1
Chapter 17Chapter 17
Magnetic FieldMagnetic Field
and and
Magnetic ForcesMagnetic Forces
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Magnetism
S
N
South
NorthSouth magnetic pole
South geographic pole
Earth’s magnetic field
North magnetic pole
North geographic pole
Opposite poles : attract each other
Like poles: repel each other
Northnorth pole
south pole
The needle of a compass aligns with the magnetic field
Earth is a magnetic. The axis of earth’s magnetic is not parallel to
its geographic axis
Magnetic declination
(Rotation axis)
(a vector field)
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N
S
N
S
F
F
Attract each other
S
N
N
S
F
F
Repel each other
S
N
S
N
F
F
Attract each other
N
S
S
N
F
F
Repel each other
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Magnetic Field
In addition to the electric field, a moving charge or a current in space can create a magnetic field.
An electric force (F = Q0E) will exert on other charge (Q) present in the electric field (E). Similarly, the magnetic field also exerts a magnetic force on other moving charge or current present in the magnetic field.
Oersted’s Experiment
N
S
I = 0 (no current)
EW
I 0I
N
S
EW
I 0
I
N
S
EW
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Magnitude of magnetic force is proportional to:
• magnitude of the charge• magnitude or strength of the magnetic field• velocity of the moving particle (for electric force, it is the same no
matter the charge is moving or not) or the component of velocity perpendicular to the field.
A charged particle at rest will have no magnetic force.
The direction of magnetic force (F) is not the same as the direction of magnetic field (B). Instead, the magnetic force is always perpendicular to both direction of magnetic field (B) and the velocity (v).
Direction of B: the north pole direction of a compass needle.
For a magnet, the direction of B is pointing out of its north pole and into its south pole.
S N
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BvQF
QBvBvQF
sin
Magnetic force on a moving charged particle:
where:
F : magnetic force [N]Q : magnitude of charge [C]v : velocity of the charge [m/s]B : magnetic field [T or Ns/Cm or N/Am (A: ampere)]
1 N/Am = 1 tesla = 1 T [Nikola Tesla (1857 – 1943)]
Right-hand rule
Magnetic field (B)+Q
Force (F)
Velocity (v)
v
Positive charge
Right hand rule
cross product
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Magnetic field (B)_-Q
Force (F)
Velocity (v)
v
negative charge
For a negative charge, the force is opposite to the case of the positive charge.
sinQBv
BvQF
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Magnetic Flux
0 AdBA
Total magnetic flux through a surface A:
AdBdABdABA
cos Sum of magnetic flux thru
areas of all elements
B
B
dA
||B
Where:
A: magnetic flux (a scalar) [weber (Wb)]
B : magnetic field [T]A : surface area [m2]
1 Wb = 1 T m2 = 1 Nm / A
Total magnetic flux through a closed surface = 0
Gauss’s Law for Magnetism
dA
dB A Magnetic field B is also
called magnetic flux density
phi"":
In = Out
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Motion of Charged Particles in a Magnetic Field
A charge particle under the action of a magnetic field only moves with a constant speed. The motion is determined by Newton’s laws of motion.
Circular motion of a positive charge in a uniform magnetic field (B):
“x” denotes that the magnetic field is pointing into the plane
v1
v2
P1
P2
R
R
0
s
R
varad
2
BQ
mvR
R
vmmavBQF
2
m : mass of the particlev : constant velocity R : radius of the circular orbit
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
F
v
+F
v+
R
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Magnetic Force on a Conductor
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
F
v
+
L
Q
A
Conductor with current
I
Magnetic force on a straight wire:
sinILBILB
BLIF
Where:
F : magnetic forceI : total currentL : length of the wire segment
Magnetic force on an infinitesimal wire (not straight wire):
BLIdFd
Divide the wire into infinitesimal straight line
IL
cos|| BB
sinBB
F
B
SOURCES OF MAGNETIC FIELD
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“Source point” is referred to the location of a charge (Q) moving with a constant velocity (v) in a magnetic field.
“Field point” is referred to the location or point where the field is to be determined, e.g. location of point “k”.
Magnetic field of a point charge moving with a constant velocity:
20
20 sin
4
ˆ
4 r
Qv
r
rvQB
+
B
v
k
r̂
r
Q
BB = 0
where:B : magnetic field Q : point chargev : velocity of the charger : distance from the charge to the field
point
runit vecto : r̂0 = 4 10-7 Ns2/C2
1 Ns2/C2 = 1 Wb/Am = 1 Tm/A = 1 n/A2
xI
B
+ Q
charge is moving into the plane
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Applying the principle of superposition, the magnetic fields of a number of moving charges can be calculated.
Total magnetic field due to a number of moving charges = vector sum of the electric fields due to the individual charges
B
k
r̂
r
dBdB = 0
dL
20
20
20
ˆ
4
sin
4
ˆ
4
r
rLIdB
r
IdL
r
rLIddB
Law of Biot and Savart for Magnetic Field of a Current Element (B):
wheredL : represents a short segment of a current-
carrying conductorI : current in the segment
runit vecto : r̂AnQvI d nQ : total charges
vd : drifting velocityA : cross-section area of segment
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Magnetic field of a straight current-carrying conductor:
conductor) carrying-currentstraight a(for 2
then x, L , If
2
4
)(4
)sin(sin
;
ˆ
4
sin
4
ˆ
4
0
22
22
0
2/3220
22
22
20
20
20
x
IB
LLxL
Lxx
LIB
dyyx
xIB
yx
x
dydLyxr
r
rLIdB
r
IdL
r
rLIddB
L
L
x
y
0x
-L
+L
dL 22 yxr
dB
I
r
- y
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Example 17.1:The figure shows an end view of two parallel wires carrying the same current I in opposite directions. Determine the magnitude and direction of magnetic flux B at point A.
Solution:Use principle of superposition of magnetic fields:
Btotal = B1 + B2
Point A is closer to wire 1 than to wire 2, the field magnitude B1 > B2
L
I
L
IB
L
I
L
IB
8)4(2
4)2(2
002
001
Use right hand rule, B1 is in the – y-direction and B2 is in the + y-direction. As B1 > B2, Btotal is in the – y-direction and the magnitude is:
L
I
L
I
L
IBBBtotal
884000
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x
2L 2L
Wire 1 Wire 2I I
B1
B2
A
y
xI I
B
x : into the plane : out of the plane
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Ampere’ Law
The line integral of magnetic field intensity around a single closed path is equal to the algebraic sum of currents enclosed.
enclILdB 0
r1
r21 23 4
B
Integration path not enclosing the conductor
rI
B
B
dL
B
B
dL
I
rr
I
dLB
dLBLdB
0
0
||
)2(2
Integration path enclosing the conductor
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Magnetic Field of a Circular Current Loop
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The Solenoid
A long wire wound in the form of a helical coil is known as a solenoid.
I
I
P
R
θ 2θ
θ 1
I
y
dy
(b)(a)
1
(y2+R2) 1/2
y2
(y2+R2) 3/2
R2
(y2+R2) 3/2
Also, the magnetic field dB due to the current dI in dy can be found as,
From the figure, the current for a length increment dy is
dyL
NIdI
dIRy
RdB
2322
20
)(2
2322
20
)(2 Ry
dy
L
NIRdB
ddyRy
R
d
dy
Ry
y
Ryd
dy
dy
d
Ry
y
cos)(
cos)()(
1sin
)(sin
2322
2
2322
2
2122
2122
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So, dB = ( ) cos θ dθ μ0 I N
2 L
=> B = ( ) ∫ cos θ dθ μ0 I N
2 L
θ2
θ1
= ( ) (sin θ2 – sin θ1 ) μ0 I N
2 L
B = (sin θ2 – sin θ1 ) jμ0 n I
2
where n = , number of turns per unit lengthN
LThis formula represents the magnetic field along the centroid axis of a finite solenoid.
For infinite long solenoid, it is assumed that θ1 = -π/2 and θ2 = π/2.
B = μ0 n I j
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Example 17.3 :
Since the length of the solenoid is quite large in comparative with its diameter, the magnetic field near its middle is approximately uniform. It is therefore reasonable to consider it as a case of infinite solenoid
B = μ0 n I j.
Solution
A solenoid has 300 turns wound around a cylinder of diameter 1.20 cm and length 14.0 cm. If the current through the coils is 0.410 A, what is the magnitude of the magnetic field inside and near the middle of the solenoid.
The number of turns per unit length (n) is
n = N/L = (300 turns) / (0.14 m)
= 2.14 × 103 turns / m
Therefore, the magnetic field inside and near the middle of the solenoid is,
B = μ0 n I j = (4π × 10-7 Tm/A) (2.14 × 103 turns / m) (0.410 A) = 1.10 × 10-3 T
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I
I
N
S
I
I
S
N
Magnetic fields of a finite solenoid
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