Transcript III-2

III–2 Magnetic Fields Due to
Currents
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1
Main Topics
•
•
•
•
Forces on Moving Electric Charges
Biot-Savart Law
Ampere’s Law.
Calculation of Some Magnetic Fields.
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Forces on Moving Electric
Charges I
• Since currents are in reality moving charges
it can be expected that all what is valid for
interaction of magnetic fields with currents
will be valid also for moving charges.


• The force F of a magnetic field B acting

on a charge q moving by a velocity v is
given by the Lorentz formula:

 
F  q (v  B )
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Forces on Moving Electric
Charges II
• Lorentz force is in fact part of a more
general formula which includes both
electric and magnetic forces:

  
F  q[ E  (v  B)]
• This relation can be taken as a definition of
electric and magnetic forces and can serve
as a starting point to study them.
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Forces on Moving Electric
Charges III
• Lorentz force is a central issue in whole
electromagnetism. We shall return to it by
showing several examples. Moreover we
shall find out that it can be used as a basis
of explanation of almost all magnetic and
electromagnetic effects.
• But at this point we need to know how are
magnetic fields created quantitatively.
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Biot-Savart Law I
• There are many analogies between
electrostatic and magnetic fields and of
course a question arises whether some
analog of the Coulomb’s law exists, which
would describe how two short pieces of
wires with current would affect themselves.
It exists but it is too complicated to use. For
this reason the generation and influence of
magnetic fields are separated.
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Biot-Savart Law II
• All what is necessary to find the mutual
forces of two macroscopic wires of various
sizes and shapes with currents is to employ
the principle of superposition, which is
valid in magnetic fields as well and
integrate.
• It is a good exercise to try to make a few
calculations then try do something better!
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Magnetic Field Due to a Straight
Wire I
• Let’s have an infinite wire which we
coincide with the x-axis. The current I flows
in the +x direction. We are interested in
magnetic induction in the point P [0, a].
• The main idea is to use the principle of
superposition. Cut the wire into pieces of
the same length dx and add contribution of
each of them.
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Magnetic Field Due to a Straight
Wire II
• For a contribution from a single piece we use
formula derived from the Biot-Savart law:


  0 I dx  rxP
dB 
3
4
rxP
• Since both vectors which are multiplied
 lie in the
x, y plane only the z component of B will be
non-zero which leads to a great simplification. We
see where the right hand rule comes from!
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Magnetic Field Due to a Straight
Wire III
• So a piece of the length dx with the coordinate x
contributes:
 0 I dx sin 
dBz 
2
4
r
• Here r is the distance of dx and P and  is the
angle between the line joining dx and P and the xaxis. We have to express all these quantities as a
function of one variable e.g. the .
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Magnetic Field Due to a Straight
Wire IV
• For r we get:
1
sin 
r sin   a  2 
2
r
a
2
and for x and dx (- is important to get
negative x at angles  <  /2 !):
x
a d
  cot  x  a cot   dx 
2
a
sin 
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Magnetic Field Due to a Straight
Wire V
• So finally we get:
0 I
Bz 
4

sin  sin  a d

2
2
0
a sin 
2

0 I
0 I
sin  d 

4a 0
2a
The conclusions we can derive from the symmetry
we postpone for later!
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Ampère’s Law
• As in electrostatics also in magnetism a law
exists which can considerably simplify
calculations in cases of a special symmetry
and can be used to clarify physical ideas in
many important situations.
• It is the Ampères
law which relates the line

integral of Bover a closed path with
currents which are surrounded by the path.
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Magnetic Field Due to a Straight
Wire VI
• As it is the case with using the Gauss’ law,
we
 have to find a path which is tangential to
B everywhere and on which the magnitude
of B is constant. So it must be a special field
line. Then we can move B out of the
integral, which then simply gives the length
of the particular integration path.
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Magnetic Field Due to a Straight
Wire VII
• Let us have a long straight wire with current I.
• We expect B to depend on r and have axial
symmetry where the wire is naturally the axis.
• The field lines, as we already know are circles and
therefore our integration path will be a circle with
a radius r equal to the distance where we want to
find the field. Then: 2rB (r )   0 I 
0 I
B(r ) 
2r
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Magnetic Field Due to a Straight
Wire VIII

• The vectors of the magnetic induction B are
tangents to circles centered on the wire,
which thereby are the field lines, and the
magnitude of B decreases with the first
power of the distance.
• It is similar as in the case of the electrostatic
field of an straight, infinite and uniformly
charged wire but there electric field lines were
radial while here magnetic are circular, thereby
perpendicular in every point.
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Magnetic Field in a Center of a
Square Loop of Current I
• Apparently by employing the Amperes law we
have obtained the same information in a
considerable easier way. But, unfortunately, this
works only in special cases.
• Let’s calculate magnetic induction in the center of
a square loop a x a of current I. We see that it is a
superposition of contributions of all 4 sides of the
square but to get these we have to use the formula
for infinite wire with appropriate limits.
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Magnetic Field in a Center of a
Square Loop of Current II
• The contribution of one side is:
0 I
Bz 
4 a2
3
4
0 I
 sin  d  2a
2
4
etc.
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Force Between Two Straight
Wires I
• Let us have two straight parallel wires in
which currents I1 and I2 flow in the same
direction separated by a distance d.
• First, we can find the directions and then
simply deal only with the magnitudes. It is
convenient to calculate a force per unit
length.
F  0 I1 I 2

l
2 d
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Force Between Two Straight
Wires II
• This is used for the definition of 1 ampere:
1 ampere is a constant current which, if
maintained in two straight parallel
conductors of infinite length, of negligible
cross section, and placed 1 meter apart in
vacuum, would produce between these
conductors a force equal to 2 10-7 N per
meter of length.
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Homework
• No homework!
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Things to read
• This Lecture Covers
Chapter 28 – 1, 2, 3, 4, 6
• Advance reading
Chapter 27 – 5; 28 – 4, 5
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Magnetic interaction of two
currents I
Let us have two currents I1 and I2 flowing in 


two short straight pieces of wire dl1 (r1 ) and dl2 (r2 )
Then the force acting on the second piece due
to the existence of the first
piece
is:






 0 I1 I 2 dl2  [ dl1  ( r2  r1 )]
dF12 ( r2 ) 

 3
4 | r2  r1 |
This very general formula covers almost all
the magnetism physics but would be hard to
use in practice.
Magnetic interaction of two
currents II
That is the reason why it is divided into the
formula using the field (we already know):




dF12 ( r2 )  I 2 dl2  dB
and the formula to calculate the field, which
particularly is the Biot-Savart law:



 
 0 I1[ dl1  ( r2  r1 )]
dB ( r2 ) 

 3
4 | r2  r1 |
Magnetic interaction of two
currents III
If we realize that:


0
( r2  r1 )
r12  

| r2  r1 |
is a unit vector pointing in the direction from


the first current r1 to the second one r2, we se
that magnetic forces decrease also with the
second power of the distance.

 
 0 I1[ dl1  r120 ]
dB ( r2 ) 

 2
4 | r2  r1 |
Magnetic interaction of two
currents IV
The “scaling” constant 0 = 4 10-7 Tm/A is
called the permeability of vacuum or of free
space. Some authors don’t use it since it is
not an independent parameter of the Nature.
It is related to the permitivity of vacuum 0
and the speed of light c by:
 0 0
1
 2
c
^
Ampère’s Law
Let us have none, one, two ore more wires
with currents I1, I2 … then:


 B  dl   0  I i
• All the current must be added but their
polarities must be taken into account !
^