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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
Chapter 28: Magnetic Field and Magnetic Force
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)
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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
Chapter 28: Magnetic Field and Magnetic Force
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 (×)
●
×
3
qvBFB =
Chapter 28: Motion of charged particle in a uniform magnetic field
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)
Chapter 28: Motion of charged particle in a uniform magnetic field
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
Chapter 28: Motion of charged particle in a uniform magnetic field
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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Chapter 28: Motion of charged particle in a uniform magnetic field
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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Chapter 28 Magnetic Force on a Current Carrying Conductor
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
Chapter 28 Magnetic Force on a Current Carrying Conductor
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
Chapter 28 Magnetic Force on a Current Carrying Conductor
( ) 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
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