Download Electromagnetic Fields Lecture

Electromagnetic Fields Lecture-13:
Static Magnetic Fields 3
Chadi Abou-Rjeily
Department of Electrical and Computer Engineering
Lebanese American University
[email protected]
March 30, 2017
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Definition (1)
Consider two neighboring closed loops C1 and C2
bounding surfaces S1 and S2 respectively.
~ 1 will be
If a current I1 flows in C1 , then a magnetic field B
created. The flux that flows through S2 due to the current
in C1 can be expressed as (flux from 1 → 2):
Z
~ 1 .d~s2
Φ12 =
B
(Wb)
S2
If C2 has N2 turns, the flux linkage from (1) to (2) is:
Λ12 = N2 Φ12
Chadi Abou-Rjeily
(Wb)
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Definition (2)
The mutual inductance between loops C1 and C2 is
defined as:
Λ12
L12 =
(H)
I1
It can be proven that L12 = L21 . In other words:
Z
Z
N2
~ 1 .d~s2 = N1
~ 2 .d~s1
B
B
I1 S2
I2 S1
The self inductance of loop C1 is defined as:
Z
Λ11
N1
~ 1 .d~s1
L11 =
=
B
I1
I1 S1
A conductor arranged in an appropriate shape to supply a
certain amount of self inductance is called an inductor.
Just as a capacitor can store electric energy, an inductor
can store magnetic energy.
For a linear medium, the inductances do not depend on the
current in the circuit.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-1 (1)
Example: Assume that N turns of a wire
are tightly wound on a toroidal frame of
rectangular cross section. Assuming that
the permeability of the medium is µ0 , find
the self inductance of the toroidal coil.
Assume a current I in the wire. Because of the symmetry
~ is along the ~aφ direction and its
of the problem, B
magnitude is constant along a circular path.
Apply Ampere’s law over a circular path with radius r
(a < r < b). This path encircles a total current NI:
I
~ ~l = µ0 NI
B.d
C
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-1 (2)
The last equation implies that:
Z 2π
Bφ~aφ r dφ~aφ = µ0 NI
0
⇒ 2πrBφ = µ0 NI
~ = µ0 NI ~aφ
⇒ B
2πr
The flux that flows through the cross section of the toroid
is:
Z
~ ~s
Φ=
B.d
S
Z µ0 NI
~aφ .(~aφ hdr )
=
2πr
S
Z
µ0 NIh b dr
µ0 NIh b
=
=
ln
2π
2π
a
a r
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-1 (3)
The flux linkage is:
Λ = NΦ =
µ0 N 2 Ih b
ln
2π
a
Finally, the self inductance (or simply inductance) is given
by:
Λ
µ0 N 2 h b
L= =
ln
I
2π
a
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-2 (1)
Example: Find the inductance per unit length of a very long
solenoid with air core having n turns per unit length.
~
For a current I in the wire, we have previously seen that B
is parallel to the axis of the solenoid and has a constant
magnitude given by:
B = µ0 nI
If S is the cross-sectional area of the solenoid, the flux
linkage per unit length is:
Λ = nΦ = n(BS) = µ0 n2 SI
Therefore, the inductance per unit length is:
L=
Λ
= µ0 n 2 S
I
Chadi Abou-Rjeily
(H/m)
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-3 (1)
Example: Two coils of N1 and N2 turns
are wound concentrically on a straight
cylindrical core of radius a and
permeability µ. The windings have
lengths l1 and l2 respectively. Find the
mutual inductance between the coils.
Assuming a current I1 in coil 1, from the previous example,
~ 1 is:
we can deduce that the magnitude of B
N1
B1 = µn1 I1 = µ
I1
l1
For coil 1, n1 is the number of turns per unit length while N1
is the number of turns.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-3 (2)
The flux of the magnetic field caused by coil 1 in coil 2 is
given by:
N1
Φ12 = µ
(πa2 )I1
l1
Since the outer coil has N2 turns, the flux linkage between
coil 1 and coil 2 is given by:
Λ12 = N2 Φ12 =
µ
N1 N2 πa2 I1
l1
Therefore, the mutual inductance is:
L12 =
Λ12
µ
= N1 N2 πa2
I1
l1
Chadi Abou-Rjeily
(H)
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-4 (1)
Example: Determine the mutual
inductance between a conducting
triangular loop and a very long straight
wire as shown in the adjacent figure.
Designate the triangular loop as circuit 1 and the long wire
as circuit 2.
Assuming a current I2 in the wire, then the magnetic flux
density at a distance r from the wire is given by:
~ 2 = µ0 I2 ~aφ
B
2πr
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-4 (2)
The flux linkage Λ21 = Φ21 is:
Z
~ 2 .d~s1
B
Λ21 =
S1
where:
d~s1 = zdr ~aφ
h
π i
= [(d + b) − r ] tan
dr ~aφ
3
√
= 3 [(d + b) − r ] dr ~aφ
Consequently:
√
3µ0 I2
2π
Z
d+b
1
[(d + b) − r ] dr
r
d
√
3µ0 I2
b
=
(d + b) ln 1 +
−b
2π
d
Λ21 =
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Inductance and Inductors: Example-4 (3)
Therefore, the mutual inductance is:
Λ21
I
√2
b
3µ0
=
(d + b) ln 1 +
−b
2π
d
L21 =
(H)
Note that the mutual inductance can be calculated from
either Λ21 /I2 or Λ12 /I1 . However, it is difficult to calculate
~ 1 caused by a current I1
Λ12 since it is difficult to calculate B
in the triangular loop everywhere.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Definition (1)
Consider a system of N loops carrying currents
I1 , I2 , . . . , IN . The magnetic energy stored in this system is
given by:
N
1X
Wm =
Ik Φk
2
k =1
where Φk is the flux linkage through the k-th circuit:
Φk =
N
X
Ljk Ij
j=1
For j 6= k , Ljk is the mutual inductance between circuits j
and k .
For j = k , Ljj is the self inductance of circuit j.
Consequently, Wm can be expressed as:
N
Wm =
N
1 XX
Ljk Ij Ik
2
j=1 k =1
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Definition (2)
For a current I flowing in a single inductor with inductance
L, the stored magnetic energy is:
Wm =
1 2
LI
2
(J)
In terms of the field quantities, Wm can be written as:
Z
1
~ Bdv
~ ′
Wm =
H.
2 V′
V ′ is the volume in which the magnetic energy is stored.
~ = µH:
~
Given that for a linear medium B
Z
1
B2 ′
Wm =
dv
2 V′ µ
Z
1
=
µH 2 dv ′
2 V′
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Example (1)
Example: Starting from the formula of
the stored magnetic energy, determine
the inductance per unit length of an air
coaxial transmission line that has a solid
inner conductor of radius a and a very
thin outer conductor of inner radius b.
Assume that a current I flows in the inner conductor and
returns via the outer conductor in the opposite sense.
~ has only a
Because of the cylindrical symmetry, B
φ-component.
Assuming that the current is uniformly distributed over the
cross section of the inner conductor, then the magnitude of
the current density is:
I
J=
πa2
Apply Ampere’s law over a circular contour of radius r .
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Example (2)
Inside the inner conductor (0 ≤ r ≤ a):
I
~ 1 .d~l = µ0 I (πr 2 )
B
πa2
C
Z 2π
r 2
⇒
Bφ1~aφ r dφ~aφ = µ0
I
a
0
r 2
⇒ Bφ1 (2πr ) = µ0
I
a
~ 1 = Bφ1~aφ = µ0 rI ~aφ
⇒ B
2πa2
The magnetic energy stored per unit length in the inner
conductor is:
Z
1
Wm1 =
B 2 dv ′
2µ0 V ′ φ1
Z a
Z
1
µ0 I 2 a 3
µ0 I 2
2
=
Bφ1 2πr dr =
r
dr
=
(J/m)
2µ0 0
16π
4πa4 0
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Example (3)
Between the inner and outer conductors (a ≤ r ≤ b):
I
~ 2 .d~l = µ0 I
B
C
⇒ Bφ2 (2πr ) = µ0 I
~ 2 = Bφ2~aφ = µ0 I ~aφ
⇒ B
2πr
The magnetic energy per unit length stored in the region
between the inner and outer conductors is:
Z
1
Wm2 =
B 2 dv ′
2µ0 V ′ φ2
Z b
1
=
B 2 2πr dr
2µ0 a φ2
Z
µ0 I 2 b 1
µ0 I 2 b
=
dr =
ln
(J/m)
4π a r
4π
a
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Energy: Example (4)
Consequently, the magnetic energy stored per unit length
of the transmission line is:
Wm = Wm1 + Wm2
µ0 I 2 1
b
=
+ ln
4π 4
a
On the other hand:
1 2
LI
2
Comparing the above expressions results in:
Wm =
L=
µ0
µ0 b
+
ln
8π 2π a
Chadi Abou-Rjeily
(H/m)
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (1)
A charge q moving with a velocity ~u in a magnetic field with
~ will experience a magnetic force F
~ m given
flux density B
by:
~ m = q~u × B
~
F
(N)
Consider a closed circuit of contour C that carries a current
~ The magnetic force
I in the presence of a magnetic field B.
on an element d~l of the circuit can be written as:
~
~ m = q~u × B
~ = (Idt) dl × B
~
dF
dt
~
= Id~l × B
Consequently, the magnetic force on the complete circuit
is:
I
~
~
Fm = I
d~l × B
(N)
C
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (2)
Consider the case of two circuits C1 and C2 carrying
currents I1 and I2 respectively.
~ 21 on C1 resulting from the magnetic field
The force F
caused by the current I2 in C2 is:
I
~
~ 21
F21 = I1
d~l1 × B
C1
~ 21 is caused by C2 . From Biot-Savart law:
where B
I
d~l2 × ~aR21
µ0 I2
~
B21 =
2
4π C2
R21
The combination of the above equations results in
Ampere’s law of forces:
I I d~l1 × d~l2 × ~a
R21
~ 21 = µ0 I1 I2
F
2
4π
R21
C1 C2
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (3)
~ 12 on circuit C2 in
From the previous equation, the force F
the presence of the magnetic field set up by the current I1
in C1 takes the following form:
I I d~l2 × d~l1 × ~a
R
12
~ 12 = µ0 I2 I1
F
2
4π
R12
C2 C1
Newton’s third law governing the forces of action and
reaction holds. It can be proven that:
~ 12 = −F
~ 21
F
even though: d~l1 × d~l2 × ~aR21 6= −d~l2 × d~l1 × ~aR12 .
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (4)
Example: Determine the force per unit
length between two infinitely long parallel
conducting wires carrying currents I1 and
I2 in the same direction. The wires are
separated by a distance d .
Let the wires be in the yz-plane and parallel to the z-axis.
By applying Ampere’s circuital law, we can directly deduce
~ 12 at wire 2 set up by the
that the magnetic flux density B
current I1 in wire 1 is:
~ 12 = µ0 I1 (−~ax )
B
2πd
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (5)
Consequently:
~ 12 = I2 d~l2 × B
~ 12
dF
µ0 I1 I2
=
(dz~az ) × (−~ax )
2πd
µ0 I1 I2
=−
dz~ay
2πd
Therefore, the force per unit length of wire 2 is:
~
~ ′ = dF12 = − µ0 I1 I2 ~ay
F
12
dz
2πd
In the same way:
~ 21 = I1 d~l1 × B
~ 21 = I1 (dz~az ) ×
dF
⇒
µ0 I2
~ax
2πd
~
~′
~ ′ = dF21 = µ0 I1 I2 ~ay = −F
F
21
12
dz
2πd
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (6)
Consider a circular loop of radius b and carrying a current I
~
in a uniform magnetic field of flux density B.
~
B can be resolved into two components:
~ =B
~⊥ + B
~k
B
~ ⊥ is perpendicular to the plane of the loop.
B
~ k is parallel to the plane of the loop.
B
~ ⊥ tends to expand the loop (or contract it if the direction
B
~ ⊥ exerts no net force to move the loop.
of I is reversed). B
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (7)
~ k produces a force dF
~ 1 on element d~l1 and a force
B
~ 2 = −dF
~ 1 on the symmetrically located element d~l2 .
dF
~ 1 is out from the paper while that of dF
~2
The direction of dF
is into the paper.
~ k is zero.
The net force on the entire loop caused by B
However, a torque exists on the loop. This torque tends to
rotate the loop about the x-axis in such a way as to align
~ k field.
the magnetic field due to I with the external B
~ 1 and dF
~ 2 is:
The differential torque produced by dF
~ = [(dF )2b sin φ] ~ax
dT
h
i
~ k 2b sin φ ~ax
= Id~l1 × B
= IdlBk sin φ 2b sin φ ~ax
h
i
= 2Ib 2 Bk sin2 φdφ ~ax
~ 1 | = |dF
~ 2 | and dl = |d~l1 | = |d~l2 |.
where dF = |dF
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (8)
The total torque acting on the loop is:
Z
Z π
2
2
~
~
T = dT = 2Ib Bk
sin φdφ ~ax
0
h i
= I πb 2 Bk ~ax
On the other hand, the magnetic dipole moment of the loop
is given by:
~ = I πb 2 ~an
m
~an is the unit vector normal to the loop (its direction verifies
the right hand rule).
~ can be written
~ , the torque T
Based on the expression of m
as:
~ =m
~
~ ×B
T
(N.m)
~ was used instead of B
~ k because:
Note that B
~
~
~
~ ⊥.
~ × (B⊥ + Bk ) = m
~ × Bk since m
~ is parallel to B
m
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (9)
~ can be generalized to any
The previous expression of T
planar loop of any shape placed in a uniform magnetic
field:
~ =m
~
~ ×B
T
(N.m)
~ = IS~an where S is the cross-sectional area of the loop, I
m
is the current carried by the loop and ~an is the unit normal
vector.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (10)
Example: A rectangular loop in the
xy-plane with sides b1 and b2 carrying a
current I lies in a uniform magnetic field
~ = Bx ~ax + By ~ay + Bz ~az . Determine the
B
force and the torque on the loop.
~ into its perpendicular and parallel components:
Resolve B
~ ⊥ = Bz ~az
B
~ k = Bx ~ax + By ~ay
B
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (11)
~ ⊥ is given by:
The force on side 1 due to B
Z
~ ⊥1 = I d~l × B
~⊥
F
Z b1 /2
=I
(dx ~ax ) × (Bz ~az ) = −Ib1 Bz ~ay
−b1 /2
~ ⊥ is:
The force on side 3 due to B
~ ⊥3 = −F
~ ⊥1 = Ib1 Bz ~ay
F
~ ⊥ is:
In the same way, the force on side 2 due to B
Z −b2 /2
~
F⊥2 = I
(dy ~ay ) × (Bz ~az ) = −Ib2 Bz ~ax
b2 /2
~ ⊥ is:
The force on side 4 due to B
~ ⊥4 = −F
~ ⊥2 = Ib2 Bz ~ax
F
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (12)
~ ⊥1 , F
~ ⊥2 , F
~ ⊥3 and F
~ ⊥4 are all directed toward the
Note that F
center of the loop.
The sum of these contracting forces is zero:
~ ⊥2 + F
~ ⊥3 + F
~ ⊥4 = 0
~ ⊥1 + F
F
The above forces produce no torque.
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (13)
~ k on side 1 is:
The force produced by B
~ k1 = I(b1~ax ) × (Bx ~ax + By ~ay )
F
= Ib1 By ~az
~ k on side 3 is:
The force produced by B
~ k3 = −F
~ k1 = −Ib1 By ~az
F
~ k on side 2 is:
The force produced by B
~ k2 = I(−b2~ay ) × (Bx ~ax + By ~ay )
F
= Ib2 Bx ~az
~ k on side 4 is:
The force produced by B
~ k4 = −F
~ k2 = −Ib2 Bx ~az
F
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3
Magnetic Forces and Torques (14)
Again, the net force on the loop is zero. However, the
above four forces result in a torque.
~ 13 due to forces F
~ k1 and F
~ k3 is:
The torque T
~ 13 = Ib1 b2 By ~ax
T
~ 24 due to forces F
~ k2 and F
~ k4 is:
The torque T
~ 24 = −Ib1 b2 Bx ~ay
T
The total torque on the loop is then:
~ =T
~ 13 + T
~ 24 = Ib1 b2 (By ~ax − Bx ~ay )
T
~ from:
Note that it was possible to calculate T
~ =m
~
~ ×B
T
= (−Ib1 b2~az ) × (Bx ~ax + By ~ay + Bz ~az )
= Ib1 b2 (By ~ax − Bx ~ay )
Chadi Abou-Rjeily
Electromagnetic Fields Lecture-13: Static Magnetic Fields 3