Download PHYS 342: Modern Physics

The Magnetic Field
• Charges produce a vector field
(the electric field E)
• Magnets also produce a vector field, the magnetic
field B, at all points in space around it
• Magnetic fields generated from a moving charge or a
collection of moving charges (an electric current)
• Moving charges and currents also respond to
magnetic fields
• Electric charges can be isolated, forming their own
intrinsic electric fields
• However, magnetic charges (called magnetic
monopoles) have not been found
– Magnets always have two poles (north and south)
Magnetic Field Lines
• Magnetic fields can be represented by field lines,
similar to electric fields
– Direction of B = direction north pole of compass needle
points at that location
– Spacing of the lines represents the magnitude of B
• The field lines for a bar magnet:
hyperphysics.phyastr.gsu.edu/hbase/magnetic/elemag.html
– Opposite magnetic poles attract each other, and like magnetic poles
repel each other
Magnetic Field Lines
• Iron filings placed
near bar magnets:
• Earth has a magnetic field produced (by mostly
unknown mechanisms) in the liquid
portion of its core
• Magnetic field lines near the surface
resemble those of a bar magnet
• Compass needle (small bar magnet)
has its north-pole end attracted to the
south magnetic pole of the Earth
(what we call a geomagnetic north pole)
Magnetic Force
• From experiments with a moving charged particle:
v
+q

B
(magnetic force is zero)
v
+q

B
X
v
X
+q  X
X
X
X
v
X
X
X
X
X
X
+q
B (directed into page)
v
+q
(magnetic force not zero)
v

B (directed out of page) +q
Magnetic Force
• From other experiments, the magnitude of magnetic
force is proportional to:
– Particle charge and velocity
– Magnitude of the magnetic field
– Component of velocity perpendicular to the field
• The direction of the magnetic force is perpendicular to
both the particle velocity and the direction of the
magnetic field
• Summary of these experiments: F  qvB sin 
– Direction of F given by the
Right-Hand Rule
– SI units for B are the tesla (T)
(1 T = Ns/Cm)
– 1 T = 104 gauss (G)
Example Problem #19.4
Determine the initial direction
of the deflection of charged
particles as they enter the
magnetic fields shown.
Solution (details given in class):
(a) Up
(b) Out of the page
(c) Zero force
(d) Into the page
CQ 1: A positively-charged particle is moving through a
magnetic field of strength B as shown below. The force
experienced by the particle due to the magnetic field is:
A)
B)
C)
D)
to the right.
to the left.
into the page.
equal to zero.
CQ 2: A positively-charged oil drop is allowed to fall
through the electric field created by the plates as
shown. In order to give the oil drop a straight
trajectory, a magnetic field should be established with
field lines pointing:
A)
B)
C)
D)
left to right.
right to left.
out of the page.
into the page.
Example Problem #19.8
An electron is accelerated through 2400 V
from rest and then enters a region where there
is a uniform 1.70–T magnetic field. What are
(a) the maximum and (b) the minimum
magnitudes of the magnetic force acting on
this electron?
Solution (details given in class):
(a) 7.90  10–12 N
(b) 0
Magnetic Force and Current
• An electrical current consists of the movement of
many charges
• Since a magnetic field exerts a force on one moving
charge, it will exert forces on many moving charges
– Net force is the sum of all the magnetic forces on the
individual charges producing the current
• Consider a flexible wire partially in a
magnetic field
– The max. force on 1 charge is qvdB
– The max. total force is (qvdB)  (nAl)
= (force/charge)  (# charges)
= BIl (see Chap. 17)  for I  B
– In general: F  BIl sin 
Application: Loudspeakers
• Schematic of one type of loudspeaker design:
• Magnetic field exerts force on current-carrying coil
(and cone) as shown
• When current reverses direction, magnetic force
also reverses direction
• Alternating force vibrates cone, which produces
sound waves
Example Problem #19.20
A conductor is suspended by two flexible wires as shown has a
mass per unit length of 0.0400 kg/m. What current must exist in
the conductor for the tension in the supporting wires to be zero
when the magnetic field is 3.60 T into the page? What is the
required direction for the current?
Solution (details given in class):
0.109 A to the right
Torque on a Current Loop
• Consider a rectangular conducting loop carrying
current I in the presence of a magnetic field:
(Area = A
= ab)
–
–
–
–
–
(Top view)
(Side view) (Later side view)
There is a magnetic force on sides of length b (in opposite
directions), but not on sides of length a
If loop can pivot about its center, these forces provide a
(clockwise) torque about the pivot point
Maximum torque produced is tmax = BIA (Figs. (a) and (b))
If field makes an angle  with a line perpendicular to the
plane of the loop (Fig. (c)) then t = BIA sin
For any loop shape having N turns: t  NBIA sin 
Application: Electric Motors
• Simplified schematic of a DC electric motor
– DC power source provides
current through conducting loop
– Stationary graphite brushes
maintain electrical contact with
split-ring contacts
(commutators)
– Circuit broken (brushes reach
gap in commutators) when
magnetic field is perpendicular
to conducting loop
– Afterwards, current in loop reverses direction and receives
another pulse of torque in same direction (otherwise loop
would not continue to rotate in same direction)
– Electrical energy converted to rotational kinetic energy
Example Problem #19.28
A rectangular loop consists of 100
closely wrapped turns and has
dimensions 0.40 m by 0.30 m. The loop
is hinged along the y–axis, and the
plane of the coil makes an angle of
30.0° with the x–axis as shown. What is
the magnitude of the torque exerted on
the loop by a uniform magnetic field of
0.80 T directed along the x–axis when
the current in the windings has a value
of 1.2 A in the direction shown? What is
the expected direction of rotation of the
loop?
Solution (details given in class):
10 Nm
Rotation is clockwise as viewed from above
Circulating Charged Particles
• The motion of a charged particle in a uniform
magnetic field directed perpendicular to the initial
particle velocity is circular:

v
= +q

X  F X  v X
F X
F
(magnetic
X
X
X
field into the
page)

v
X

F
X
R

v
X
X


• Since F is perpendicular to v , the particle speed
remains constant
• From Newton’s 2nd Law: F = ma = mv2 / R
• Plugging in for F: F = qvB = mv2 / R
Circulating Charged Particles
• Solving for R, the radius of the circular path:
mv
R
qB
• When a charged particle has velocity components
both perpendicular and parallel to a uniform
magnetic field, particle moves in a helix (no force
parallel to field)
Circulating Charged Particles
• Above Earth’s surface, charged particles from
cosmic rays and the solar wind are trapped by
Earth’s magnetic field
– Particles spiral back and forth along magnetic field lines
• Near the poles, the magnetic field is stronger
– As field strength increases, the radius of a spiraling
particle’s path gets smaller
– Thus there is a concentration of
particles near the poles
• When particles collide with and
ionize air molecules, aurorae can
result
Application: Cyclotrons
• A cyclotron is a charged-particle accelerator with a
circulating beam of particles
• A magnetic field circulates the charged particles
• An alternating potential difference between 2
electrodes (“dees”) accelerates the particles
• The period (or frequency) of one revolution remains
constant as the particle speed
increases
– Alternating potential difference
matches this frequency
• Cyclotrons have several
medical uses
– Produce radioisotopes used in
nuclear medicine at hospitals
– Proton beam radiosurgery
(from College Physics, Giambattista et al.)
Magnetic Fields from Currents: Long Straight
Wire
• From experiments, the magnetic field
lines around a long, straight wire form
circles (top figure)
• The direction of B follows another
right-hand rule (bottom figure)
• The magnitude of B is given by
m0 I
B
2p r
– m0 = constant = permeability of free space
= 4p  10–7 Tm / A
– I = current in the wire
– r = radial distance from the wire
Example Problem #19.51
A wire carries a 7.00-A
current along the x-axis, and
another wire carries a 6.00-A
current along the y-axis as
shown. What is the magnetic
field at point P, located at
x = 4.00 m, y = 3.00 m?
Solution (details given in class):
0.167 mT out of the page (or screen)
Magnetic Force Between 2 Parallel
Conductors
• A magnetic force acts on a current-carrying conductor
when the conductor is in a magnetic field
• Current-carrying conductors create a magnetic field
• Thus, 2 current-carrying conductors placed close
together exert magnetic forces on each other
• Consider 2 wires in close proximity:
– Magnetic field due to wire #2: B2  m0 I 2
2pd
– Magnetic force on wire #1:
m 0 I1 I 2 l
 m0 I 2 
F1  B2 I1l  
 I1l 
or
2pd
 2pd 
F1 m 0 I1 I 2

l
2pd
– Wires with currents in same (opposite)
direction attract (repel) each other
– Newton’s 3rd Law holds (F1 = –F2)
Example Problem #19.58
In the setup shown, the current in
the long, straight wire is I1 = 5.00 A,
and the wire lies in the plane of the
rectangular loop, which carries 10.0
A. The dimensions shown are c =
0.100 m, a = 0.150 m, and l = 0.450 m.
Find the magnitude and direction of
the net force exerted by the
magnetic field due to the straight
wire on the loop.
Solution (details given in class):
–2.70  10–5 N (to the left)
Magnetic Fields from Currents: Current
Loop
• A current loop is just a straight wire curved around to
form a closed loop
– Creates an enhanced magnetic field at the center
– The field is very similar to a bar magnet
– Current loop is referred to as a magnetic dipole
Magnetic Fields from Currents: Solenoids
• A solenoid is a wire bent into a coil of several closely
spaced loops
• When a current passes through the coil, a magnetic
field is produced (“electromagnet”)
• The field inside a solenoid is strong and nearly
uniform, while outside it is weak and nonuniform
– The field inside becomes
stronger with more turns of the
wire
– The larger the number of turns,
and the closer the spacing of
the loops, the more the field
approximates a bar magnet
– Inside solenoid:
B  m0 nI
(n = # turns/unit length of the solenoid)
(loosely wrapped)
(tightly wrapped)
Applications of Solenoids
• TV picture tubes
• Magnetic Resonance
Imaging (MRI) devices
(from College Physics, Giambattista et al.)
Example Problem #19.62
A single-turn square loop of wire 2.00 cm on a side
carries a counterclockwise current of 0.200 A. The
loop is inside a solenoid, with the plane of the loop
perpendicular to the magnetic field of the solenoid.
The solenoid has 30 turns per centimeter and carries
a counterclockwise current of 15.0 A. Find the force
on each side of the loop and the torque acting on the
loop.
Solution (details given in class):
F = 2.26  10–4 N (away from loop interior)
t=0
CQ 3: Interactive Example Problem:
Mapping Magnetic Field Lines and Forces Part 1
For Configuration 1, what could be the source of
the magnetic field?
A) A bar magnet with north (south) end on the
left (right).
B) A bar magnet with north (south) end on the
right (left).
C) A long, straight wire with current into the
page.
D) A long, straight wire with current out of the
page.
(PHYSLET Physics Exploration 27.1, copyright Pearson Prentice Hall, 2004)
CQ 4: Interactive Example Problem:
Mapping Magnetic Field Lines and Forces Part 2
For Configuration 2, what could be the source of
the magnetic field?
A) A bar magnet with north (south) end on the
left (right).
B) A bar magnet with north (south) end on the
right (left).
C) A long, straight wire with current into the
page.
D) A long, straight wire with current out of the
page.
(PHYSLET Physics Exploration 27.1, copyright Pearson Prentice Hall, 2004)