Transcript CHAPTER 29 : MAGNETIC FIELDS 29.1) THE MAGNETIC FIELD • The symbol B has been used to represent a magnetic field. • The.

CHAPTER 29 : MAGNETIC FIELDS
29.1) THE MAGNETIC FIELD
• The symbol B has been used to represent a magnetic field.
• The direction of the magnetic field B at any location = the direction in which a
compass needle points at that location (Refer Figure (29.1) pg. 906).
• Figure (29.1) – shows how the magnetic field of a bar magnet can be traced with
the aid of a compass.
• The magnetic field lines outside the magnet – point away from north poles and
toward south poles.
• We can define a magnetic field B at some point in space in terms of the magnetic
force FB that the field exerts on a test object, for which we use a charged particle
moving with a veocity v.
• Experiments on various charged particles moving in a magnetic field give the
following results :
 The magnitude FB of the magnetic force exerted on the particle is proportional
to the charge q and to the speed v of the particle (FB q, and FB  v).
 The magnitude and direction of FB depend on the velocity of the particle and on
the magnitude and direction of the magnetic field B.
 When a charged particle moves parallel to the magnetic field vector, the
magnetic force acting on the particle is zero.
 When the particle’s velocity vector makes any angle   0 with the magnetic
field, the magnetic force acts in a diretion perpendicular to both v and B; that is,
FB is perpendicular to the plane formed by v and B (Figure (29.3a)).
FIGURE (29.3a)
FIGURE (29.3b)
 The magnetic force exerted on a positive charge is in the direction opposite the
direction of the magnetic force exerted on a negative charge moving in the same
direction (Figure (29.3b)).
 The magnitude of the magnetic force exerted on the moving particle is
proportional to sin  , where  is the angle the particles’s velocity vector makes
with the direction of B.
• We can summarize these observations by writing the magnetic force in the form
:

 
FB  qv  B
(29.1)
where the direction of FB is in the direction of v x B if q is positive – which by
definition of the cross product is perpendicular to both v and B.
• The magnetic field is defined in terms of the force acting on a moving charged
particle.
FIGURE (29.4)
• Figure (29.4) – Reviews the right-hand rule for determining the direction of the
cross product v x B.
• You point the four fingers of your right hand along the direction of v with the
palm facing B and curl them toward B.
• The extended thumb, which is at a right angle to the fingers, points in the
direction of v x B, that is the direction of FB.
• Because FB =qv x B, FB is in the direction of v x B if q is positive (Figure
(29.4a).
• And, FB is in the opposite direction of v x B if q is negative (Figure (29.4b)).
• The magnitude of the magnetic force on a charged particle moving in a
magnetic field :
FB  q vBsin 
(29.2)
where  is the smaller angle between v and B.
• From this expressin, we see that F is zero when v is parallel or antiparallel to B
(=0, or 180o).
• And maximum (FB,max = |q|vB) when v is perpendicular to B (=90o.
Differences between electric and magnetic force
Electric Force
Magnetic Force
1) Acts in the direction of the electric
field
1) Acts perpendicular to the magnetic
field.
2) Acts on a charged particle regardless 2) Acts on a charged particle only when
of whether the particle is moving.
the particle is in motion.
3) Does work in displacing a charged
particle
3) Does no work when a particle is
displaced
• From the last statement : When a charged particle moves with a velocity, v
through a magnetic field, the field can alter the direction of the velocity vector
but cannot change the speed or kinetic energy of the particle.
• From Equation (29.2) – the SI unit of magnetic field is the newton per
coulomb-meter per second = tesla (T).
1T 
N
C.m / s
• Because a coulomb per second is define to be an ampere,
N
1T  1
A.m
• 1T = 104 G (gauss)
• Gauss (G) = a non-SI magnetic-field unit in common use.
Example (29.1) : An Electron Moving in a Magnetic Field
An electron in a television picture tube moves toward the front of the tube with a
speed of 8.0 x 106 m/s along the x axis (Figure (29.5)). Surrounding the neck of
the tube are coils of wire that create a magnetic field of magnitude 0.025T,
directed at an angle of 60o to the x axis and lying in the xy plane. Calculate the
magnetic force on and acceleration of the electron.
FIGURE (29.5)
Solution
•Using Equation (29.2), we can find the magnitude of the magnetic force :
FB  q vBsin   (1.6 1019 C)(8.0 106 m / s)(0.025T)(sin60o )  2.81014 N
•Because v x B is in the positive z direction (from the right-hand rule) and the
charge is negative, FB is in the negative z direction.
•The mass of the electron is 9.11 x 10-31 kg, and so its acceleration is :
FB
2.8 1014 N
16
2
a


3
.
1

10
m
/
s
me 9.111031 kg
in the negative z-direction.
29.2) MAGNETIC FORCE ACTING ON A CURRENT-CARRYING
CONDUCTOR
•A current-carrying wire experiences a force when placed in a magnetic field.
•The current is a collection of many charged particles in motion.
• Hence, the resultant force exerted by the field on the wire is the vector sum of
the individual forces exerted on all the charged particles making up the current.
• The force exerted on the particles is transmitted to the wire when the particles
collide with the atoms making up the wire.
• To indicate the direction of B in illustrations :
 We depict a magnetic field directed into the page with blue crosses, which
represent the tails of arrows shot perpendicularly and away from you. In this
case, we call the field Bin (means into the page)
 If B is perpendicular and directed out of the page, we use a series of blue dots,
which represent the tips of arrows coming toward you. In this case, we call the
field Bout.
 If B lies in the plane of the page, we use a series of blue field lines with
arrowheads.
















 Bin




























Bout
B
FIGURE (29.6a)
FIGURE (29.6c)
The magnetic force acting on a
current-carrying conductor can
be demonstrate by hangging a
wire between the poles of a
magnet as shown in Figure
(29.6a)
When the current directed
upward flows in the wire, as
shown in Figure (29.6c), the
wire deflects to the left.
FIGURE (29.6b)
•The magnetic field is directed
into the page and covers the
region within the shaded
circles.
•When the current in the wire is
zero, the wire remains vertical
as shown in Figure (29.6b).
FIGURE (29.6d)
If we reverse the current, as
shown in Figure (29.6d), the
wire deflects to the right.
FIGURE (29.7)
•Consider a straight segment of wire of length L and cross-sectional area A,
carrying a current I in a uniform magnetic field B (Figure (29.7)).
•The magnetic force exerted on a charge q moving with a drift velocity vd is
qvd x B.
•To find the total force acting on the wire, we multiply the force qvd x B exerted
on one charge by the number of charges in the segment.
•Because the bolume of the segment is AL, the number of charges in the segment
is nAL, where n is the number of charges per unit volume.
•Hence, the total magnetic force on the wire of length L is :

 
FB  (qvd  B)nAL
•From Equation (27.4), the current in the wire is I = nqvdA.
•Therefore,

 
FB  IL  B
(29.3)
Force on a segmetn of a wire
in a uniform magnetic field
where L = a vectro that points in the direction of the current I and has a
magnitude equal to the length L of the segment.
• This expression applies only to a straight segmetn of wire in a uniform
magnetic field.
• Now let us consider an arbitrarily shaped wire segment of uniform crosssection in a magnetic field as shown in Figure (29.8)
FIGURE (29.8)
• From Equation (29.3) – the magnetic force exerted on a small segment of
vector length ds in the presence of a field B is :

 
dFB  Ids  B
(29.4)
where dFB is directed out of the page for the directions assumed in Figure
(29.8)
• The total force FB acting on the wire (Figure (29.8)), is – intefrate Equation
(29.4) over the length of the wire :


Where a and b represent the end
b 
(29.5)
FB  Ia ds  B
points of the wire
• The magnitude of the magnetic field and the direction of the field makes with the
vector ds – may differ at different points.
• Consider two special case involving Equation (29.5).
• Both cases – the magnetic field is taken to be constant in magnitude and direction.
Case 1
• A curved wire carries a current I and is located in a uniform magnetic field B, as
shown in Figure (29.9a) page 912.
• Because the field is uniform, we can take B outside the integral in Equation (29.5),
and we obtain :
 


b 
FB  I a ds  B
• But the quantity
to b.
(29.6)
 represents the vector sum of all the length elements from a

 ds
b
a
• From the law of vectro addition, the sum equals the vector L’, directed form a to b.
• Therefore, Equation (29.6) reduces to :

 
FB  IL'B
(29.7)
Case 2
• An arbitrarily shaped closed loop carrying a current I is placed in a uniform
magnetic field (Figure (29.9b) page 912).
• Express the force acting on the loop in the form of Equation (29.6).
• This time we must take the vector sum of the length elements ds over the entire
loop :

 
FB  I ds  B
• Because the set of length elements form a closed polygon, the vector sum must be
zero.
• This follows from the graphical procedure for adding vectors by the polygon
method.


• Because  ds  0 , we conclude that FB  0 :
• Conclusion : The net magnetic force acting on any closed current loop in a
uniform magnetic field is zero.
Example (29.2) : Force on a Semicircular Conductor
A wire bent into a semicircle of radius R forms a closed circuit and carries a
current I. The wire lies in the xy plane, and a uniform magnetic field is directed
along the positive y axis, as shown in Figure (29.10) page 913. Find the
magnitude and direction of the magnetic force acting on the straight portion of
the wire and on the curved portion.
Solution
•The force F1 acting on the straight portion has a magnitude F1 = ILB = 2IRB
because L = 2R and the wire is oriented perpendicular to B.
•The direction of F1 is out of the page because L x B is along the positive z axis.
(That is L is to the right, in the direction of the current; thus, according to the
rule of cross products, L x B is out of the page in Figure (29.10) pg 913.
•To find the force F2 acting on the curved part, we first write an expression for the
force dF2 on the length element ds shown in Figure (29.10) pg 913.
•If  is the angle between B and ds, then the magnitude of dF2 is :
 
dF2  I d s  B  IBsin ds
• To integrate this expression, we must express ds in terms of .
• Because s = R, we have ds = Rd, and we can make this substitution for dF2 :
dF2  IRBsin d
• To obtain the total force F2 acting on the curved portion, we can integrate this
expression to account for contributions from all elements ds.
• The direction of the force on every element is the same; into the page (because
ds x B is into the page).
• Therefore, the resultant force F2 on the curved wire must also be into the page.
• Integrating our expression for dF2 over the limits  = 0 to  =  (that is, the
entire semicircle) gives :

F2  IRB0 sin d
 IRB cos0

 IRBcos   cos0 
 IRB(1  1)
 2IRB
•Because F2 with a magnitude of 2IRB, is
directed into the page and because F1, with a
magnitude of 2IRB, is directed out of the
page, the net force on the closed loop is zero.
•The result is consistant with case 2 described
earlier.
29.3) TORQUE ON A CURRENT LOOP IN A UNIFORM MAGNETIC
FIELD
• A force – exerted on a current-carrying conductor placed in a magnetic field.
• A torque – exerted on any current loop placed in a magnetic field.
FIGURE (29.12a)
FIGURE (29.12b)
• Consider a rectangular loop carrying a current I in the presence of a uniform
magnetic field directed parallel to the plane of the loop (Figure (29.12a)).
• No magnetic forces act on sides  and  - because these wires are parallel to
the field, L x B = 0 for these sides.
• Magnetic force act on sides  and  - because these sides are perpendicular
to the field.
• The magnitude of these forces is, from Equation (29.3) :
F2  F4  IaB
• The direction of F2 : The force exerted on wire  is out of the page (Figure
(29.12a).
• The direction of F4 : The force exerted on wire , is into the page.
• If we view the loop from side  and sight along sides  and  - we see the
view in Figure (29.12b).
• And the two forces F2 and F4 are directed as shown.
• The two forces point in opposite directions.
• These two forces produce about O a torque that rotates the loop clockwise,
about point O.
• The magnitude of this torque max is :
 max  F2
b
b
b
b
 F4  (IaB )  (IaB )  IabB
2
2
2
2
• The area enclosed by the loop is A = ab, the maximum torque is :
max  IAB
(29.8)
• Valid only when the magnetic field is parallel to the plane of the loop.
• The rotation is clockwise viewed from side  (Figure (29.12b).
• If the current direction were reversed, the force directions would also reverse.
• So the rotation is counterclockwise.
• If the uniform magnetic field makes an angle  < 90o with a line perpendicular
to the plane of the loop.

F1

F1

A

b
a

O


F2
(a)

F4

A

a/2


I

a
sin 
2


F3

B
FIGURE (29.13)


B
O

(b)

F3
• For convenience, we assume that B is perpendicular to sides  and .
• In this case, the magnetic forces F2 and F4 exerted on sides  and  cancel each
other and produce no torque, because they pass through a common origin.
• The forces acting on sides  and , F1 and F2 , form a couple and produce a
torque about any point.
• Figure(29.13b) – the moment arm of F1 about the point O is equal to (a/2)sin
- the moment arm of F3 about O is also (a/2)sin .
• F1 = F3 = IbB, the net torque about O has the magnitude :
a
a
a

a

  F1 sin   F3 sin   IbB sin    IbB sin    IabBsin   IABsin 
2
2
2

2

• where A = ab is the area of the loop
• The torque has its maximum value IAB when the field is perpendicular to the
normal to the plane of the loop ( = 90o).
• The torque is zero when the field is parallel to the normal to the plane of the loop
( = 0o).
• From Figure (29.13b) – the loop tends to rotate in the direction of decreasing
values of  (the area vector A rotates toward the direction of the magnetic
field).
• The torque exerted on a loop placed in a uniform magnetic field B :
 
  IA  B
(29.9)
• A is perpendicular to the plane of the loop and has a magnitude equal to the area
of the loop (Figure (29.13)).
• The direction of A can be determined by using the right-hand rule.
FIGURE (29.14)
• When you curl the fingers of your right hand in the direction of the current in
the loop, your thumb points in the direction of A.
• The product IA is defined to be the magnetic dipole moment  (or magnetic
moment) of the current loop :


  IA
(29.10)
SI unit = ampere-meter2 (A.m2)
• The torque exerted on a current-carrying loop in a magnetic field B :
  
   B
(29.11)
•If a coil consists of N turns of wire.
•Each carrying the same current and enclosing the same area.
•So the total magnetic dipole moment of the coil is N times the magnetic dipole
moment for one turn.
•Then the troque on an N-turn coil is N times that on a one-turn coil.
•Thus we write :
 



  Nloop  B  coil  B
•The potential energy of a magnetic dipole in a magnetic field depands on the
orientation of the dipole in the magnetic field and is given by :
 
U    B
(29.12)
•
A magnetic dipole has its lowest energy Umin = - B when  points in the
same direction as B.
•
The dipole has its highest energy Umax = + B
direction opposite B.
when  points in the
Example (29.3) : The Magnetic Dipole Moment of a Coil
A rectangular coil of dimensions 5.40cm x 8.50cm consisits of 25 turns of wire
and carries a current of 15.0 mA. A 0.350 T magnetic field is applied parallel
to the plane of the loop.
(a) Calculate the magnitude of its magnetic dipole moment.
Solution
Because the coil has 25 turns, we modify Equation (29.10) to obtain
coil  NIA  (25)(15.0 103 A)(0.0540m)(0.0850m)  1.72103 A.m2
(b) What is the magnitude of the torque acting on the loop?
Solution : Because B is perpendicular to coil, Equation (29.11) gives :
  coilB  (1.72103 A.m2 )(0.350T)  6.02104 N.m
Example (29.4) : Satellite Attitude Control
Many satellites use coils called torquers to adjust their orientation. These devices
interact with the Earth’s magnetic field to create a torque on the space craft in the
x, y, or z direction. This major advantage of this type of attitude-control system is
that it uses solar-gnerated electricity and so does no consume any thruster fuel.
If a typical device has a magnetic dipole moment of 250 A.m2, what is the
maximum torque applied to a satellite when its torquer is turned on at an altitude
where the magnitude of the Earth’s magnetic field is 3.0 x 10-5 T?
Solution
•Apply Equation (29.11)
•The maximum torque is obtained when the magnetic dipole moment of the torquer
is perpendicular to the Earth’s magnetic field :
max  B  (250A.m2 )(3.0 105 T)  7.5 103 n.m
29.4) MOTION OF A CHARGED PARTICLE IN A UNIFORM MAGNETIC
FIELD
• Consider the special case of a positively charged particle moving in a uniform
magnetic field with the initial velocity vector of the particle perpendicular to the
field.
• Assume that the direction of the magnetic field is into the page.
• Figure (29.17) – the particle moves in a circle in a plane perpendicular to the
magnetic field.

q




v


+
r  
FB
+ q

FB

FB



+

q

v

v

 Bin



FIGURE (29.17)
• Thhe particle moves in this way because the magnetic force FB is at right angles
to v and B and has a constant magnitude qvB.
• As the force deflects the particle, the directions of v and FB change continuously
(Figure (29.17)).
• Because FB always points toward the center of the circle, it changes only the
direction of v and not its magnitude.
• Figure (29.17) – the rotation is counterclockwise for a positive charge.
• If q were negative, the rotation would be clockwise.
• From Equation (6.1) :
r = the radius of the path (is
propotional to the linear
momentum (mv) of the
particle and inversely
propotional to the magnitude
of the charge on the particle
and to the magnitude of the
magnetic field.
 F  mar
mv2
FB  qvB 
r
r
mv
qB
(29.13)
• The angular speed of the particle (from Equation (10.10)) is :
v qB
 
r m
(29.14)
• The period of the motion (the time that the particle takes to complete one
revolution) is equal to the circumference of the circle devided by the linear speed
of the particle :
2r 2 2m
T


v

qB
(29.15)
• Angular speed of the particle and the period of the circular motion do not depend
on (i) the linear speed of the particle, and (ii) the radius of the orbit.
Example (29.6) : A proton Moving Perpendicular to a Uniform Magnetic Field
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 linear speed of the proton.
Solution
From Equation (29.13), we have :
qBr (1.601019 C)(0.35T)(14102 m)
6
v


4
.
7

10
m/s
 27
mp
1.6710 kg
Example (29.7) : Bending an Electron Beam
In an experiment designed to measure the magnitude of a uniform magnetic field,
electrons are accelerated from rest through a potential difference of 350V. The
electrons travel along a curved path because of the magnetic force exerted on them,
and the radius of the path is measured to be 7.5 cm (Figure (29.19) shows such a
curved beam of electrons). If the magnetic field is perpendicular to the beam,
(a) What is the magnitude of the field?
(b) What is the angular speed of the electrons?
Solution for (a)
• First we must calculate the speed of the electron.
• We can use the fact that the increase in their kinetic energy must equal the
decrease in their potential energy |e|V (because of conservation of energy).
• Then we can use Equation (29.13) to find the magnitude of the magnetic field.
2
m
v
• Because Ki = 0, and K f  e
we have :
2
1
m e v 2  e V
2
v
2 e V
me
2(1.601019 C)(350V)
7


1
.
11

10
m/s
31
9.1110 kg
me v (9.111031 kg)(1.11107 m / s)
4
B


8
.
4

10
T
19
er
(1.6010 C)(0.075m)
Solution for (b)
Using Equation (29.14), we find that :
v 1.11107 m / s
 
 1.5 108 rad / s
r
0.075m