Transcript III–3 Applications of Magnetism 22. 5. 2016 1

III–3 Applications of Magnetism
22. 5. 2016
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Main Topics
• Magnetic Dipoles
• The Fields they Produce
• Their Behavior in External Magnetic Fields
• Calculation of Some Magnetic Fields
• Solenoid
• Toroid
• Thick Wire with Current
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Magnetic Dipoles I
• In electrostatics we defined electric dipoles.
We can imagine them as solid rods which
hold one positive and one negative charge
of the same absolute values some distance
apart. Although their total charge is zero
they are sources of fields with special
symmetry which decrease faster than fields
of point sources. External electric field is
generally trying to orient and shift them.
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Magnetic Dipoles II
• Their analogues in magnetism are either
thin flat permanent magnets or loops of
current. These also are sources of fields
with a special symmetry which decrease
faster than fields from straight currents and
by external magnetic field they are affected
similarly as electric dipoles. Later we shall
describe magnetic behavior of matter using
the properties of magnetic dipoles.
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Magnetic Dipoles III
• Let us have a circular conductive loop of
the radius a and a current I flowing in it. Let
us describe the magnetic field at some
distance b on the axis of the loop.
• We can “cut” the loop into little pieces
dl = ad and vector add their contribution
to the magnetic induction using the BiotSavart law.
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Magnetic Dipoles IV
• For symmetry reasons the direction of B is
the same as the direction of the axis z perpendicular to the loop and integration in
this case means only to add the projections
dBz = dB sin . And from the geometry:
sin = a/r  1/r2 = sin2 /a2
r2 = a 2 + b 2
• Let us perform the integration.
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Magnetic Dipoles V
• Since magnetic dipoles are sources of
magnetic fields they are also affected by
them.
• In uniform magnetic field they will
experience a torque.directing them in the
direction of the field.
• We shall illustrate it using a special case of
rectangular loop a x b carrying current I.
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Magnetic Dipoles VI
• Form the drawing we see that forces on the
sides a are trying to stretch the loop but if it
is stiff enough they cancel.
• Forces on the sides b are horizontal and the
upper acts into the blackboard and the lower
from the blackboard. Clearly they are trying
to stretch but also rotate the loop.
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Magnetic Dipoles VII
• To find the contribution of each of the b
sides to the torque we have to find the
projection of the force Fb perpendicularly to
the loop: T/2 = Fbsin a/2
• Since both forces act in the same sense:
T = BIabsin
• We can generalize this using the magnetic
dipole moment m = Iabm0 :
T=mxB
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Magnetic Field of a Solenoid I
• Solenoid is a long coil of wire consisting of
many loops.
• In the case of finite solenoid the magnetic
field must be calculated as a superposition
of magnetic inductions generated by all
loops.
• In the case of almost infinite we can use the
Ampere’s law in a very elegant way.
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Magnetic Field of a Solenoid II
• As a closed path we choose a rectangle
whose two sides are parallel to the axis of
the solenoid.
• From symmetry we can expect that the field
lines will be also parallel to the axis
direction.
• Since the closed field lines return through
the whole Universe outside the solenoid we
can expect they are infinitely “diluted”.
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Magnetic Field of a Solenoid III
• Only the part of the path along the side
inside the solenoid will make non-zero
contribution to the loop integral.
• If the rectangle encircles N loops with
current I and its length is l then:
Bl = 0NI
• And if we introduce the density of loops
n = N/l  B = 0nI
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Magnetic Field of a Solenoid IV
• For symmetry reason we didn’t make any
assumptions about how deep is our
rectangle immersed in the solenoid. We
didn’t have to since the magnetic field in the
long solenoid is expected to be uniform or
homogeneous.
• A reasonably uniform magnetic field can be
obtained if we shorten thick solenoid and
cut it into halves - Helmholtz coils.
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Magnetic Field of a Toroid I
• We can think of the toroid as of a solenoid bent
into a circle. Since the field lines cant escape we
do not have to make any assumptions about the
size.
• If the toroid has a radius R to its central field line
and N loops of current I, we can simply show that
all the field is inside and what is the magnitude on
a particular field line.
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Magnetic Field of a Toroid II
• Let’s us choose the central filed line as our
path then the integration simplifies and:
B(2r) = 0NI  B(r) = 0NI/2r
• His is also valid for any r within the toroid.
• The field:
• is not uniform since it depends on r.
• is zero outside the loops of the toroid
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Magnetic Field of a Thick Wire I
• Let’s have a straight wire of a diameter R in
which current I flows and let us suppose
that the current density is constant.
• We use Ampere’s law. We use circular paths
one outside and one inside the wire.
• Outside the field is the same as if the wire
was infinitely thin.
• Inside we get linear dependence on r.
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Magnetic Field of a Thick Wire II
• If we take a circular path of the radius r
inside the wire we get:
B(2r) = 0Ienc
• The encircled current Ienc depends on the
area surrounded by the path
Ienc = I r2/R2 
B = 0Ir/2R2
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Homework
• No homework today!
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Things to read
• Chapter 27 – 5 and 28 – 4, 5
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Circular Loop of Current I
 0 Idl
 0 Iad
dB 

2
2
4r
4r
 0 Iad
dBz 
sin 
2
4r
 0 Ia sin 
Bz   dBz 
d 
2

4r
2
 0 Ia sin  2
 0 2a I


2
3
4r
4r
0
2 IA
4 ( a 2  b 2 )
3
2
Circular Loop of Current II
A = a2 is the area of the loop and its normal
has the z direction. We can define a magnetic
dipole moment m = IA and suppose that we
are far away so b>>a. Then:


 0 2m
B (b) 
4 b 3
Magnetic dipole is a source of a special
magnetic field which decreases with the third
power of the distance.
^
Circular Loop of Current II
A is the area of the loop and its normal has
the z direction. We can define a magnetic
dipole moment  = IA and suppose that we
are far away so b>>a then we can write:
 0 Idl
 0 Iad 
dB 

2
2
4r
4r
 0 Iad 
sin
and dB
the formula
to calculate
thefield, which
z 
2
4

r

is the Biot-Savart law:


 
 0 I1[ dl1  ( r2  r1 )]
dB ( r2 ) 

 3
4 | r2  r1 |
Circular
Current
I
 0 IdlLoop
 0of
Iad

dB 

4r 2
4r 2
 0 Iad
dBz 
sin 
2
4r
 0 Ia sin 
Bz   dBz 
4r 2
 d
That is the reaso
an



 
 0 I1[ dl1  ( r2  r1 )]
dB ( r2 ) 

 3
4 | r2  r1 |
Circular Loop of Current II
 0 Idl
 0 Iad 
dB 

2
4r
4r 2
 0 Iad 
dBz 
sin 
2
4r
That is the reaso
and the formula to calculate the field, which

is the Biot-Savart law:


 
 0 I1[ dl1  ( r2  r1 )]
dB ( r2 ) 

 3
4 | r2  r1 |
The vector or cross product I
Let c=a.b
Definition (components)
ci   ijk a j bk
The magnitude |c|

 
c  a b sin 
Is the surface of a parallelepiped made by a,b.
The vector or cross product II
The vector c is perpendicular to the plane
made by the vectors a and b and they have to
form a right-turning system.



ux
uy
uz

c  ax
ay
az
bx
by
bz
ijk = {1 (even permutation), -1 (odd), 0 (eq.)}
^