# Hour Exam

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```					            Examples of Magnetic Field
r        Calculations
 Bgd
— l  0 I
dl
I

xxxxx
xxxxxxxx
xxxxxxxxx
r
xxxxxxxx
1
a   xxxxx
B Field by Ampere’s Law

"High symmetry"
r
 Bgd
— l  0 I

Integral around a path …           Net Current ―enclosed‖ by
hopefully a simple one             that path

   I

2
B-field of  Straight Wire, reprise

• Calculate field at distance R
from wire using Ampere's Law:                                       dl

Choose loop to be circle of radius R centered on the          I        R
wire in a plane ^ to wire.
Why?
Magnitude of B is constant (function of R only)
Direction of B is parallel to the path.
r
Evaluate line integral in Ampere’s Law:                Bg
— dl  B(2š R)
Current enclosed by path = I
Apply Ampere’s Law:                               0 I
B
2 RB  0 I                           2š R
Ampere's Law simplifies the calculation thanks to
3
symmetry of the current! (axial/cylindrical)
Question

Two identical loops are placed in
proximity to two identical current
carrying wires.
A    B
r
 Bgd
For which loop is — l the greatest?

A             B      Same

4
Question
Now compare loops B and C.
r
For which loop is
 Bgd
— l              C
the greatest?                   B

B        C    Same

5
Question
y
A current I flows in an infinite straight                    a
wire in the +z direction as shown. A                         x
x   b   x
concentric infinite cylinder of radius R
carries current 2I in the -z direction.              x               x 2I
– What is the magnetic field Bx(a) at                   x
I
x
point a, just outside the cylinder as                     x
shown?
x
(a) Bx(a) < 0        (b) Bx(a) = 0        (c) Bx(a) > 0
• This situation has cylindrical symmetry
• By applying Ampere’s Law, we see that the field at point
a must just be the field from an infinite wire with current I
flowing in the -z direction!
B

B         x
I
B               6

B
Question
y
A current I flows in an infinite straight wire                 a
in the +z direction as shown. A concentric                     x
x   b   x
infinite cylinder of radius R carries current 2I
in the -z direction.                                   x               x 2I
– What is the magnetic field Bx(b) at point               x
I
x
b, just inside the cylinder as shown?                      x

x
(a) Bx(b) < 0         (b) Bx(b) = 0         (c) Bx(b) > 0

• This time, the Ampere loop only encloses current I in the
+z direction — the loop is inside the cylinder!

• The current in the cylindrical shell does not contribute to
at point b.

7
B Field Inside a Long Wire
xxxxx
• Suppose a total current I flows through
the wire of radius a into the screen as    xxxxxxxx
shown.                                    xxxxxxxxx
• Calculate B field as a function of r, the        r
xxxxxxxx
distance from the center of the wire.
a     xxxxx
• B field is only a function of r  take path
r
to be circle of radius r:                          Bgd
— l  B(2 š r)
r2
• Current passing through circle:     I enclosed    2I
a
r r                              0 I r
• Ampere's Law:
 Bgd
— l  o Ienclosed           B
2 š a2
8
B Field of a Long Wire

• Inside the wire: (r < a)

0 I r                 a   a
B
2 a2

• Outside the wire:
B   B
(r>a)

0 I
B
2 r                       r   r

9
Question
• Two cylindrical conductors each                   I                   I
carry current I into the screen as
shown. The conductor on the
left is solid and has radius R=3a.
The conductor on the right has a                                          a
hole in the middle and carries               3a                  3a   2a
current only between R=a and
R=3a.
–What is the relation between the magnetic field at R = 6a
for the two cases (L=left, R=right)?

(a) BL(6a)< BR(6a)         (b) BL(6a)= BR(6a)         (c) BL(6a)> BR(6a)
• Use Ampere’s Law in both cases by drawing a loop in the plane of
the screen at R=6a
• Both fields have cylindrical symmetry, so they are tangent to the
loop at all points, thus the field at R=6a only depends on current
enclosed                                                        10
• Ienclosed = I in both cases
Question
• Two cylindrical conductors each carry           I                        I
current I into the screen as shown. The
conductor on the left is solid and has
radius R=3a. The conductor on the
right has a hole in the middle and                                           a
carries current only between R=a and 3a                        3a        2a
R=3a.
– What is the relation between the
magnetic field at R = 2a for the two
cases (L=left, R=right)?
(a) BL(2a)< BR(2a)      (b) BL(2a)= BR(2a)           (c) BL(2a)> BR(2a)
Again, field only depends upon current enclosed...
LEFT cylinder:                   RIGHT cylinder:
 (2a)2
4                       (2a)2  a 2      3
IL          I I              IR                       I I
 (3a)2
9                       (3a)2  a 2      8
11
B Field of  Current Sheet
• Consider an  sheet of current described by                 x
n wires/length each carrying current i into the             x
screen as shown. Calculate the B field.
x
w x
x
• What is the direction of the field?
x
x
• Symmetry  vertical direction                             x
x
•   Calculate using Ampere's law for a square of              x
x
side w:                                                   x
r                                                     constant
 Bgd
constant
• — l  Bw  0  Bw  0  2Bw
•   I  nwi          r                   0 ni
 Bgd
therefore, — l  0 I          B
2                     12
B Field of a Solenoid
• A constant magnetic field can (in principle) be produced by an 
sheet of current. In practice, however, a constant magnetic field is
often produced by a solenoid.

• A solenoid is defined by a current i flowing                   L
through a wire that is wrapped n turns per unit
length on a cylinder of radius a and length L.
a
To correctly calculate the B-field, we should use
Biot-Savart, and add up the field from the different loops.

• If a << L, the B field is to first order contained within the
solenoid, in the axial direction, and of constant magnitude.
In this limit, we can calculate the field using Ampere's Law.
13
B Field of an  Solenoid
• To calculate the B field of the solenoid using Ampere's Law, we
need to justify the claim that the B field is nearly 0 outside the
solenoid (for an  solenoid the B field is exactly 0 outside).

• To do this, view the  solenoid from the     xxxxxxxxxxx
side as 2  current sheets.
••••••••••••••

• The fields are in the same direction in the region
between the sheets (inside the solenoid) and
cancel outside the sheets (outside the solenoid).

• Draw square path of side w:
xxxxxxxx
r
• ••••••••••
 Bgd
— l  Bw
       B  0 ni     Note: B 
Amp
I  nwi                                         Length
14
Question
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?
a) left    b) right    c) up    d) down
e) out of the screen    f) into the screen

Which side of the solenoid should be labeled as the magnetic north
pole?
a) right
b) left
15
Right Hand Rule
Use the right hand rule to find the B-field: wrap your fingers in the
direction of the current, the B field points in the direction of the
thumb (to the left). Since the field lines leave the left end of
solenoid, the left end is the north pole.

16
Solenoids
The magnetic field of a solenoid is essentially identical to
that of a bar magnet.

The big difference is that we can turn the solenoid on and
off ! It attracts/repels other permanent magnets; it attracts
ferromagnets, etc.
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Toroid                               •          •     •
•                           •
• Toroid defined by N total turns                  •                xx x               •
x               x
with current i.                                          x                   x
• B=0 outside toroid! (Consider          •                 x                r x            •
x                 x
integrating B on circle outside toroid) •                  xx              x         •
x   x
•                           •
• To find B inside, consider circle of radius r,
centered at the center of the toroid.
•      • B•
r
 Bgd
— l  B(2 š r)
I  Ni
r                                 0 Ni
Apply Ampere’s Law:
 Bgd
— l  0 I                        B
2š r
18
Magnetic Materials
Material is made up of atoms.                         

2 r
For an electron v                                          e
T
q e ev                              L
I         
t    T     2 r
r ev              e            e r
r
  IA 
2 r
 
r  
2

2m
rmv    L
2m
r                          r r
where L is the angular momentum r  p

So a material object is full of magnetic moments ---
-- which generally cancel due to no preferred direction.
19
Magnetization & H
When a magnetic field is applied the symmetry is broken
And there is a direction along which the  line up
For our toroid let B0 be the magnetic field when the
windings are empty and B is the field when the space is
filled with material.

B B0  Bm                     M is the magnetization or
r

total magnetic moment per
unit volume
Bm  0                0 M
i

V
r   r
Define another vector H such that           
B  0 H  M     
H is due to the current and does not depend on the
presence of material.
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Toroid                                 •     •     •
•                      •
• Toroid defined by N total turns with        •             xx x              •
x          x
current i.                                           x              x
• B=0 outside toroid! (Consider integrating •          x           r x            •
B on circle outside toroid)                           x            x
•           xx       x x         •
x
•                      •
• To find B inside, consider circle of radius r,
centered at the center of the toroid.
•     • B•
r
 Bgd
— l  B(2 š r)
I  Ni  2 rni
 0 Ni
 B  2š r   0 ni
r
 Bgd
Apply Ampere’s Law: — l   I
21
Materials
r
Recall there are three types of materials   M  H

•Diamagnetic           0 and small : 10 5
•Paramagnetic          0 and small : 10 5
•Ferromagnetic         0 very large : 5000

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