FIZ1112 - Physics 2
CHAPTER 29 MAGNETIC FIELDS
Magnetism has a long history. The stone magnet (Fe3O4) found in nature. The experiments
showed that every magnet, regardless of its shape, has two poles, called north (N) and south (S)
poles. Similar poles (N–N or S–S) repel each other, and opposite poles (N–S) attract each other.
29.1 Magnetic Fields and Forces
In our study of electricity (Chp. 23), we described the
interactions between charged objects in terms of electric fields.
A charge creates E field around.
And any moving electric charge also contains a magnetic field.
The symbol B has been used to represent a magnetic field.
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Properties of the magnetic force on a charge moving in a magnetic field B
•
The magnitude of the magnetic force (FB) on the particle is proportional to
the charge q and to the speed v of the particle.
•
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 θ ≠ 0 with the magnetic
field, the magnetic force acts in a direction perpendicular to both v and B; that
is, FB is perpendicular to the plane formed by v and B (Fig. 3).
•
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 (Fig. 4).
•
The magnitude of the magnetic force exerted on the moving particle is
proportional to sinθ, where θ is the angle the particle’s velocity vector makes
with the direction of B
Figure 3:The magnetic force is
perpendicular to both v and B.
We can summarise these observations by
writing the magnetic force in the form
FB = q v x B
Figure 4: The direction of the magnetic force F
acting on a charged particle moving with a
velocity v in the presence of a magnetic field B.
B
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CHAPTER 29 MAGNETIC FIELDS
The right-hand rules for determining the direction of F
B
In this rule, the fingers point in the direction of v, with B
coming out of your palm, so that you can curl your fingers in
the direction of B. The direction of v x B, and the force on a
positive charge, is the direction in which the thumb points.
In this rule, the vector v is in the direction of your
thumb and B in the direction of your fingers. The force
FB on a positive charge is in the direction of your palm,
as if you are pushing the particle with your hand
FB = q v x B
FB = q vB sinθ
From this expression,
FB is zero when v is parallel or antiparallel to B (θ = 0 or 180°)
FB is maximum when v is perpendicular to B (θ = 90°).
the SI unit of magnetic field B is the newton per coulombmeter per second, which is called the tesla (T):
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Homework PROBLEMS
Q1: Determine the initial direction of the deflection of charged particles as they enter
the magnetic fields as shown in Figure. (Use right hand rule)
Answer:
UP
out of page
ZERO
in to page
Q2: An electron moving along the positive x axis perpendicular to a magnetic field experiences
a magnetic deflection in the negative y direction. What is the direction of the magnetic field?
Q3: A proton moves perpendicular to a uniform magnetic field B at 1.00 x 107 m/s and experiences
an acceleration of 2.00 x 1013 m/s2 in the +x direction when its velocity is in the +z direction.
Determine the magnitude and direction of the field.
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CHAPTER 29 MAGNETIC FIELDS
29.2 Magnetic Force Acting on a Current-Carrying Conductor
The magnetic force (FB) acting on
magnetic field notation
think of a current-carrying conductor wire I,
the wire is in the magnetic field B, which is directed into the page
The magnetic force (FB) acting on single charge (q) is FB = q v x B
Total Magnetic Force Acting on a piece wire
Total FB = (force on a single charge) x (total charge within volume)
B out of page is shown by dots,
and B into page shown by crosses
the volume of the segment is AL,
n is the number of charges per unit volume
the number of charges in the segment is nAL,
FB = (qvd x B) nAL
the current in the wire is I=nqvdA
FB = I LxB
Note that this expression applies only to a straight segment of wire in a uniform magnetic field
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CHAPTER 29 MAGNETIC FIELDS
29.2 Magnetic Force Acting on a Current-Carrying Conductor
A wire suspended
vertically between the
poles of a magnet.
When there is no current
in the wire, (no force) it
remains vertical.
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When the current
is upward, the wire
deflects to the left.
When the current is
downward, the wire
deflects to the right
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29.2 Magnetic Force Acting on a Current-Carrying Conductor
When the wire is not straight (arbitrary shape)
then the force on any segment ds ==>
dFB = I ds x B
Note that, force is
==>
FB = I L x B
==>
if the force FB for on length L ==>
maximum when B is perpendicular to ds
zero when B is parallel to ds
To calculate the total force FB acting on the wire
we integrate over the length of the wire
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CHAPTER 29 MAGNETIC FIELDS
two interesting special cases involving
Case 1.
Case 2.
An arbitrarily shaped closed loop
carrying a current I is placed in a
uniform magnetic field B
this time we must take the vector
sum of the length elements ds
over the entire loop:
A curved wire carries a current I and is located in
a uniform magnetic field B,
Because the field is uniform,
we can take B outside the integral
But the quantity
represents the vector sum
of all the length elements from a to b.
the sum equals the vector L’ , directed from a to b.
Therefore, the equation ==>
a closed polygon,
the vector sum must
be zero
Because
we conclude that FB = 0
FB = I L’ x B
the magnetic force on a curved current-carrying wire in a
uniform magnetic field is equal to that on a straight wire
connecting the end points and carrying the same current
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the net magnetic force acting on any closed
current loop in a uniform magnetic field is zero
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CHAPTER 29 MAGNETIC FIELDS
Example: 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 in figure.
θ
ds
dθ
Find the magnitude and direction of the magnetic force acting on
(1) the straight portion of the wire and on (2) the curved portion.
θ
Solution:
The magnetic force F1 acting on the straight portion has a magnitude F1=ILB=2IRB because L=2R.
Considering the cross product LxB (right-hand rule), we find the direction of F1 is out of page.
To find the magnetic force F2 acting on the curved part, we use the results of Case 1.
The magnetic force on the curved portion is the same as that on a straight wire of length 2R carrying
current I to the left. Thus, F2=ILB=2IRB. The direction of F2 is into the page based on the right-hand
rule for the cross product LxB.
or 2. method: We can calculate the dFB force acting on ds segment by using dFB = I ds x B, where ds is the
displacement along the curve. The arc length s is related to the angle θ through the relationship s=Rθ.
Therefore ds is related with dθ as ds =Rdθ. Thus, the force on ds segment is dFB = IRdθxB = IBR(sinθdθ).
If we integrate over the curved part;
1st force is out of the page, F1 =
2IRB k
2nd force is in to the page, F2 = - 2IRB k
Question: is the system in equilibrium?
Consider net force, and net torque.
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Σ F = F1 + F2 = 2IRB k - 2IRB k = 0
The total net force is ZERO
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29.3 Torque on a Current Loop in a Uniform Magnetic Field
Consider a rectangular loop carrying a current I in the presence of
a uniform magnetic field directed parallel to the plane of the loop,
as shown in Figure
No magnetic forces act on sides 1 and 3 because these wires are
parallel to the field; hence, L x B = 0 for these sides.
However, magnetic forces do act on sides 2 and 4 because these
sides are oriented perpendicular to the field. The magnitude of
these forces is F2 = F4 = IaB
(a) Overhead view of a rectangular current
loop in a uniform magnetic field. No magnetic
forces are acting on sides 1 and 3 because
these sides are parallel to B. Forces are acting
on sides 2 and 4, however.
If we view the loop from side 3
The direction of F2, exerted on wire 2, is out of the page
The direction of F4, exerted on wire 4, is into the page
The two magnetic forces F2 and F4 are directed as shown in left
Figure (the view of the loop from side 3).
The two forces point in opposite directions but are not directed
along the same line of action. If the loop is pivoted so that it can
rotate about point O, these two forces produce about O a torque
that rotates the loop clockwise. The torque is
(b) Edge view of the loop sighting down sides 2 and 4
shows that the magnetic forces F2 and F4 exerted on
these sides create a torque that tends to twist the
loop clockwise. The purple dot in the left circle
represents current in wire 2 coming toward you; the
purple cross in the right circle represents current in
wire 4 moving away from you.
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τmax = F2 b/2 + F4 b/2 = (IaB) b/2 + (IaB) b/2 = I(ab)B
the area enclosed by the loop is A=ab, ==>
τmax = IAB
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CHAPTER 29 MAGNETIC FIELDS
29.3 Torque on a Current Loop in a Uniform Magnetic Field
Because F2 = F4 = IaB, the magnitude of the net torque about O is
τ = F2 (b/2 sinθ) + F4 (b/2 sinθ)
τ = IaB (b/2 sinθ) + IaB (b/2 sinθ) = IabB sinθ
τ = I AxB
τ = IAB sinθ
As the loop rotated through an angle θ
with respect to the magnetic field B.
A convenient expression for the torque exerted on a loop
placed in a uniform magnetic field B is τ = I AxB,
where A is (perpendicular to the plane of the loop and has a
magnitude equal to) the area of the loop.
The product IA is defined to be the magnetic dipole moment μ
(often simply called the “magnetic moment”) of the loop:
The SI unit of magnetic dipole moment is ampere-meter2 (A m2)
μ = IA
Magnetic dipole moment of a current loop
τ = IA x B => τ = µ x B
Remember : (in Chp.26) τ = pxE for the torque exerted on an electric dipole in
the presence of an electric field E, where p is the electric dipole moment.
HomeWork: A rectangular coil of dimensions 5.40 cm x 8.50 cm consists of 25
turns of wire and carries a current of 15.0 mA. A magnetic field of B=0.350-T is
applied parallel to the plane of the loop.
(A) Calculate the magnitude of its magnetic dipole moment.
(B) What is the magnitude of the torque acting on the loop?
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CHAPTER 29 MAGNETIC FIELDS
29.4 Motion of a Charged Particle in a Uniform Magnetic Field
As the particle moves a centripetal force (FB) is created
and FB forms a centripetal acceleration (aC)
ΣF=ma
C
FB = q v x B = m(v2/r)
(If q were negative, the rotation would be clockwise)
The angular speed of the particle
Figure: When the velocity of a charged particle is perpendicular
to a uniform magnetic field, the particle moves in a circular path
in a plane perpendicular to B. The magnetic force FB acting on
the charge is always directed toward the center of the circle.
The angular speed is often referred to as
the cyclotron frequency
The period of the motion
(the time interval to complete one turn)
If a charged particle moves in a uniform
magnetic field with its velocity at some arbitrary
angle with respect to B, its path is a helix.
Since there is no force along x-direction (Fx=0)
x-component of velocity remains constant.
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CHAPTER 29 MAGNETIC FIELDS
Homework PROBLEMS
Q4: A uniform magnetic field of magnitude 0.150 T is
directed along the positive x axis. A positron moving at
5.00 x 106 m/s enters the field along a direction that makes
an angle of 85.0° with the x axis. The motion of the particle
is expected to be a helix, Calculate (a) the pitch p and (b)
the radius r of the trajectory.
Solution:
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29.5 Applications Involving Charged Particles Moving in a Magnetic Field
A charge moving with a velocity v in the
presence of both an electric field E and a
magnetic field B experiences both an electric
force qE and a magnetic force qvxB.
The total force (called the Lorentz force)
acting on the charge is
If q is positive and the velocity v is to the right, the magnetic force qvxB is upward and the electric
force qE is downward. When the magnitudes of the two fields are chosen so that qE = qvB, the particle
moves in a straight horizontal line through the region of the fields.
From the expression qE = qvB, we find that
v = E/B
The Mass Spectrometer
Positively charged particles are sent first through a velocity selector and
then into a region where the magnetic field B0 causes the particles to
move in a semicircular path and strike a detector array at P.
We just showed the radius of the circular motion of a charged is
Then mass to charge ratio becomes m/q=rB0/v
If we use velocity v=E/B in the equation, then we find m/q=rB0B/E .
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29.6 The Hall Effect
To observe the Hall effect, a magnetic field is applied to a current carrying
conductor. When I is in the x-direction and B in the y-direction, both
positive and negative charge carriers are deflected upward in the
magnetic field.
The Hall voltage is measured between points a and c.
the Hall voltage, we first note that the magnetic force
exerted on the carriers has magnitude qvdB.
In equilibrium, this force is balanced by the electric force qEH, where EH is the magnitude
of the electric field due to the charge separation (sometimes referred to as the Hall field).
(a)When the charge carriers in a Hall-effect apparatus are negative, the upper edge of the conductor becomes negatively
charged, and c is at a lower electric potential than a.
(b)When the charge carriers are positive, the upper edge becomes positively charged, and c is at a higher potential than a. In
either case, the charge carriers are no longer deflected when the edges become sufficiently charged that there is a balance on
the charge carriers between the electrostatic force qEH and the magnetic deflection force qvB.
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