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Physics 1502: Lecture 17
Today’s Agenda
• Announcements:
– Midterm 1 distributed today
• Homework 05 due Friday
• Magnetism
Trajectory in Constant B Field
• Suppose charge q enters B field with velocity v as
shown below. (vB) What will be the path q follows?
x x x x x x x x x x x x
x x x x x x x x x x x v
x B
x x x x x x x x x x x x
v
F q
F
R
• Force is always  to velocity and B. What is path?
– Path will be circle. F will be the centripetal force needed to keep
the charge in its circular orbit. Calculate R:
Radius of Circular Orbit
• Lorentz force:
• centripetal acc:
• Newton's 2nd Law:
x x x x x x x x x x x x
x B
x x x x x x x x x x x v
x x x x x x x x x x x x
v
F
F q
R


This is an important result,
with useful experimental
consequences !
Ratio of charge to mass
for an electron
e-
1) Turn on electron ‘gun’
R
2) Turn on magnetic field B
DV
‘gun’
3) Calculate B … next week; for now consider it a measurement
4) Rearrange in terms of measured values, V, R and B
&

Lawrence's Insight
"R cancels R"
• We just derived the radius of curvature of the trajectory of a
charged particle in a constant magnetic field.
• E.O. Lawrence realized in 1929 an important feature of this
equation which became the basis for his invention of the
cyclotron.

•

Rewrite in terms of angular velocity w ! 
• R does indeed cancel R in above eqn. So What??
– The angular velocity is independent of R!!
– Therefore the time for one revolution is independent of the
particle's energy!
– We can write for the period, T=2p/w or T = 2pm/qB
– This is the basis for building a cyclotron.
The Hall Effect
l
Force balance
c
B
qvd B  qE H
I
vd
F
-
d
I
qEH
B
a
Hall voltage generated
across the conductor
vd B  E H
DVH  E H d  vd Bd
Using the relation between drift velocity and current we can write:
DVH  vd Bd 
IBd
IB RH IB


, RH  1 / nq - Hall
nqA nql
l
coefficien t
The Laws of Biot-Savart &
Ampere
P
q
r
R
I
q
dx
I
x

dl
Calculation of Electric Field
• Two ways to calculate the Electric Field:
• Coulomb's Law:
"Brute force"
• Gauss' Law
"High symmetry"
• What are the analogous equations for the Magnetic Field?
Calculation of Magnetic Field
• Two ways to calculate the Magnetic Field:
• Biot-Savart Law:

I
"Brute force"
• Ampere's Law
"High symmetry"
• These are the analogous equations for the Magnetic Field!
Biot-Savart Law…bits and pieces
dl
q
A
r
X
dB
B in units of Tesla (T)
I
0= 4pX 10-7 T m /A
So, the magnetic field “circulates” around the wire
Magnetic Field of
 Straight Wire
P
• Calculate field at point P using
Biot-Savart Law:
q
r
R
q
Which way is B?
dx
• Rewrite in terms of R,q:

,
\

I
x
Magnetic Field of
 Straight Wire
P
q
r
R
q
dx
x
I

\
1
Lecture 17, ACT 1
• I have two wires, labeled 1 and 2, carrying equal
current, into the page. We know that wire 1
produces a magnetic field, and that wire 2 has
moving charges. What is the force on wire 2 from
wire 1 ?
Wire 1
Wire 2
X
X
I
F
I
B
(a) Force to the right
(b) Force to the left (c) Force = 0
Force between two conductors
• Force on wire 2 due to B at wire 1:
• Force on wire 2 due to B at wire 1:
• Total force between wires 1 and 2:
• Direction:
attractive for I1, I2 same direction
repulsive for I1, I2 opposite direction
Circular Loop
• Circular loop of radius R carries
current i. Calculate B along the axis
of the loop:
•
Rq
R
x

dB
q
z
• Magnitude of dB from element dl:
• What is the direction of the field?
• Symmetry  B in z-direction.
r
z
r
dB
Circular Loop
•
Rq
r
q
z
R
z
r
x

• Note the form the field takes for z>>R:
• Expressed in terms of the magnetic moment:

dB
note the typical
dipole field
behavior!
dB
Circular Loop
y=
f(x)
R
1
B
z

0
0
0
0
3
z
x=
x
1
z3
Lecture 17, ACT 2
• Equal currents I flow in identical
circular loops as shown in the
diagram. The loop on the right (left)
carries current in the ccw (cw)
direction as seen looking along the
+z direction.
– What is the magnetic field Bz(A)
at point A, the midpoint between
the two loops?
(a) Bz(A) < 0
(b) Bz(A) = 0
I
o
I
x
B
A
x
(c) Bz(A) > 0
o
z
Lecture 17, ACT 3
• Equal currents I flow in identical
circular loops as shown in the
diagram. The loop on the right
(left) carries current in the ccw
(cw) direction as seen looking
along the +z direction.
I
o
I
x
B
A
x
o
– What is the magnetic field Bz(B) at point B, just to the right of
the right loop?
(a) Bz(B) < 0
(b) Bz(B) = 0
(c) Bz(B) > 0
z
Magnetic Field of  Straight Wire
• Calculate field at distance R
from wire using Ampere's
Law:
dl
• Choose loop to be circle of radius R
centered on the wire in a plane  to wire.
I R
– Why?
» Magnitude of B is constant (fct of R only)
» Direction of B is parallel to the path.
–
Evaluate line integral in Ampere’s Law:
–
Current enclosed by path = I
–
Apply Ampere’s Law:

• Ampere's Law simplifies the calculation thanks to symmetry of
the current! ( axial/cylindrical )
B Field inside a Long Wire ?
• What is the B field at a distance R,
with R<a (a: radius of wire)?
• Choose loop to be circle of radius R,
whose edges are inside the wire.
– Why?
» Left Hand Side is same as before.
– Current enclosed by path = J x Area of Loop
– Apply Ampere’s Law:

Radius a
I
R
Review: B Field of a
Long Wire
• Inside the wire: (r < a)
y=
a
b1 (x);b2(x)
1
0 I r
B=
2p a2
B
• Outside the wire: (r>a)
0 I
B=
2 pr
0
0
4
r
x=
x
Lecture 17, ACT 4
• A current I flows in an infinite straight
wire in the +z direction as shown. A
2A
concentric infinite cylinder of radius R
carries current I in the -z direction.
– What is the magnetic field Bx(a) at
point a, just outside the cylinder as
shown?
(a) Bx(a) < 0
(b) Bx(a) = 0
y
x
x
a
x
b
x
x 2I
I
x
(c) Bx(a) > 0
x
x
x
Lecture 17, ACT 4
• A current I flows in an infinite straight wire
in the +z direction as shown. A concentric
infinite cylinder of radius R carries current
I in the -z direction.
2B
– What is the magnetic field Bx(b) at
point b, just inside the cylinder as
shown?
(a) Bx(b) < 0
(b) Bx(b) = 0
y
x
x
a
x
b
x
x 2I
I
x
x
x
x
(c) Bx(b) > 0
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.
L
• A solenoid is defined by a current I flowing
through a wire which is wrapped n turns per
unit length on a cylinder of radius a and
length L.
• 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.
a
B Field of a
 Solenoid
• To calculate the B field of the  solenoid using Ampere's Law,
we need to justify the claim that the B field is 0 outside the
solenoid.
• To do this, view the  solenoid from the
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:
xxxxx
• •• • •
xxxxx
• •• • •

(n: number of
turns per unit
length)
Toroid
• Toroid defined by N total turns with
current i.
•
x
x
x
x
xx
•
• B=0 outside toroid! (Consider
integrating B on circle outside toroid) •
Apply Ampere’s Law:

•
•
xx x
•
• To find B inside, consider circle of radius
r, centered at the center of the toroid.
•
•
•
•
x
•
x
x
r xx
xx
• B•
•
•
•
Magnetic Flux
Define the flux of the magnetic field
through a surface (closed or open) from:
dS
B
B
Gauss’s Law in Magnetism
Magnetism in Matter
• When a substance is placed in an external magnetic field Bo,
the total magnetic field B is a combination of Bo and field due to
magnetic moments (Magnetization; M):
–
B = Bo + oM = o (H +M) = o (H + cH) = o (1+c) H
» where H is magnetic field strength
 c is magnetic susceptibility
• Alternatively, total magnetic field B can be expressed as:
– B = m H
» where m is magnetic permeability
» m = o (1 + c)
• All the matter can be classified in terms of their response to
applied magnetic field:
– Paramagnets
– Diamagnets m < o
– Ferromagnets
m > o
m >>> o