chapter 28: magnetic fields

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Chapter 28: Magnetic fields

Historically, people discover a stone (Fe3O4) that attract pieces of iron these stone was called magnets.

two bar magnets can attract or repel depending on their orientation this is due to non-equivalent poles. One pole called North (N) and other pole called South (S)

FF

Opposite poles attractFF

FF

Like poles repels

Chapter 28: Magnetic fields

If the bar magnet is suspended on a thread (like compass)

N pole search geographic north (earth magnetic S pole)

S pole search geographic south (earth magnetic N pole)

magnetic poles are always found in pairs, we cannot isolate N from S

Cut in two pieces

N

N

N

N

N N N

S

S

SS

S

SS

Chapter 28: Magnetic Field and Magnetic Force

Like E-field, Magnets has a magnetic field (B - field) can be represented by lines away from N - pole towards S – pole. These lines represent the direction of force that would exert on pieces of iron

Analogous to electric dipole

Magnetic field lines can be traced by the aid of compass or with iron fillings (برادة حديد).

N

N

NN

S

S

θ

B

v+q


vqFBrr

,∝

( )BvFF BB

rrrr,=

0)0(// =⇒°= BFBvrrr θ

vBFBrrr

,0 ⊥⇒≠θ

( )vqFBrr

,0< ( )vqFBrr

,0>

θsin∝BFr

FB

Opposite direction to

vBqFBv B ==⇒°=⊥ max)90(rrr θ

B-force is proportional to q magnitude and its speed

(When particle moves // B-field)

B-force is a funtion of velocity v and B-field B

When a charged particle move with velocity v through a magnetic field B, it will experience a magnetic force FB.A series of experiements shows that

(Direction of B-force ┴ to B and v plane)

2

BvqFBrrr

×=

θsinvBqFF BB == The B-field SI unit isTesla (T)=N/(C.m/s)=N/(A.m)

Above results can be summarize by:

Magnitude of force on moving charged particle is

Direction of force on +ve charge

Direction of force on -ve charge


The direction of FB can be determined using right hand rule

thumb is in direction of FB on +ve charge

thumb is opposite to the direction of FB on –ve charge

E-forces act in the direction of the E-field, B-forces are perpendicular to the B-field.A B-force exists only for charges in motion. But E-force act on moving or steady chargesThe B-force of a steady magnetic field does no work when displacing a charged particle (B ┴ ds). But, E-force do work when displacing charged particle (E // ds)The B-field can alter the direction of a moving charged particle but not its speed or its kinetic energy. But, E-field alter direction, speed or kinetic energy of charged particle.

BvqFBrrr

×= EqFErr

=and

If we compare between B-force and E-force

Chapter 28: Magnetic Field and Magnetic Force: comparison between magnetic and electric forces

Ex: An electron in TV moves toward screen with a speed of 8×106

m/s along the x Axis. Surrounding the neck of the tube are coils of wire that create a magnetic field of magnitude 0.025 T, directed at an angle of 60° to the x axis and lying in the xy plane. (A) Calculate the magnetic force on the electron. (B) Find a vector expression for the magnetic force on the electron

Chapter 28: Magnetic Field and Magnetic Force:

By right hand rule, thump is in +ve z-direction FB on electron is in –ve z-direction

Chapter 28: Motion of charged particle in a uniform magnetic field

If we assume the plane of the page to be the xy plane, the perpendicular to the plane (z-direction) will be out of the page or into the page .

Similarly, If the magnetic field B out of the page, it can be represented by dots (●). If B-field is into the page, it can be represented by (×)

●

×

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qvBFB =


The force FB is always at right angles to v and its magnitude is,

Consider a positive charge moving perpendicular to a magnetic field with an initial velocity, v.

towards the center

So, as q moves, it will rotate about a circle and FB and v will always be perpendicular. The magnitude of v will always be the same, only its direction will change (uniform circular motion).

To find the radius and frequency of the rotation:

∑ = rmaF

rvmqvBFB

2

==

mqBrvand

qBmvr

=

=⇒

mqB

rv==ω

qBm

vrT π

ωππ 222===

The radial force

The angular speed

The period for this rotation

(Cyclotron frequency)

Chapter 28: Motion of charged particle in a uniform magnetic field: the cyclotron frequency

(Newton 2nd law in radial direction)

(radius of rotation)

(charge speed)


If v makes an arbitrary angle (θ≠90◦) with B (in the x-direction)

θax=0 vx = constant

We have ac due to force towards center which changes between y and z

vy and vz change in time

The equations for the cyclotron frequency and rotation period still apply provided that v replaced by 22

zy vvv +=⊥

The resulting motion is a helix

FB must be ┴ to x-axis no force component in x-direction we have only force components in y and z direction


rvmqvBFB

2

==

Ex: A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of the proton. Find the speed of the proton.

mqBrv =⇒

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Ex: Electrons are accelerated from rest through a potential difference of 350 V. The electrons travel along a curved path of radius 7.5 cm due to B-field. If the magnetic field is perpendicular to the beam. (A) What is the magnitude of the magnetic field? (B) What is the angular speed of the electrons?

qrvmB

rvmqvBF e

B =⇒==2

To find v we may use conservation of energy

(A)

(B)

Chapter 28: Applications Involving Charged Particles Moving in a Magnetic Field

If both FE and FB have same magnitude but opposite directions as shown charged particle will move vertically straight through the region of fields with constant velocity (velocity selector application)

From the expression

If charged particle move under influence of E-field and B-field it will be affected by both FE and FB on it (Lorentz force):

Shown directions for Fe and FB are on +vecharged particle

Chapter 28: Applications Involving Charged Particles Moving in a Magnetic Field

qrmvB

rvmqvBFB =⇒== 0

2

0

BEvbut

vrB

qm

==⇒ ,0

Other application includes the separation of ions according to their mass-to-charge ratio (Mass Spectrometer device). A beam of ions first passes through a velocity selector and then enters a second area of uniform magnetic field as shown

From eq.

ErBB

qm 0=⇒

when entering , the ions move in a semicircle of radius r before striking a detector array at P.

Chapter 28 Magnetic Force on a Current Carrying Conductor

Magnetic force acts upon charges moving in a conductor.

The total force on the current is the integral sum of the force on each charge in the current.

In turn, the charges transfer the force on to the wire when they collide with the atoms of the wire.

For a conductor with current passing through it and placed in a magnetic field

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consider a straight segment of wire of length L and cross-sectional area A carrying a current I in a uniform magnetic field as in figure

Vertical suspended wire between magnet poles

When there is no current (I = 0)

wire remains vertical

When I is upward wire deflect to

the left (FB to the left)

When I is downward wire deflect to the

right (FB to the right)

Flat plane view ( )BqF dqB

rrr×= v,

( ) ( )nALBqnVBqNFF ddqBB

rrrrrr×=×== vv,

qnAvI d=

For the total number of charges N moving inside the wire

Current – carrying wire will experience FB if it exist in a B-field. Inside the wire, a charges move with vd along the length L will have a magnetic force

Where n = N/V

But


BLIFBrrr

×=For a straight wire in a uniform B-field.L is a vector of magnitude equal to length and directed in the direction of current

Total force acting on the wire will be:

BsIdFd B

rrr×=

∫ ×=b

aB BsdIF

rrrFor an arbitrary wire in an arbitrary field.a and b are end points of the wire

Chapter 28 Magnetic Force on a Current Carrying ConductorIf we consider an arbitrary shaped wire of

uniform cross sectional area in an arbitrary B-Field part of total magnetic force (dFB) will occur for the small segment length vector (ds )

a

b


( ) 0=×= ∫ BsdIFBrrr

For a closed loop wire in a uniform magnetic field starting point is same as end point

For arbitrary shaped wire in uniform B-field.

BLIBsdIFb

aB

rrrrr×=×⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∫ '

L’ is displacement vector between the end points)

Net magnetic force acting on any closed current loop in a uniform magnetic field is zero.

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A semicircle wire in the xy plane of radius R forms a closed circuit and carries a current I. A uniform magnetic field is directed along the positive y axis. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion.

On the straight wire

IRBILBFF 211 ===⇒r

BLIFrrr

×=1

dsIBFd

BsdIFd

θsin2

2

=⇒

×=r

rrrOn the curved wire

θθ RddsRs =⇒=

θθdIRBdF sin2 =

[ ]π

π

θ

θθ

0

02

cos

sin

IRB

dIRBF

−=

= ∫

IRBF 22 =⇒Directed out of the board )ˆ(k

IRBF

BILFF

BLIF

2

'

'

2

22

2

=⇒

==⇒

×=r

rrror

into the board )ˆ( k−

into the board )ˆ( k−

The total force on the closed loop is

0

ˆ2ˆ221

=−=

+=

kIRBkIRB

FFFrrr

Chapter 28 Magnetic Force on a Current Carrying Conductor: Semicircular conductor loop

00 =⇒=×⇒ BFBLrrr

IaBFF ==⇒ 42

Consider a rectangular current loop in a uniform B-field, which is parallel to the plane of the loop

Chapter 28: Torque on a Current Loop in a Uniform Magnetic Field

Although the net B-force on a current loop must be zero, we may have a net torque ( دوران عزم ) on it.

For 1 and 3, L//B

For 2 and 4, L ┴ B

F2 out of board and F4 into board Fnet = 0. But F2and F4 act on different line of action they will produce a torque τ (look at side 3 to see other view)

Loop view from side 3

eye

Current out of the board

Current into the board

( ) ( ) ( )bIaBbIaBbIaBbFbF =+=+=2222 42maxτ

IAB=⇒ maxτ

Chapter 28: Torque on a Current Loop in a Uniform Magnetic Field

θττ sinrFwithFr =×=rrr

If loop is fixed on pivot (محور) O, Loop will rotate clockwise a bout y-axis due to the torque τ

At θ = 90° max. torque42 τττ rrr

+=net

Where A = ab is the area of the loop

If the B-field makes an angle with a line perpendicular to the plane of the loop, then the torque is:

θθτ

θθθτ

sinsin

sin)22

(sin2

sin2 42

IABIabB

bbIaBbFbF

==

+=+=

BAIrrr

×=⇒τ Torque on current loop in uniform B-field

A is a vector ┴ loop plane and its magnitude equal to A

Concept QuestionA rectangular loop is placed in a uniform magnetic field with the plane of the loop perpendicular to the direction of the field.If a current is made to flow through the loop in the sense shown by the arrows, the field exerts on the loop:

1. a net force.2. a net torque.3. a net force and a net torque.4. neither a net force nor a net torque.

B

(*)F1F2

F3

F4

A

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AIrr

=µ

Brrr

×=⇒ µτ

(Amperes.m2) is defined as magnetic dipole moment or often called the magnetic moment (العزم المغناطيسي)

If the wire makes N loops around A,

BBN coilloop

rrrrr×=×= µµτ

BUrr

•−= µ

The potential energy of the loop is:

µ has same direction as A

NIAwith coil =µ In the direction of A

Chapter 28: Torque on a Current Loop in a Uniform Magnetic Field: Magnetic dipole moment

Chapter 28: Torque on a Current Loop in a Uniform Magnetic Field: Magnetic dipole moment

A rectangular coil of dimensions 5.4 cm×8.5 cm consists of 25 turns of wire and carries a current of 15 mA. A 0.35 T magnetic field is applied parallel to the plane of the coil. (A) Calculate the magnitude of the magnetic dipole moment of the coil. (B) What is the magnitude of the torque acting on the loop?A)

B) BBN coilloop

rrrrr×=×= µµτ

θ = 90°

SummaryThe right hand ruleMagnetic forces on charged particles and current carrying wiresCharged particle motion in B-fieldTorque on loops and magnetic dipole moments

chapter 28: magnetic fields

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