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Springr2004 OSU Sources of Magnetic Field
Chapter 28
1
Magnetic Field of a Wire
• Fundamental Laws for Calculating B-field
– Biot-Savart Law (long method, but works always)
– Ampere’s Law (high symmetry)
• B-Field of a Straight Wire
– For a thin straight conductor carrying current the
magnetic field is: We expect B from wire to be
proportional to I/R.
Springr2004 OSU Sources of Magnetic Field
Chapter 32
2
Springr2004 OSU Sources of Magnetic Field
Chapter 32
3
Magnetic Field of a Wire
• B-Field of a Straight Wire
• Where
• is the Permeability of free space
• Use the right hand rule to determine the
direction. (Place thumb in direction of
current and B Field is in direction of fingers
grabbing the wire.)
7
0 4 10 .T mA
0
2
IB
r
dl
r I
Springr2004 OSU Sources of Magnetic Field
Chapter 32
4
Ampere’s Law
I
r dl
Springr2004 OSU Sources of Magnetic Field
Chapter 32
5
Today
Example B-field Calculations (Application of
Ampere’s Law)
– Inside a Long Straight Wire
– Infinite Current Sheet
– Solenoid
– Toroid
– Biot-Savart Law
– Circular Loop
Springr2004 OSU Sources of Magnetic Field
Chapter 32
6
Ampere’s Law
A) Evaluate the integral dl around each loop
1) 0 3) ab
2) 2(a + b) 4) Not enough info
R
dl
1) 0 3) 2R
2) R2 4) not enough information
b
a
dl
Springr2004 OSU Sources of Magnetic Field
Chapter 32
7
Ampere’s Law - Examples
Two identical loops are placed in
proximity to two identical current
carrying wires.
1) A 2) B 3) Same
3) For which loop is B • dl the greatest?
A B
Springr2004 OSU Sources of Magnetic Field
Chapter 32
8
Two identical loops are
placed in proximity to
two identical current
carrying wires.
1) B 2) C 3) Same
D) Now compare loops B and C. For which loop is B • dl
the greatest?
C B
Springr2004 OSU Sources of Magnetic Field
Chapter 28
9
B Field Inside a Long Wire
Suppose a total current I flows
through the wire of radius a
into the screen as shown.
To calculate B field as a function
of r, from center of the wire:
Take an amperian loop of radius r outside the wire,
using Ampere’s Law:
The enclosed current is all of current through wire:
The B-field diminishes as 1/r outside the wire
r
R
r
0 encB dl I 02 encB rd rB I
0
2
IB
r
Springr2004 OSU Sources of Magnetic Field
Chapter 28
10
B Field Inside a Long Wire
Now B-field inside the wire:
We choose an amperean loop
of radius r inside the wire:
But the enclosed current is a fraction of total; since
current is uniform:
Inside the wire the B-field is linear with r.
r
R
r 0
02
enc
enc
B dl I
B rd rB I
2
2enc
rI I
R
0
22
IrBtherefo e
Rr
Springr2004 OSU Sources of Magnetic Field
Chapter 32
11
B Field Inside a Long Wire
Inside the wire: (r < R)
B
r x
b1(x);b2(x)
0 4
0
1
x =
y =
• Outside the wire:( r>R )
B = 0 I
2 r
2
0
π 2
μ
R
r I B
R
Springr2004 OSU Sources of Magnetic Field
Chapter 32
12
B Field of Current Sheet
Consider an sheet of current described
by n wires/length each carrying current I I
into the screen as shown. Calculate the B field.
What is the direction of the field?
From the Symmetry +/ y direction w
square of side w
• BwBwBwldB 200
• nwiI
therefore, IμldB 0
2
0niμB
x
x
x x
x
x
x x
x
x x
x
x
y
constant constant
Calculate using Ampere's law for a
Springr2004 OSU Sources of Magnetic Field
Chapter 32
13
B-Field of A Solenoid
A uniform magnetic field can
be produced by a solenoid A solenoid is defined by a current
I flowing through a wire that is
wrapped n turns per unit length
on a cylinder of radius a and
length L.
dl l
0B dl Bl NI 0
0
NIB nI
l
Springr2004 OSU Sources of Magnetic Field
Chapter 32
14
B-Field of A Toroid
• Toroid defined by a solenoid of N total turns with
current I connected at both ends. It becomes a donut
shape with a coil wrapped around it.
outside Toroid B = 0
integrating B on circle outside
toroid, enclosed current = I +(- I) = 0
Inside the
Toroid
Get a concentrated field inside the toroid.
dl
r
r
NIB
2
0
02B dl B r NI
Springr2004 OSU Sources of Magnetic Field
Chapter 32
15
Examples
Two cylindrical conductors, one solid and the other hallow in middle, each carry current I into the screen as shown. The conductor has radius R=4a. The conductor on the right has a hole in the middle and carries current only between R=a and R=4a.
At R = 5a which conductor produces stronger B-field?
1) Left conductor 3) Both are the same
2) Right Conductor 4) both are zero
a
4a
4a
I I
Springr2004 OSU Sources of Magnetic Field
Chapter 32
16
Examples
Two cylindrical conductors, one solid and the other hallow in middle, each carry current I into the screen as shown. The conductor has radius R=4a. The conductor on the right has a hole in the middle and carries current only between R=a and R=4a.
At R = 2a which conductor produces stronger B-field?
1) Left conductor 3) Both are the same
2) Right Conductor 4) both are zero
a
4a
4a
I I
Springr2004 OSU Sources of Magnetic Field
Chapter 32
17
Examples
Use Ampere’s Law in both cases by drawing a loop in the plane of the screen at R = 5a and R = 2a. Both fields have cylindrical symmetry, so the fields are tangent to the loop at all points, thus the field at R=5a only depends on current enclosed
Ienc = I in both cases
Field only depends on
enclosed current
a
4a
4a
I I
Springr2004 OSU Sources of Magnetic Field
Chapter 32
18
Examples
A current carrying wire is wrapped around an iron core,
forming an electro-magnet.
Which direction does the
magnetic field point inside
the iron core?
1) left 4) right 2) up 5) down
3) out of the screen 6) into the screen
Which side of the solenoid should be labeled as the
magnetic north pole?
1) left 3) right
2) up 4) down
Springr2004 OSU Sources of Magnetic Field
Chapter 32
19
B Field Inside a Long Wire
Inside the wire: (r < R)
R
r x
b1(x);b2(x)
0 4
0
1
x =
y =
• Outside the wire:( r>R )
B = 0 I
2 r
2
0
π 2
μ
R
r I B
Springr2004 OSU Sources of Magnetic Field
Chapter 28
20
B Field of Current Sheet
Consider an sheet of current described
by n wires/length each carrying current I I
into the screen as shown. Calculate the B field.
What is the direction of the field?
From the Symmetry +/ y direction w
square of side w
• BwBwBwldB 200
• nwiI
therefore, IμldB 0
2
0niμB
x
x
x x
x
x
x x
x
x x
x
x
y
constant constant
Calculate using Ampere's law for a