sf026 ch19 magnetic field

Topic19--- Magnetic Field

19 MAGNETIC FIELD7 hours

19.1 Magnetic Field (1/2 Hour)19.2 Magnetic Field Produced By Current-Carrying Conductor (1 Hour)19.3 Force on a moving charged particle in a uniform magnetic field (1 Hour)19.4 Force on a current carrying conductor in a uniform magnetic field (1 Hour)19.5 Forces between two parallel current-carrying conductors (1 1/2 Hour)19.6 Torque on a coil (1 Hour)19.7 Motion of charged particle in magnetic field and electric field (1 Hour)


19.1 MAGNETIC FIELD(a)Define magnetic

field(b)Identify

magnetic field sources and sketch their magnetic field lines


Magnetic field

a region around a magnet where a magnetic force

can be experienced

Magnetic field lines

•used to represent a magnetic field•magnetic field lines leave the north pole and enters the south pole of a magnet

19.1 MAGNETIC FIELD


Magnetic field lines

UniformParallel lines

The number of lines passing perpendicularly through unit area at all

cross-sections in a magnetic field are the

same

Non-uniformNon-parallel linesThe number of

magnetic field lines varies at different unit

cross-sections

unit cross-sectional area

stronger field in A1

A1 A2weaker field

in A2

19.1 MAGNETIC FIELD


• Magnetic field can also be represented by crosses or by dotted circles

B into the page B out of page

Magnetic field lines enter the page perpendicularly

Magnetic field lines leave the page perpendicularly

direction of magnetic field at

point P

P

The tangent to a curved field line at a point

indicates the direction of the magnetic field at that

point

19.1 MAGNETIC FIELD


• The magnetic field lines pattern can be obtained by using iron filings or a plotting compass

• By spreading fine iron filings or dust on a piece of paper laid on top of a magnet, you can see the outline of the magnetic lines of force or the magnetic field.

6

Magnetic Field Line Pattern

19.1 MAGNETIC FIELD


(a) A bar magnet (b) Horseshoe or U magnet

Magnetic Field Source

19.1 MAGNETIC FIELD


(d) Two bar magnets (like poles) - repulsive

Neutral point: point where the resultant magnetic force/field strength is zero

(c) Two bar magnets (unlike pole) - attractive

(e) A circular coil

19.1 MAGNETIC FIELD


(f) A long straight wire (g) A solenoid (h) Earth Magnetic Field Extra Reading

19.1 MAGNETIC FIELD

Topic19--- Magnetic FieldUse magnetic field:

(i) for a long straight wire

(ii) at the centre of a circular coil

(iii) at the centre of a solenoid

(iv) at the end of a solenoid

rIμ

B

2

0

RIμ

B2

0

nIμB 0

nIμB 021

10

19.2 MAGNETIC FIELD

PRODUCED BY CURRENT-CARRYING

CONDUCTOR


• is defined as the scalar product between the magnetic flux density, B and the vector of the surface area, A

• where• B: magnetic field strength/ flux density

• A: area that field lines pass through• : angle between B and A

• Scalar quantity• Unit: Weber (Wb)/ T m2

• 1 T m2 = 1 Wb

θBAAB cos=•=

BAAB

area, A

θA

B

Magnetic Flux,

19.2 MAGNETIC FIELD PRODUCED BY CURRENT-CARRYING CONDUCTOR


Magnetic Flux Density,

Magnetic induction

Magnetic field intensity/ strength

is defined as the magnetic flux per unit area across an area at right angles to the magnetic field

Mathematically

A

B Φ : magnetic fluxA: area at right angles

to the magnetic field

Vector quantity (follows the direction of the magnetic field)

Unit: tesla (T) OR weber per metre squared (Wb m2)

)G(gauss 10 m Wb1T 1 42



A long straight wire

A circular coil

A solenoid

Types ofCurrentCarrying

conductor



B

B

I

Current out of the page

View from the top

A long straight wire I = currento = permeability of free space

r = distance from the wirerIμB

20



2 cm

6 cm

X

Y

I = 5 A

Determine the magnetic field strength at point X and Y from a long, straight wire carrying a current of 5 A as shown below.

Example 19.1



A circular coil

I IX

S

N

S

N

I I

I

RNIμ

B20

o = permeability of free spaceI = current

N = number of coils (loops)R = radius of the circular coilMagnetic field strength at the center given as



A circular coil having 400 turns of wire in air has a radius of 6 cm and is in the plane of the paper. What is the value of current must exist in the coil to produce a flux density of 2 mT at its center?

Example 19.2



A solenoid

I I

I

IX X X X

I

I

I

I

I

I

SN

The magnitude of magnetic field intensity at the centre (mid-point/ inside) of N turn solenoid is given by

A solenoid is an electrical device in which a long wire has been wound into a succession of closely spaced loops with geometry of a helix

lNIμ

B 0

nlN

and

nIμB 0

The magnitude of magnetic field intensity at the end of N turn solenoid is given by

nIμB 021



A solenoid is constructed by winding 400 turns of wire on a 20 cm iron core. The relative permeability of the iron is 13000. What current is required to produce a magnetic induction of 0.5 T in the center of the solenoid?

Example 19.3



Two long straight wires are placed parallel to each other and carrying the same current I. Sketch the magnetic field lines pattern around both wires when the currents are in the same direction.

Example 19.4(a) – parallel wires

I I

I I



OR

I IX

I

I

I

I

Two long straight wires are placed parallel to each other and carrying the same current I. Sketch the magnetic field lines pattern around both wires when the currents are in opposite direction.

Example 19.4(b)



A long wire (X) carrying a current of 50 A is placed parallel to and 5.0 cm away from a similar wire (Y) carrying a current of 10 A. (a) Determine the magnitude and direction of the magnetic flux

density at a point midway between the wires:(i) when the current are in the same direction(ii)when they are in opposite direction

(b) When the currents are in the same direction there is a point somewhere between X and Y at which the magnetic flux density is zero. How far from X is this point?

(Given 0 = 4 107 H m1)

Example 19.5



Two long straight wires are oriented perpendicular to the page as shown. The current in one wire is I1 = 3.0 A pointing into the page and the current in the other wire is I2= 4.0 A pointing out of page. Determine the magnitude and direction of the nett magnetic field intensity at point P.

(Given 0 = 4 107 H m1)

Example 19.6




(a) A closely wound circular coil of diameter 10 cm has 500 turns and carries a current of 2.5 A. Determine the magnitude of the magnetic field at the centre of the coil.

(b) A solenoid of length 1.5 m and 2.6 cm in diameter carries a current of 18 A. The magnetic field inside the solenoid is 2.3 mT. Calculate the length of the wire forming the solenoid.

(Given 0 = 4 107 T m A1)

Example 19.7


26

A 00.5 A 00.5

cm 0.10

cm 0.15cm 0.52P

1P

1. The two wires shown in Figure 19.21 carry currents of 5.00 A in opposite directions and are separated by 10.0 cm.

(a) Sketch the magnetic field lines pattern around both wires.

(b) Determine the nett magnetic flux density at points P1and P2.ANS: 1.33105 T, out of page;

2.67106 T, out of page

Exercise 19.2

Given 0 = 4 107 T m A1



27

X X

2. Four long, parallel power wires each carry 100 A current. A cross sectionaldiagram for this wires is a square,20.0 cm on each side as shown.

(a) Sketch the magnetic field lines pattern on the diagram.(b) Determine the magnetic flux density at the centre of the square.

ANS: 4.0 104 T , to the left (180)



(a) Use magnetic force,

(b) Describe circular motion of a charge in a uniform magnetic field

(c) Use relationship

28

19.3 FORCE ON A MOVING CHARGED

PARTICLE IN A UNIFORM MAGNETIC

FIELD

BvqF

CB FF


whereF: magnetic force

B: magnetic flux densityv: velocity of a charge

q: magnitude of the charge: angle between v and B

• In vector form,

A stationary electric charge in a magnetic field will not experience a magnetic force

But if the charge is moving with a velocity, v in a magnetic field, B then it will experience a magnetic force

The magnitude of the magnetic force can be calculated by using the following equation

Magnetic Force (Lorentz Force)

θqvBF sin

BvqF

19.3 FORCE ON A MOVING CHARGED PARTICLE IN A UNIFORM MAGNETIC FIELD


Right Hand Rule The direction of the magnetic force can be determined by using the Right Hand RuleApplication from vector

product BvqF



Determine the direction of the magnetic force, F exerted on a charge in the following problems:a. b.

c. d. e.

B

v

B v

X X X X

X X X X

X X X X

v

II

v

B

v

Example 19.8



Calculate the magnitude of the force on a proton travelling 5.0107 m s1 in the uniform magnetic flux density of 1.5 Wb m2, if:(a) the velocity of the

proton is perpendicular to the magnetic field.

(b) the velocity of the proton makes an angle 50 with the magnetic field.

Example 19.9



• Consider a charged particle moving in a uniform magnetic field with its velocity perpendicular to the magnetic field

• As the particle enters the region, it will experiences a magnetic force which the force is perpendicular to the velocity of the particle. Hence the direction of its velocity changes but the magnetic force remains perpendicular to the velocity

• This magnetic force, FB makes the path of the particle is a circular

Motion of Charged Particle In a Uniform Magnetic Field

v v

BF

v

BF

X X X X

X X X X

X X X X

X X X X

v v

BF

v

BF



Since the path is a circle therefore the magnetic force FB contributes the centripetal force Fc (nett force) in this motion. Thus

Motion of Charged Particle In a Uniform Magnetic Field

v v

BF

v

BF

X X X X

X X X X

X X X X

X X X X

v v

BF

v

BFcB FF

rmvθqvB

2

sin 90θand

rmvqvB

2



wherem = mass of charged particlev = magnitude of the velocityr = radius of the circular path

q = magnitude of the charged particle

The period of the circular motion, T makes by the particle is given by

And the frequency of the circular motion is

Bqmvr

rmvqvB

2

rv

Trv 2

T 2

and

BqmT 2

Bqmvr and

vrT 2

Tf 1



An electron at point A in Figure19.26 has a speed v of 2.50 106 m s-1. Determine(a) the magnitude and direction of

the magnetic field that will cause the electron to follow the semicircular path from A to B.

(b) the time required for the electron to move from A to B. (Given e=1.601019 C and me=

9.111031 kg)

Example 19.10

v

BA cm 0.20



Q1A proton is moving with

velocity 3 x 10 5 m/s vertically across a magnetic field 0.02 T. Calculate(a) kinetic energy of the proton(b) the magnetic force exerted on the proton(c) the radius of the circular path of the proton7.52 x 10-17 J , 9.6 x 10 -16 N,

0.16 m

Q2An electron is projected

from left to right into a magnetic field directed into the page. The velocity of the electron is 2 x 10 6 ms-1 and the magnetic flux density of the field is 3.0 T. Find the magnitude and direction of the magnetic force on the electron. (charge of electron = 1.6 x 10-19 C)

9.6 x 10-13 N, downwards

Q3A proton with a mass of

1.67 x 10-27 kg is moving in a circular orbit perpendicular to a magnetic field. The angular velocity of the proton is 1.96 x 104 rad s-1. Determine;(a) the period of revolution,(b) the magnetic field strength of the field

T = 3.2 x 10-4 s, B = 2.05 x 10-4 T

Exercise 19.3



Use magnetic force,

19.4 FORCE ON A CURRENT CARRYING

CONDUCTOR IN A UNIFORM MAGNETIC

FIELD

BlIF


• When a current-carrying conductor is placed in a magnetic field B, thus a magnetic force will acts on that conductor.

• The magnitude of the magnetic force exerts on the current-carrying conductor:

• In vector form,

F: magnetic forceB: magnitude of magnetic flux density

I: current: length of the conductor

: angle between direction of I and B

• Direction of F: Right Hand Rule• The right-hand rule: Pointing the thumb of

the right hand in the direction of the conventional current and the fingers in the

direction of B the force on the current points out of the palm. The force is reversed for a

negative charge.

Force On a Current Carrying ConductorIn a Uniform Magnetic Field

θIlBF sin

BlIF

19.4 FORCE ON A CURRENT CARRYING CONDUCTOR IN A UNIFORM MAGNETIC FIELD

https://en.wikipedia.org/wiki/Right-hand_rule

https://en.wikipedia.org/wiki/Conventional_current

https://en.wikipedia.org/wiki/Conventional_current


• F = 0 when Ө = 0 • F is maximum when Ө = 90o

0BILF sin0F

B

0I

90BILF sinmax BILF max

B

90I



Determine the direction of the magnetic force, exerted on a current-carrying conductor in the following cases.

a. b.

B I

X X X X

X X X X

X X X X

F

(to the left)

B I

X X X X

X X X X

X X X X

F

(to the right)

Example 19.11



A wire of 100 cm long is placed perpendicular to the magnetic field of 1.20 Wb m2.(a) Calculate the magnitude of the force on the wire when a current of

15 A is flowing.(b) For the same current in (a), determine the magnitude of the force

on the wire when its length is extended to 150 cm.(c) If the force on the wire in part (b) is 60102 N and the current flows

is 12 A, calculate the magnitude of magnetic field was supplied.

Example 19.12



A straight horizontal rod of mass 50 g and length 0.5 m is placed in a uniform magnetic field of 0.2 T perpendicular to the rod. The force acting on the rod just balances the rod’s weight.(a) sketch a labelled diagram shows

the directions of the current, magnetic field, weight and force.

(b) Calculate the current in the rod. (Given g = 9.81 m s2)

Example 19.12



(a) Derive magnetic force per unit length of two parallel current-carrying conductors

(b) Use magnetic force per unit length,

(c) Define one ampere44

19.5 FORCES BETWEEN TWO

PARALLEL CURRENT-CARRYING

CONDUCTORSπdIIμ

lF

2210


• Consider two identical straight conductors 1 and 2 carrying currents I1 and I2 with length l are placed parallel to each other

• The conductors are in vacuum and their separation is d

• The magnitude of the magnetic flux density, B1 at point P on the conductor 2 due to the current in the conductor 1 is given by

• Conductor 2 carries a current I2 and in the magnetic field B1 thus the conductor 2 will experiences a magnetic force, F12

Force per Unit Length

d

1 2

1I 2I

P1B

12F

Q

dIB

2

101 Direction: into the page

19.5 FORCES BETWEEN TWO PARALLEL CURRENT CARRYING CONDUCTORS


• The magnitude of F12 is given by

sin1212 lBIF 90and

90sin2

102

dIlI

dlIIF

2210

12


Topic19--- Magnetic Field• The magnitude of F21 is given by

• Conclusion :

type of the force is attractive

• Rearrange the equation, thus the force per unit length is given by

sin2121 lBIF 90and90sin

220

1

dIlI

dlIIF

2210

21

dlIIFFF

2210

2112

d

1 2

1I 2I

P1B

12F

21FQ

2B

πdIIμ

lF

2210


Topic19--- Magnetic Field• If the direction of current in the conductor 2 is change to upside down

as shown below2I1I

d

1 2

21F

Q 2B

12F1B

P

Note: The currents are in the same

direction – 2 conductors attract each other.

The currents are in opposite direction – 2 conductors repel each other

• The magnitude of F12 and F21 can be determined and their direction can be determined by applying Fleming’s left hand rule

Conclusion: Type of the force is repulsive



• If I1 = I2 = 1 A and d = 1 m , then • One ampere is defined as the constant current, which flowing in each of two infinitely long parallel straight conductors of negligible of cross sectional area separated by a distance of 1.0 metre in vacuum, would produce a force per unit length between the conductors of 2.0107 N m 1

The Ampere

πdIIμ

lF

2210

12

11104 7

ππ

17 m N100.2 lF



Two very long parallel wires are placed 2.0 cm apart in air. Both wires carry a current of 8.0 A and 10 A respectively. Find the magnitude of the magnetic force in newton, on each metre length of wire.

Two long parallel wires are 5.0 cm apart. They each exerts a force of attraction per unit length on the other of 6 x 10 -7 Nm-1. The current in one wire is 400 mA.(i) Calculate the current in the

second wire.(ii) In which direction is the

current in the second wire, relative to the first?

Example 19.13 Example 19.14

dII

LF

2

210= * I2 = 0.375 A (same direction)



Exercise 19.51. Two long straight parallel wires are placed 0.25 m apart in a vacuum.

Each wire carries a current of 2.4 A in the same direction. (a) Sketch a labelled diagram to show clearly the direction of the force

on each wire.(b) Calculate the force per unit length between the wires.(c) If the current in one of the wires is reduced to 0.64 A, calculate the

current needed in the second wire to maintain the same force per unit length between the wires as in (b).

(Given 0 = 4 107 T m A1)



2

A vertical straight conductor Y of length 0.5 m is situated in a uniform horizontal magnetic field of 0.1 T.(a) Sketch a labelled diagram to show the directions of the current, field and force.(b) Calculate the force on Y when a current of 4 A is passed into it.(c) Through what angle must Y be turned in a vertical plane so that the force on Y is halved?

(Advanced level physics, 7th edition, Nelkon&Parker, Q6, p.336) ANS: 0.2 N; 60

3

A current-carrying conductor experiences no magnetic force when it is placed in a uniform

magnetic field. Explain the statement.



53

Force in a uniform magnetic field, FB

A moving chargeFB = qvB sin

On a current carrying

conductorFB = IlB sin



Consider the three long, straight, parallel wires C, D and G. Find the force experienced by a 25 cm length of wire C.

0.30 mN

EXTRA



(a) Use torque

where N = number of turns

(b) Explain the working principles of a moving coil galvanometer

55

19.6 TORQUE ON A COILS

BANIτ

Electrical energy from the current is converted to mechanical energy as the loop and shaft rotate, and this mechanical energy is then used to power another device


• When a steady current I passes round the coil, a magnetic force acts on each side of the coil

• No magnetic forces act on sides 1 and 3 because these wires are parallel to the field

Formula of Torque



The two magnetic forces on sides 2 and 4 each of length a are equal and opposite and have the value F where,

F2 = F4 = BIL = BIa

The forces exerted a torque that tends to rotate the coil clockwise


Topic19--- Magnetic Field The magnitude of this torque

for each side is

turnsNfor sinsinsin

sin2

sin2

sin2

sin2 42

NBIABIABIab

bBIabBIa

bFbF

Fd

ab=A (area of the coil)Ө = angle between B andthe normal to plane of the coil



A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is placed vertically in a uniform horizontal magnetic field of magnitude 1.0 T. If the current flows in the coil is 5.0 A, determine the torque acting on the coil when the plane of the coil is(a) perpendicular to the field,(b) parallel to the field,(c) at 60 to the field.

Example 19.15



A rectangular loop of wire has an area of 0.30 m2. The plane of the loop makes an angle of 30o with a 0.75 T magnetic field. What is the torque on the loop if the current is 7.0 A?

Example 19.16



• A galvanometer consists of a coil of wire suspended in the magnetic field of a permanent magnet

• The coil is rectangular shape and consists of many turns of fine wire

A Moving Coil Galvanometer


Topic19--- Magnetic Field• When the current I flows

through the coil, the magnetic field exerts a torque on the coil as given by

• This torque is opposed by a spring which exerts a torque, s given by

k: torsional (spring) constant : rotation angle of the coil

in radian

• The coil and pointer will rotate only to the point where the spring torque balances the torque due to magnetic field, thus

NIABτ

kθτ s sττ kθNIAB NAB

kθI



The moving coil of a galvanometer has 100 turns and an area of 1.5 x 10-4 m2. It is suspended by a wire with a torsional constant of 2.6 x 10-8 Nm rad-1. The coil is placed in a radial magnetic field of 0.1 T. Calculate the current flowing in the coil if a deflection of 1.2 rad is observed.

I = 2.08 x 10-5 A

Example 19.17



Exercise 19.61. A rectangular coil of 10 cm

4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 102 T. The resistance of the coil is 40 and a potential difference of 12 V is applied across the galvanometer, calculate the maximum torque on the coil.

max = 3.0 x 10-3 Nm

2. A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is held in a radial magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0106 N m rad1. Determine the current is required to give a deflection of 0.5 rad.

1.0103 A


Topic19--- Magnetic Field3. FIGURE 2 shows

two long parallel wires, each carrying a 5 A current in opposite direction. Determine the magnetic field, at point P, 3 cm from both wires.

1.0103 A



(a) Explain the motion of a charged particle in both magnetic field and electric field

(b) Derive and use velocity,

in a velocity selector67

19.7 MOTION OF CHARGED

PARTICLE IN A MAGNETIC

FIELD & ELECTRIC FIELD

BEv


• Consider a positively charged particle with mass m, charge q and velocity v enters a region of space where the electric and magnetic fields are perpendicular to the particle’s velocity and to each other (velocity selector) as shown below

Motion of Charged Particle in Magnetic Field & Electric Field

E

X X X X X X

X X X X X X

X X X X X X

X X X X X X

B

v v v

BF

EF

19.7 MOTION OF CHARGED PARTICLE IN A MAGNETIC FIELD & ELECTRIC FIELD

Topic19--- Magnetic Field• The charged particle will

experiences the electric force FE is downwards with magnitude qE and the magnetic force FB is upwards with magnitude qvB

• If the particle travels in a straight line with a constant velocity hence the electric and magnetic forces are equal in magnitude. Therefore

• Only the particles with velocity equal to E/B can pass through without being deflected by the fields

• This equation also works for electron or other negatively charged particles

EB FF qEqvB 90sin B

Ev

What is the velocity of protons (+1 e) injected through a velocity selector if E = 3 x 105 V/m and B = 0.25 T?

Example 19.18


Topic19--- Magnetic Field• After the charged particle passing

through the velocity selector it will enter the next region consist of a uniform magnetic field only

• This apparatus known asmass spectrometer

Mass spectrometer - an instrument which can measure the masses and relative

concentrations of atoms and molecules

Mass Spectrometer



E

X X X X X X

X X X X X X

X X X X X X

X X X X X X

B

v

EF

X X X X X

X X X X X

X X X X X

X X X X X

X X X X X

X X X X X

X X X X X

X X X X X

X X X X X

v

X X X X X

BF

v

BFr

v

When the charged particle entering the region consist of magnetic field only, the particle will make a semicircular path of radius r

CB FF

rBv

mq

andr

mvqvB2

BEv

2rBE

mq

From the equation, the mass spectrometer can be used to determine the value of q/m for any charged particle



An electron with kinetic energy of 8.01016 J passes perpendicular through a uniform magnetic field of 0.40103 T. It is found to follow a circular path. Calculate(a) the radius of the circular path.(b) the time required for the

electron to complete one revolution.

(Given e/m = 1.761011 C kg-1, me = 9.111031 kg)

Example 19.19



Q1A velocity selector is

to be constructed to select ions (positive) moving to the right at 6.0 kms-1. The electric field is 300 Vm-1 upwards. What should be the magnitude and direction of the magnetic field?

0.05 T, out of pageQ

2An electron moving at a steady speed of 0.50106 m s1 passes between two flat, parallel metal plates 2.0 cm apart with a potential difference of 100 V between them. The electron is kept travelling in a straight line perpendicular to the electric field between the plates by applying a magnetic field perpendicular to the electron’s path and to the electric field. Calculate :(a) the intensity of the electric field.(b) the magnetic flux density needed.

ANS: 0.50104 V m1; 0.010 T

Q3A proton moving in a

circular path perpendicular to a constant magnetic field takes 1.00 s to complete one revolution. Determine the magnitude of the magnetic field.

(Physics for scientist and engineers, 6th edition,

Serway&Jewet, Q32, p.921)(mp = 1.671027 kg and

charge of the proton, q = 1.601019 C)

ANS: 6.56102 T

Exercise 19.7



74

Next Chapter…CHAPTER 20 :Electromagnetic induction

sf026 ch19 magnetic field

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