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Fall 2008Physics 231 Lecture 8-1
Sources of the Magnetic Field
Fall 2008Physics 231 Lecture 8-2
Magnetic Field of a Point Charge
Given a point charge, q, we know that it generates an electric field regardless of whether it is moving or not
If the charge is in fact moving then the charge also generates a magnetic field
The magnetic field generated by the charge however does not behave in the same way the electric field of the charge does
Fall 2008Physics 231 Lecture 8-3
Magnetic Field of a Point ChargeIt is found that the magnetic field is perpendicular to both the velocity of the charge and the unit vector from the charge to the point in question
The magnitude of the fieldis given by
20 sin
4 rvq
Bφ
πµ
=
where φ is the angle betweenthe velocity and unit vector
The magnetic field lines form concentric circles about the velocity vector
Fall 2008Physics 231 Lecture 8-4
Magnetic Field of a Point ChargeThe direction of the magnetic field is given by one of the right-hand rules
Point your thumb, of your right hand, in the direction of the velocity and curl your fingers. The direction of the field is then the direction your fingers curl. The magnetic field for a negative charge will be in the opposite direction.
The vector equation for this is
20 ˆ
4 rrvqB ×
=rr
πµ
Fall 2008Physics 231 Lecture 8-5
Magnetic Field of a Current Element
If instead of a single charge, we have a current what is the magnetic field then given by
We start with the principle of superposition – we add the magnetic fields of the individual charges that make up the current
Given n moving charges per unit volume each carrying a charge q, then the total charge in this element is
dlAqndQ =where A is the cross sectional area
Fall 2008Physics 231 Lecture 8-6
Magnetic Field of a Current Element
We then have for dB
20
20 sin
4sin
4 rdlAvqn
rvdQ
dB dd φπµφ
πµ
==
AvqnI d=But from our work on currents we also have that
20 sin
4 rdlIdB φ
πµ
=Therefore
20 ˆ
4 rrldIBd ×
=r
r
πµ
Or in vector form
Fall 2008Physics 231 Lecture 8-7
Biot-Savart LawFor a complete circuit we then use the integral form of the previous equation
∫×
= 20 ˆ
4 rrldIB
rr
πµ
This can be a very difficult integral to evaluate
The level of difficulty depends upon the variable chosen to integrate over and the presence of any symmetries that can be exploited
We will develop another, much simpler method, shortly
Fall 2008Physics 231 Lecture 8-8
Field of a Straight WireWe set up the problem as shown
The integral becomes
( )∫−
⎟⎟⎠
⎞⎜⎜⎝
⎛
+=
a
a yxdyxIB 2/322
04πµ
The result of this integral is
220 2
4 axxaIB+
=π
µ
In the limit that a >> xxIB
πµ2
0=
Fall 2008Physics 231 Lecture 8-9
Magnetic Force Between Two Current Carrying Wires
We have two wires carrying currents Iand I’ separated by a distance r
The first wire carrying current I sets up a magnetic field at the second wire given by
rIB
πµ2
0=
Fall 2008Physics 231 Lecture 8-10
Magnetic Force Between Two Current Carrying Wires
From before we know that a wire carrying a current that is placed in a magnetic field will experience a force per unit length that is given by r
IILF
πµ
2'
0=
This is in fact what the second wire will experience
BLIFrrr
×= 'And from
This force is directed towards the other wire – the force is attractive
Fall 2008Physics 231 Lecture 8-11
Magnetic Force Between Two Current Carrying Wires
If we had started with the wire carrying the current I’, we would have ended up with the same result
If the wires were carrying currents in opposite directions, thenthe force between the wires would be repulsive
Fall 2008Physics 231 Lecture 8-12
Circular Current LoopAnother circuit for which it is worthwhile to calculate the magnetic field is the circular current loop
We start by utilizing theBiot-Savart law
20 ˆ
4 rrldIBd ×
=r
r
πµ
222 axr += a being the radius of the loopwith
Fall 2008Physics 231 Lecture 8-13
Circular Current LoopA current element at the top will yield the magnetic field as shown
A current element at the bottom will yield a magnetic field of the same magnitude, but directed downward
The vertical components will cancel and all that will be left is a horizontal magnetic field of incremental value
( ) 2/3220
220
4cos
4 ax
dlaIax
dlIdBx+
=+
=π
µθπ
µ
Fall 2008Physics 231 Lecture 8-14
Circular Current LoopThe integral is over just the circumference of the loop as all other variables are constant
( ) ( ) 2/322
20
2/3220
24 ax
aIdlax
aIB x+
=+
= ∫µ
π
µ
If there are N loops, turns, then the result is just N times above results
( ) 2/322
20
2 ax
aINB x+
=µ
Fall 2008Physics 231 Lecture 8-15
Circular Current Loop
An important point for the field of a current loop is at the center of the loop where x = 0
aINBx 2
0µ=
A current loop consisting of N turns is often referred to as aSolenoid
A final caution: These results for a current loop only apply on the axis of the loop
Fall 2008Physics 231 Lecture 8-16
Ampere’s LawJust as we developed an easy way to calculate the electric field in symmetric situations, Gauss’ Law, there is a similar technique we can use for calculating magnetic fields
∫ =⋅ IldB 0µrr
This is known as Ampere’s Law
where the integral is around a closed path and I is the total current passing through the area bounded by this path
Fall 2008Physics 231 Lecture 8-17
Ampere’s LawThere is one big difference to remember between the two methods
Gauss’ Law is an integral over a closed surface
∫ =⋅0ε
QAdErr
Whereas Ampere’s Law is an integral over a closed path
∫ =⋅ IldB 0µrr
Fall 2008Physics 231 Lecture 8-18
Ampere’s LawGiven a path around which the integration is to be done, there is a sign convention for the current
Determine the direction of the normal, as defined by the sense of movement around the integration path
If the current is going in the same direction as the normal, the current is considered to be positive otherwise it is negative
Fall 2008Physics 231 Lecture 8-19
Long Straight WireCurrent in same direction as normal of integration path
rIB
IrB
IdlB
IldB
πµ
µπ
µ
µ
2
2
0
0
0
0
=
=
=
=⋅
∫∫
rr
This is the same as we derived from the Biot-Savart Law
Fall 2008Physics 231 Lecture 8-20
Long Straight WireCurrent in opposite direction to normal of integration path
rIB
IrB
IdlB
IldB
πµ
µπ
µ
µ
2
2
0
0
0
0
=
−=−
−=−
=⋅
∫∫
rr
Again this is the same as we derived from the Biot-Savart Law
Fall 2008Physics 231 Lecture 8-21
Caution
Ampere’s Law gives you only the magnetic field due to any currents that cut through the area bounded by the integration path
For the indicated integration path, the result of Ampere’s Law is zero
Fall 2008Physics 231 Lecture 8-22
A B C
Example
Given three currents and their associated loops
1) For which loop is ∫ B • dl the smallest?→ →
A B C Same
The currents that are used in Ampere’s Law are the currents that are within the loop that is used for integration
Loop A has no current going through it whereas the other two loops do have currents going through them, the integral for Loop A is zero
Fall 2008Physics 231 Lecture 8-23
B C Same
2) Now compare loops B and C. For which loop is ∫ B • dl the greatest?
→ →
A B C
Example
Given three currents and their associated loops
The right hand side of Ampere’s law is the same for both loops, therefore the integral for both loops is the same
Fall 2008Physics 231 Lecture 8-24
Example
(a) Bx(a) < 0 (b) Bx(a) = 0 (c) Bx(a) > 0
xx
xx
xx
x
x
2II
a
b
x
y
1) What is the magnetic field Bx(a) at point a, just outside the cylinder as shown?
A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius Rcarries current 2I in the -z direction.
The loop used for application of Ampere’s Law will be outside the cylinderThe total current going through this loop is the sum of the two current yielding a current of –I. This gives a magnetic that points in the positive x direction at a
xI
B
BB
B
Fall 2008Physics 231 Lecture 8-25
Example
(a) Bx(b) < 0 (b) Bx(b) = 0 (c) Bx(b) > 0
2) What is the magnetic field Bx(b) at point b, just inside the cylinder as shown?
xx
xx
xx
x
x
2II
a
b
x
y
A current I flows in an infinite straight wire in the +z direction as shown. A concentric infinite cylinder of radius Rcarries current 2I in the -z direction.
This time, the Ampere loop only encloses current I which is in the +z direction — the loop is inside the cylinder!
The current in the cylindrical shell does not contribute to at point b.
Fall 2008Physics 231 Lecture 8-26
Long Cylindrical Conductor
The problem can be divided into two regions
Outside the conductor
Inside the conductor
Fall 2008Physics 231 Lecture 8-27
Long Cylindrical ConductorOutside the conductor
We start with an integration path around the conductor at an arbitrary distance rfrom the central axis of the conductor
∫ =⋅ IldB 0µrr
Then using Ampere’s Law
IrBdlB 02 µπ ==∫we then have
rIB
πµ2
0=or
Fall 2008Physics 231 Lecture 8-28
Long Cylindrical ConductorInside the conductor
Since we have not been told otherwise, we assume that the current is uniformly distributed across the cross section of the conductor
We define a current density 200
RI
AIJ
π==
We now apply Ampere’s Law around an integration path centered on the cylindrical axis and radius r < R
∫ =⋅ IldB 0µrr with I being the current going through
the area defined by the integration path
Fall 2008Physics 231 Lecture 8-29
Long Cylindrical ConductorInside the conductor
2
2
02
2
02
RrI
RrIrJI ===
πππ
2
2
000 RrII µµ =rBldB π2=⋅∫
rrand
200 2 RrIBπ
µ=So after equating, we then have
The magnetic field increases linearly with the distance from the axis of the cylinder
Fall 2008Physics 231 Lecture 8-30
Inside the wire: (r < a)
Outside the wire: ( r > a )
r
Ba
Long Cylindrical Conductor
200 2 RrIBπ
µ=
rIB
πµ2
0=
Fall 2008Physics 231 Lecture 8-31
Solenoid
We choose a path of integration as shown with one part of the path inside the solenoid and one part of the path outside
We assume that the path on the outside, cd, is sufficiently far from the solenoid, compared to the diameter of the solenoid, that the field is approximately equal to zero
Fall 2008Physics 231 Lecture 8-32
We use Ampere’s LawSolenoid
∫ =⋅ IldB 0µrr
First the Left-Hand-Side
∫∫
∫∫ ∫
⋅+⋅+
⋅+⋅=⋅
a
d
d
c
c
b
b
a
ldBldB
ldBldBldB
rrrr
rrrrrr
a b: Using the right hand rule for currents, the B field is in the same direction as the path and it is constant, we then have
∫ =⋅b
a
BLldBrr
Fall 2008Physics 231 Lecture 8-33
Solenoidb c: Along the part of the path that is inside the solenoid, B is perpendicular to dl, so that contribution is zero, while for the part of the path that is outside of the solenoid, B ~0, so we then have
0=⋅∫c
b
ldBrr
0=⋅∫a
d
ldBrr
0=⋅∫d
c
ldBrr
The same is true for d a
c d: Here B is zero so we again have
Fall 2008Physics 231 Lecture 8-34
SolenoidNow for the Right-Hand-SideThe total current passing through the area defined by the integration path is NI where N is the number of turns and I is the current in the loop, so the right-hand-side is
IN0µ
Now equating the left and right-hand-sides we haveINLB 0µ=
Or solving for BI
LNB 0µ=
Fall 2008Physics 231 Lecture 8-35
SolenoidThe magnetic field of a solenoid looks like the following with the exact shape of the field lines being dependent upon the the number of turns, the length, and the diameter of the solenoid