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Chapter 7 MAGNETIC FIELD
Recommended Problems:
1,3,5,7,9,11,13,14,17,20,23,55,57,72.
In this chapter we are going to discuss the following topics:
Magnetic Force on a charged Particle
Magnetic Force on a Current-Carrying Wire
Torque on a Current Loop
INTRODUCTION
It was found when a magnet suspended from its center, it tends to
line itself up in a north-south direction (the compass needle). The
north end is called the North Pole (N-pole), and the south end is
called the South Pole (S-pole). These two poles cannot be
isolated, that is, poles always appear in pairs. Opposite poles
attract each other, while like poles repel each other.
As for the e. field, we represent the m. field by field lines called the
magnetic field lines. These lines can be drawn such that
(i) they point out from the north pole toward the south pole of a
magnet,
(ii) the direction of the magnetic field at any point is tangent to
these lines at that point, and
(iii) the number of lines per unit area is proportional to the strength
of the magnetic field.
Notice that the magnetic field lines continue inside the magnet
forming closed loops, unlike electric filed lines that begin on
positive charges and terminate on negative charges.
The earth is considered as a magnet with its geographic south
pole is the N-pole. This explains the direction of a suspended
magnet.
(a) (b)
Figure 13.1 (a) Magnetic field lines for a bar magnet. (b) Magnetic filed pattern for a bar magnet as displayed by iron filings.
MAGNETIC FIELD AND FORCES
A moving charge, or a current, creates a magnetic field B in the
space around it.
Any moving charge q exists in the region of B will be affected by a
magnetic force according to
sinqvBBvqF
where v is the velocity of the charge q and is the angle between
v and B.
Note that F is zero if v and B are parallel and maximum if v and B
are perpendicular.
The SI unit of B is N.s/C.m, which is called Tesla (T). The cgs unit
of B is the Gauss (G), which is related to the Tesla through
1 T = 104 G
Let the outstretched fingers of your
right hand to point along the
direction of v and then bend them
toward the direction of B. The thumb
then gives the direction of the
magnetic force. This is true if q is
positive only, and if q is negative the
direction must be reversed.
In contrary to the electrostatic force which is parallel to E, the
direction of F is always normal to both v and B. It is given by the
right hand rule:
A dot is adopted for a direction out of page which represents
the tip of arrow coming toward you.
A cross is adopted for a direction into the page which
represents the feathered tails of arrow fired away from you.
A positive charge moving to the right. After
entering a region of uniform magnetic field
directed into the page, as shown, the
electron:
B
v
• Test Your Understanding (1)
a) Deflects up. b) Deflects down.
a) Deflects backward. d) Stays unaffected.
A negative charge moving up. After
entering a region of uniform magnetic field
directed into the page, as shown, the
electron:
B
v
• Test Your Understanding (2)
a) Deflects right. b) Deflects left.
a) Deflects backward. d) Stays unaffected.
B
jkiBvqF ˆˆˆ
jkiBvqF ˆˆˆ
Consider two opposite
charged particles
moving to the right and
entering a m.field into
the page.
The +ve charge will deflect upward while the –ve charge will
deflect downward.
Since F and v are perpendicular, then
0dt
dsFvF
This means that the magnetic force is always perpendicular to the
displacement ds, and hence it does not do any work in a
displaced particle.
The work-energy theorem states that the net work exerted on a
particle must equal the change in its kinetic energy, i.e.,
KWnet
the magnetic force cann’t change the speed of the particle.
Example 29.1 An electron moves with a
speed of 8.0106 m/s along the x-axis in a region
of uniform m. field of 0.025 T, directed at an angle
60o to the x-axis. Calculate the magnitude and the
direction of the magnetic force acting on the
electron as it enters the region of B. x
y
v
B
60o
Solution: Expressing v and B in vector notations we have
m/sˆ100.8 6 iv Tˆ022.0ˆ013.0ˆ60sin025.0ˆ60cos025.0 jijiB o
jiiBvqF ˆ022.0ˆ013.0ˆ100.8106.1 619
Nkji ˆ108.2ˆˆ022.0100.8106.1 14619
• Test Your Understanding (3)
A magnetic field exerts a force on a charged particle:
a) always
b) ) if the particle is at rest
c) if the particle is moving along the field lines.
d) if the particle is moving across the field lines.
MOTION OF A CHARGED PARTICLE IN
MAGNETIC FIELD
In the previous section we said that the static magnetic force
could not alter the speed of the particle.
Hence, the only thing the magnetic force can do is to change the
direction of the velocity.
The acceleration, therefore, is centripetal, and so the force. This
means that a moving particle in a magnetic field follows a circular
path.
If the magnetic is uniform and the initial velocity of the particle is
normal to the direction of B, the particle will move in a circle.
Since the magnetic force is always perpendicular to both v and B,
the circle in a plane normal to the direction of the magnetic field.
Now we have r
vmBvqF
2
If the velocity of the particle is perpendicular to the direction of B,
r
vmqvB
2
qB
mvr
The angular frequency and the period of the motion are
m
qB
r
v
qB
mT
22
v
F
+q
The sense of rotation can be determined
again from the right hand rule. The
thumb must now point toward the center,
the direction of the force. If B is directed
into the page. A positive charge will
rotate counterclockwise, while a negative
charge will rotate clockwise.
If the velocity of the charged particle and the magnetic field are
not perpendicular the path is a helix whose axis is parallel to B.
Example 29.7 An electron is accelerated from rest by a
potential difference of 350 V. It then enters a region of uniform
magnetic field and follows a circular path of radius 7.5 cm. If its
velocity is perpendicular to B. Calculate
a) The magnitude of B, b) the angular speed of the electron.
Solution a) To find the speed of the electron we use
UK eVmv221
m/s101.1
1011.9
350106.122 7
31
19
m
eVv
T104.8075.0106.1
1011.11011.9Now 4
19
731
qr
mvB
b) The angular speed is
sadr
v/r105.1
075.0
1011.1 87
• Test Your Understanding (4)
A magnetic field CAN’T:
a) exert a force on a charged particle
b) change the velocity of a charged particle
c) change the speed of a charged particle.
d) change the momentum of a charged particle.
MAGNETIC FORCE ON CURRENT-CARRYING
WIRE
It is known that current is a collective of moving charges of the
same sign. This means that any current-carrying wire experiences
a magnetic force when placed in a magnetic field. This force is the
vector sum of the individual forces on the charged particles.
Let us consider a straight portion of
wire of length l and cross-sectional
area A. The wire is assumed to carry a
current I and to be placed in a uniform
magnetic field B as shown.
× × × × × × × × × × ×
× × × × × × × × × × ×
× × × × × × × × × × ×
Bin vd
L
A
The magnetic force on a single charge dq moving with drift
velocity vd is
BvdqFd
IdtdqBut
BvIdtFd
lddtv
thatKnowing BlIdFd
For straight wires we can integrate the last equation to obtain
BlIF
If the wire is not straight we have to find the force on small
segment and then integrate, i.e.,
f
i
BLdIF
If B is uniform
BLiBLdIFf
i
is the straight line from i to f.
Lwith
Note that for closed loop
0L
The m. force on any closed current-loop in a uniform m. field
must be zero.
The direction of the magnetic force on a wire is determine by the
right hand rule described before with the four fingers now point
toward the direction of the current instead of the velocity.
L’
Example 29.2 A wire bent into a
semicircle of radius R and carries a
current I. The wire lies in the x-y plane,
and a uniform m.f. is directed along the
+ve y-axis. Find Fm on the straight portion
and on the curved portion of the wire.
R
I
B
Solution
For the straight portion we have
BLIF
jBiRI ˆˆ2 kIRBjiIRB ˆ2ˆˆ2
And for the curved portion we have
BLIF
kIRBjBiRI ˆ2ˆˆ2
Not that the net force for the whole wire is zero as expected.
• Test Your Understanding (5)
The four wires shown all carry the same current from point A to
point B through the same magnetic field. In all four parts of the
figure, the points A and B are 10 cm apart. Rank the wires
according to the magnitude of the magnetic force exerted on
them, from greatest to least.
a) (a), (b)= (c), (d)
TORQUE ON A CURRENT LOOP
a
b I
B
FR FL
Consider a rectangular loop of sides a
and b carrying a current I and immersed
in a uniform magnetic field B.
For the horizontal sides of the loop the
direction of current is parallel to B, so
the forces on the two sides is zero.
For the vertical sides we have.
BLIF LL
BLIF RR
kIbBijIbB ˆˆˆ
That is the two forces are opposite and equal in magnitude
If the loop is assumed to rotates about its center, these two forces
will produce a torque that tends to rotate the loop
counterclockwise. The magnitude of this torque is
kbBijIbB ˆˆˆ
22
aIbB
aIbB
Knowing that A=ab is the area of the loop
IAB
If B is not parallel to the plane of the loop the last equation can be
generalized to be
BBA
I A
Iwith
Is called the magnetic dipole moment.
The direction of is perpendicular to the plane of the loop and its
sense is determined by the right hand rule: Curling the four
fingers of the right hand in the direction of the current, the thumb
points in the direction of .
Example 29.3 A rectangular loop of 25 turns has dimensions
5.4 cm × 8.5 cm and carries a current of 15 mA. The loop exists in
a uniform magnetic field of 0.35 T and initially parallel to the plane
of the loop. What is the magnitude of the initial torque acting on
the loop?
Solution
Since the coil has 25 turns, we have
2334 A.m1072.11015105.84.525 NIA
Now noting that and B are perpendicular we have
N.m1002.635.01072.1 43 B