnetic fields associated with electromagnetic radiation have their source in moving electric charge.

The SI unit for the magnetic field is the tesla, T. A more common unit of magnetic field strength is the gauss, G. Earth’s magnetic field at the surface is about 0.5 G. 104G 1T

The magnetic field is characterized by what we call a north and a south pole. Similar poles repel and opposite poles attract. Outside of magnetic materials magnetic field lines extend away from north poles and toward south poles. However, magnetic field lines, unlike electric field lines, are continuous. Each magnetic field line forms a complete loop. The simplest of all magnetic fields is the magnetic dipole, like a bar magnet.. Magnetic monopoles, such as an isolated north pole, do not exist.

THE FORCE ON A MOVING CHARGE IN A MAGNETICFIELD:

ˆ sin

F qv B

F nqv B

F = Force on moving charge

q = charge on particle / object

v = velocity of moving charge

B = magnetic field

× = vector cross-product

= angle between v and B

v

B

FI

n = unit vector in correct direction

v

B

FI

1. Use Right Hand2. Point Fingers In Direction Of Current3. Point Open Palm In direction Of Magnetic Field4. Extended Thumb Is Direction Of Force On Charge

THE RIGHT-HAND RULE:

EXAMPLE: A proton traveling horizontally to the right with a speed of 1.5x104 m/s encounters a uniform magnetic field of 0.55T oriented 60o above the horizontal. What is the magnitude of the force on the proton and what is its initial direction?

Bv

+e

19

5

4

1

sin 1.6 10 1.5 10 0.55 sin(60 )

1.14 10

omsF qv C

F

B T

N

According to the right-hand-rule this force will be directed outward from the diagram.

B

v

F

r

q, m

MOTION OF A CHARGED PARTICLE WHOSE VELOCITY ISPERPENDICULAR TO B:

F qvB qvB

Fm v

rqvB

m v qrB

rm vqB

v r

rm rqB

qBm

o

s in 9 02

According to the vectors shown in the diagram, and the right-hand-rule, q must be a negative charge.

Using the right-hand-rule, can you determine the sign of q in the diagram? Click for the answer.

B

q

Is the charge q of the moving particle positive or negative?

q is positive.

B out of diagram

view from above

v

FB

v

v(perpendicular)

v B

B

q

q

EXAMPLE: An alpha particle, q = 6.4x10-19C and m = 6.68x10-27 kg, is following a circular path of radius 0.5m in a uniform magnetic field of 1.2 T. (a) What is the angular speed of the alpha particle? (b) What is the linear speed of the alpha particle?

19

8

8

2

7

7

1.15 10

5.

6.4 10 1.2

6.68 10

0.5 1.15 1

1

0

75 0

rads

r

ms

ads

C Tq

v

B

m kg

v r m

FORCE ON A CURRENT-CARRYING WIRE:

wire

B

I

l

sin

sin

sin

drift

drift

drift

F qvB

q neAl v v

F neAv Bl

I neAv

F IBl

n = density of mobileelectrons in wire.

EXAMPLE: Calculate the net force on a wire of length 20m and carrying a current of 12A through a magnetic field of 0.5 Gauss directed 70o to the wire.

B

12A

20m

4

2

sin( )

12 20 0.5 10 sin(70

1.13 10

)o

F I

F

lB

F A

N

m T

According to the right-hand-rule, this force will be directed outward from the diagram.

I

B

F

F

a

b

m

TORQUE ON A CURRENT-CARRYING LOOP IN AMAGNETIC FIELD:

2 sin sin2

sin sin

sin

F NIaB

bF Fb

NIaBb NIAB

NIA m

mB

2

I

mB

I I

mB

I

EXAMPLE: A coil of wire composed of 100 turns of radius 6cm carries a current of 5A. It is in a uniform magnetic field 0f 0.65T oriented 75o to the plane of the coil. (a) What is the magnetic moment of the coil? (b) What is the torque on the coil?

2

2

25.

1

65

00 5 .06

m NIA NI r

m A m

m A m

2

sin

5.65 0.65 sin 15

0.95

o

mB

A m T

N m

is the angle between B and m. = 90o – 75o

AMPERE’S LAW:

cos oB l I

I = the current that is encircled by a closed path.

l = small piece of the closed path.

B = local magnetic field at l

= angle between B and direction of l

Sum is over all l comprising the closed path

MAGNETIC FIELD OF A LONG, STRAIGHT WIRE:

I

WIRE

B outward B inward

BIr

oo

T mA

24 1 0 7

Right-Hand Rule:1-fingers in direction of I2-Palm toward field point3-Thumb points indirection of B

SUPERPOSITION OF MAGNETIC FIELDS: B B B BP P P P , , , . . .1 2 3

Calculate the magnetic field mid way between the two wires.

,2 ,1

7

,1

5,1

7

,2

5,2

5 5

6

4 10 3

2 0.025

1.6 10

4 10 4

2 0.025

2.1 10

2.1 10 1.6

5.0 1

1

0

0

P P P

T mA

P

P

T mA

P

P

P

P

B B B

AB

m

B T

AB

m

B

B T

T

B T T

d = 5cmI1 = 3A

I2 = 4A

P

2.5cm

outward is

inward is

2

1,

p,

p

B

B

EXAMPLE: The central conducting wire of a coaxial cable carries a current of 3A upward. The outer, cylindrical conducting wire of radius 3mm carries a uniformly distributed current of 5A downward. What are the magnetic field strengths at 2mm from the center and 5cm from the center of this wire?

7

01

1

7

02

2

41

62

4 10 3

2 2 .

3.00 10

8.00

002

4 10 2

2 2 .05

10

TmAenclosed

TmAenclosed

B T

AIB

r m

AI

r

B

B

T

m

Only the net enclosed current is used. Ienc=5A-3A. The net current for B1 is upward and the net current for B2 is downward so these two fields are oriented in opposite directions.

MAGNETIC FORCE BETWEEN PARALLEL WIRES:

I1 I2

1 22,1

11 2

2,1 2 1 2

2

2o

oof at

of at

IB

rF I B

F

l

I I l

r

r

l

The force between parallel wires isattractive if the currents are in thesame direction.

F2,1

EXAMPLE: A long straight, vertical wire carries a current of 5A upward. Near the wire is a rectangular loop of wire carrying a current of 3A. The loop has dimensions 5cm by 10cm and is positioned so that its long side is parallel to the long wire at a distance of 5cm with its current also directed upward. What is the net force on the loop?

I1=5A

I2=3A

5cm

5cm

10cm

1

2

3

4

wire

loop

F2 F4

The magnetic field produced by the wire in the plane of the loop is directed inward, into the diagram. The strength of this field decreases as you move further from the wire.The force on side 1 is downward and the force on side 3 is upward. Use the right-hand-rule to confirm this. These forces are of equal magnitudes so cancel each other out. The force on side 2 is toward the wire and the force on side 4 is away from the wire. These are also in opposite directions but the force on side 2 is stronger because it is closer to the wire.

B from the wire is inward on this side of the wire

Solution continues on next slide.

6

o 1 2 o 1 22 4

2 4

7

= =2 2

4 10 5 3 .1 1 1

2 .0

3.00 10

5 .10

net

TmA

net

net

I I l I I lF F F

r r

A A mF

m

F N

m

continued from previous slide

Fnet is directed toward the wire.

THE BIOT-SAVART LAW:

I

l

P

r

B

I lr

n

B B

024

sin

MAGNETIC FIELD OF A LOOP OR COIL:

x-axisx

a

r

B

Bx

By

I

P

2

32oa I

Br

2 2

2

2 3 3 3

cos

sin

4 4

24 4 4

x

o o

o o o o

aB B B B B B

rI l I

B lr r

I a Ia Ia a IB l l a

r r r r r

If there are N turns of wire in the coil, then:2

32oa NI

Br

If x>>a, then:

2

32oa NI

Bx

MAGNETIC FIELD AT THE CENTER OF A LOOP OR COIL:

a

Px-axis

I

2

32

2o

oa NIB

rr

N

a

IB

a

a = radius

I

I

BPalm toward center

EXAMPLE: Calculate the magnetic field strength at the center of a coil of radius 20cm, 200 turn of wire, and carrying a current of 2.5A.

74 10 200 2.5

0.001

2

6

2 .2

TmAo

ANIB

a m

B T

MAGNETIC FIELD OF A SOLENOID:

I

I

oo

NIB nI

l

N = total number of turnof wire

n = number of turns ofwire per unit length,n = N/l

I

I1

2

3

4

1 21 2

3 43 4

1 4 2

44

cos

cos cos

cos cos

90 , 0 , 0

cos

o

o

o

o

o

o

o

NI B l

NI B l B l

B l B l

B

NI B l B l

l

Bl

NIB nI

EXAMPLE: A solenoid is to be made so that a magnetic field of 0.5T is at its center. If the solenoid is to be 10cm in length and wrapped with wire carrying a current of 3A, how many turns of wire are required?

7

0.5 0.1

4 10

13263 turn

3

s

o

Tmo A

T mNI BlB N

l I A

N

End Of Presentation