Download Lecture 10 - UConn Physics

Physics 1202: Lecture 10
Today’s Agenda
• Announcements:
– Lectures posted on:
www.phys.uconn.edu/~rcote/
– HW assignments, solutions etc.
• Homework #3:
– On Masterphysics today: due Friday this week
– Go to masteringphysics.com
• Midterm 1:
– Friday Oct. 2
– Chaps. 15, 16 & 17.
Bar Magnet
• Bar magnet ... two poles: N and S
Like poles repel; Unlike poles attract.
• Magnetic Field lines: (defined in same way as electric
field lines, direction and density)
You can see this
field by bringing
a magnet near a
sheet covered
with iron filings
• Does this remind you of a similar case in electrostatics?
Forces due to Magnetic Fields?
•
Electrically charged particles come under various sorts of forces.
•
As we have already seen, an electric field provides a force to a
charged particle, F = qE.
•
Magnets exert forces on other magnets.
•
Also, a magnetic field provides a force to a charged particle, but
this force is in a direction perpendicular to the direction of the
magnetic field.
Definition of Magnetic Field
Magnetic field B is defined operationally by the magnetic
force on a test charge.
(We did this to talk about the electric field too)
• What is "magnetic force"? How is it distinguished from
"electric" force?
Start with some observations:
• Empirical facts: a) magnitude:  to velocity of q
b) direction: ^ to direction of q
q
v
F mag
Lorentz Force
• The force F on a charge q moving with velocity v
through a region of space with electric field E and
magnetic field B is given by:
B
x x x x x x
B

x x x x x x
v
x x x x x x
q
F
v

´ q
F
Units:
1 T (tesla) = 1 N / Am
1G (gauss) = 10-4 T
B
v
q
F=0
1
Lecture 10, ACT 1
•
1A
Two protons each move at speed v (as shown in the
diagram) toward a region of space which contains a
constant B field in the -y-direction.
– What is the relation between the magnitudes of the
forces on the two protons?
y
1
B
2
z
(a) F1 < F2
(b) F1 = F2
v
(c) F1 > F2
v
x
Circular motion
• Force is perp. to v
 q= 90o so sinq = 1 or F=qvB
• Work proportional to cos f(recall 1201)
 f :angle between F and Dx
– cos f =0 (perpendicular)
• W=0  DK=0
– Kinetic energy not changed
– Velocity constant: UCM !
R
Lecture 10, ACT 2
• Cosmic rays (atomic nuclei stripped bare of their
electrons) would continuously bombard Earth’s
surface if most of them were not deflected by
Earth’s magnetic field. Given that Earth is, to an
excellent approximation, a magnetic dipole, the
intensity of cosmic rays bombarding its surface is
greatest at the (The rays approach the earth
radially from all directions).
A) Poles
B) Equator
C) Mid-lattitudes
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’
2) Turn on magnetic field B
R
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 coefficien t
nqA nql
l
Magnetic Force on a Current
• Consider a current-carrying wire in the
presence of a magnetic field B.
• There will be a force on each of the charges
moving in the wire. What will be the total force
DF on a length Dl of the wire?
• Suppose current is made up of n
charges/volume each carrying charge q and
moving with velocity v through a wire of crosssection A.
• Force on each charge =
• Total force =

• Current =
Simpler: For a straight length of wire L carrying
a current I, the force on it is:
or
N
S
Lecture 10, ACT 3
y
• A current I flows in a wire which is formed in the
shape of an isosceles triangle as shown. A
constant magnetic field exists in the -z direction.
B
– What is Fy, net force on the wire in the ydirection?
(a) Fy < 0
(b) Fy = 0
(c) Fy > 0
x
x
x
x
x
x
x
x
x
x
xI
x
x
x
x
x
x x
x x
x x
x Ix
x x
x x
xIx
x x
x
x
x
x
x
x
x
x
x