Download Some More Slides on Magnetism

A Kingdom for a Magnet!
Finishing the topic
Next few days

Today
◦ QUIZ
◦ Pulling the loose ends together.
◦ Working some problems.
Next week – next chapter.
 Don’t forget the WA that is due Sunday
night.

Exam Number 2
Average=67%
Loose End – Force And Wires
The magnetic force on the
moving charges pushes the
wire to the right.
Look at the direction of the force
and the velocity
Magnetism
5
RECALL
v2
Centripeta l Accelerati on 
r
mv 2
Centripeta l Force 
r
The magnetic force is qvB
mv 2
qvB 
r
mv
r
Bq
Recall : v  r or r  v/ 
v Bq
 
r m
This is called the cyclotron
Magnetism
angular
frequency
6
Bq

  2f
m
1
mv T  period  f
r
Bq
Magnetism
7
Off Angle
PITCH
P  v parallelT
P
Magnetism
8
PROBLEM
An electron at point A in the figure has a speed v0 of 1.4 x 106 m/s. Find
(a) the magnitude and direction of the magnetic field that will cause the electron
to follow the semicircular path from A to B and
(b) the time required for the electron to move from A to B.
(c) What magnetic field would be needed if the particle were a proton instead of
an electron?
m=9.1E-31 Kg
e=1.6E-19 C
Magnetism
9
Consider a section of wire.
In time t, a charge Q
passes through the wire.
F=Bqv=BQv
In time t, a charge I t passes
through the first surface.
Q
I=
; So Q=It
t
Force
F  BQv  B
Q
tv  BIL
t
Or:
F=BIL Sin(I,B)
Same Thing but angular:
F  qvB sin 
 q 
F   
vt B sin 

t L

I
F  ILB sin 
1+1=2 Wires
0 I
B
2d
0 I
0 I 2 L
F  BIL 
IL 
2d
2d
or
L
 0 I1 I 2 L
F
2d
F
I
I
d
LET’S TALK ABOUT
TORQUE
Current Loops in Magnetic Fields
Current Loop
What is force
on the ends??
Loop will tend to rotate due to the torque the field applies to the loop.
Magnetism
14
The Loop (From the top)
OBSERVATION
Force on Side 2 is out
of the paper and that on
the opposite side is into
the paper. No net force
tending to rotate the loop
due to either of these forces.
The net force on the loop is
also zero,
pivot
Magnetism
15
The other sides
t1=F1 (b/2)Sin()
=(B i a) x (b/2)Sin()
total torque on
the loop is: 2t1
Total torque:
t=(iaB) bSin()
=iABSin()
(A=Area)
Magnetism
16
Application: The motor
If the conductor is a loop, the torque can create an
electric motor.

A circular coil of wire 8.6 cm in
diameter has 15 turns and
carries a current of 2.7 A. The
coil is in a region where the
magnetic field is 0.56 T.
What orientation of the coil
gives the maximum torque on
the coil.
What is this maximum torque?
t  NiABSin( )
Magnetism
18
PROBLEM SOLVING
A charge particle enters a uniform magnetic field and follows the circular
path shown in the drawing.
Is the particle positively or negatively charged?
Two isotopes of carbon, carbon-12 and carbon-13, have masses of
1.993 10-26 kg and 2.159 10-26 kg, respectively. These two isotopes
are singly ionized (+e) and each is given a speed of 7.38 105 m/s.
The ions then enter the bending region of a mass spectrometer where
the magnetic field is 0.5100 T. Determine the spatial separation
between the two isotopes after they have traveled through a half-circle.
.03 m
0
When a charged particle moves at an angle of 250
with respect to a magnetic field, it experiences a
magnetic force of magnitude F. At what angle (less
than 900) with respect to this field will this particle,
moving at the same speed, experience a magnetic
force of magnitude 2F?
A horizontal wire of length 0.49 m, carrying a current of 7.0 A, is
placed in a uniform external magnetic field. When the wire is
horizontal, it experiences no magnetic force. When the wire is tilted
upward at an angle of 19° it experiences a magnetic force of 4.60 103 N. Determine the magnitude of the external magnetic field.
0
.00412 T
A square coil of wire containing a single turn is placed in a uniform 0.20
T magnetic field, as the drawing shows. Each side has a length of 0.24
m, and the current in the coil is 12 A. Determine the magnitude of the
magnetic force on each of the four sides.
top
bottom
left
right
10.576 N
20.576 N
30 N
40 N
The drawing shows a thin, uniform rod that has a length of
0.35 m and a mass of 0.080 kg. This rod lies in the plane of
the screen and is attached to the floor by a hinge at point P. A
uniform magnetic field of 0.27 T is directed perpendicularly into
the plane of the screen. There is a current I = 3.8 A in the rod,
which does not rotate clockwise or counter-clockwise. Find the
angle θ. (Hint: The magnetic force may be taken to act at the
center of gravity.)
2.7°
6
A copper rod of length 0.80 m is lying on a frictionless table (see
the drawing). Each end of the rod is attached to a fixed wire by
an unstretched spring that has a spring constant of k = 70 N/m. A
magnetic field with a strength of 0.17 T is oriented perpendicular
to the surface of the table.
(a) What must be the direction of the current in the copper rod that causes the
springs to stretch?
The current flows eft-to-right in the copper rod.
l
(b) If the current is 11 A, by how much does each spring stretch?
.0107 m
0
Two coils have the same number of circular turns and carry the same
current. Each rotates in a magnetic field as in the figure below. Coil 1 has a
radius of 5.2 cm and rotates in a 0.14-T field. Coil 2 rotates in a 0.42-T field.
Each coil experiences the same maximum torque. What is the radius (in cm)
of coil 2?
cm
3
A wire has a length of 7.00 10-2 m and is used to make a
circular coil of one turn. There is a current of 2.50 A in the wire.
In the presence of a 3.00-T magnetic field, what is the maximum
torque that this coil can experience?
.00292 N · m
0
mc
The rectangular loop in the drawing consists of 80 turns and carries a current
of I = 2.5 A. A 1.7-T magnetic field is directed along the +y axis. The loop is
free to rotate about the z axis. (θ = 31°)
(a) Determine the magnitude of the net torque exerted on the loop.
02 N · m
1
(b) State whether the 31° angle will increase or decrease.
the angle increases the angle decreases