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PHYS 1442 – Section 004
Lecture #17, Review for test 2
Monday March 242014
Dr. Justin Griffiths for Dr. Brandt
Loose ends and review for test 2
Mar 24, 2014
PHYS 1442-004
Inductor
• An electrical circuit always contains some inductance but it is often
negligible
– If a circuit contains a coil of many turns, it could have a large inductance
• A coil that has significant inductance, L, is called an inductor and is
express with the symbol
– Precision resistors are normally wire wound
• Would have both resistance and inductance
• The inductance can be minimized by winding the wire back on itself in opposite
direction to cancel magnetic flux
• This is called a “non-inductive winding”
• For an AC current, the greater the inductance the less the AC current
– An inductor thus acts like a resistor to impede the flow of alternating current (not
to DC, though. Why?)
– The quality of an inductor is indicated by the term reactance or impedance
3/17/2014
PHYS 1442-004, Dr. Andrew Brandt
2
Energy Stored in a Magnetic Field
• When an inductor of inductance L is carrying current
I which is changing at a rate I/ t, energy is
supplied to the inductor at a rate
I
– P  I   IL
t
• What is the work needed to increase the current in an
inductor from 0 to I?
– The work, dW, done in time dt is W  Pt LII
– Thus the total work needed to bring the current from 0 to I
in an inductor is

W
W 

I
3/17/2014
1 2
LI I  LI
2
3
PHYS 1442-004, Dr.
Andrew Brandt
Energy Stored in a Magnetic Field
• The work done to the system is the same as the
energy stored in the inductor when it is carrying
current I
–
1 2
U  LI
2
Energy Stored in a magnetic
field inside an inductor
– This is compared to the energy stored in a capacitor, C,
when the potential difference across it is V U  1 CV 2
2
– Just like the energy stored in a capacitor is considered to
reside in the electric field between its plates
– The energy in an inductor can be considered to be stored
in its magnetic field
3/17/2014
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PHYS 1442-004, Dr.
Andrew Brandt
Stored Energy in terms of B
• So how is the stored energy written in terms of magnetic field
B?
– Inductance of an ideal solenoid without a fringe effect
L  m0 N 2 A l
– The magnetic field in a solenoid is B  m0 NI l
– Thus the energy stored in an inductor is
2
2
1 B2
1 2 1 m0 N 2 A  Bl  1 B
U
Al E
U  LI 

  2 m Al
2 m0
2
2
l  m0 N 
0
– Thus the energy density is
What is this?
2
2
1
B
U U 1B
E density
u
u


Volume V
2 m0
V Al 2 m0
– This formula is valid to any region of space
– If a ferromagnetic material is present, m0 becomes m.
What volume does Al represent?
3/17/2014
The volume inside a solenoid!!
5
PHYS 1442-004, Dr.
Andrew Brandt
Example
Energy stored in a coaxial cable. (a) How much
energy is being stored per unit length in a coaxial cable
whose conductors have radii r1 and r2 and which carry a
current I? (b) Where is the energy density highest?
L m0 r2
(a) The inductance per unit length for a coaxial cable is
ln

l 2 r1
2
2
m
I
r2
Thus the energy stored U 1 LI
0
ln


per unit length is
4
r1
l 2 l
m0 I
(b) Since the magnetic field is B 
2 r
2
1B
And the energy density is u 
2 m0
3/17/2014
The energy density is highest
where B is highest. B is
highest close to r=r1, near the
surface of the inner conductor.
PHYS 1442-004, Dr. Andrew Brandt
6
Why do we care about AC circuits?
• The circuits we’ve learned so far contain resistors, capacitors and
inductors and have been connected to a DC source or a fully charged
capacitor
– What? This does not make sense.
– The inductor does not work as an impedance unless the current is changing. So
an inductor in a circuit with DC source does not make sense.
– Well, actually it does. When does it impede?
• Immediately after the circuit is connected to the source so the current is still changing.
So?
– It causes the change of magnetic flux.
– Now does it make sense?
• Anyhow, learning the responses of resistors, capacitors and inductors in
a circuit connected to an AC emf source is important. Why is this?
– Since most the generators produce sinusoidal current
– Any voltage that varies over time can be expressed in the superposition of sine and
cosine functions
3/17/2014
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PHYS 1442-004, Dr.
Andrew Brandt
AC Circuits – the preamble
• Do you remember how the rms and peak values for
current and voltage are related?
V0
I0
Vrms 
I rms 
2
2
• The symbol for an AC power source is
• We assume that the voltage gives rise to current
I  I 0 sin 2 ft  I 0 sin  t
– where   2 f
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PHYS 1442-004, Dr.
Andrew Brandt
AC Circuit w/ Resistance only
• What do you think will happen when an ac source
is connected to a resistor?
• From Kirchhoff’s loop rule, we obtain
• Thus
V  IR  0
V  I 0 R sin  t  V0 sin  t
– where V0  I 0 R
• What does this mean?
– Current is 0 when voltage is 0 and current is in its
peak when voltage is in its peak.
– Current and voltage are “in phase”
• Energy is lost via the transformation into heat at
an average rate
2
2
P  I V  I rms
R Vrms R
•
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PHYS 1442-004, Dr.
Andrew Brandt
Review ch 20
Magnets, magnetic fields
Force on current carrying wire due to external field
F  Il  B
F  qv  B
  NIAB sin 
Mar 24, 2014
Force on moving charge due to external field
Torque on a current loop
PHYS 1442-004
Example
Electron’s path in a uniform magnetic field. An electron
travels at a speed of 2.0x107m/s in a plane perpendicular to a
0.010-T magnetic field. Describe its path.
v2
What is the formula for the centripetal force? F  ma  m
r
Since the magnetic field is perpendicular to the motion
of the electron, the magnitude of the magnetic force is
Since the magnetic force provides the centripetal force,
we can establish an equation with the two forces
Solving for r
Mar 24, 2014
F  evB
v
F  evB  m
r
mv
9.1  1031 kg    2.0  107 m s 


 1.1  102 m
r
eB
1.6  1019 C    0.010T 
PHYS 1442-004
2
Conceptual Example: Velocity selector
Some electronic devices and experiments
need a beam of charged particles all moving
at nearly the same velocity. This can be
achieved using both a uniform electric field
and a uniform magnetic field, arranged so
they are at right angles to each other.
Particles of charge q pass through slit S1 If
the particles enter with different velocities,
show how this device “selects” a particular
velocity, and determine what this velocity is.
Figure 27-21: A velocity selector: if v = E/B, the particles passing through S1 make it
through S2. Solution: Only the particles whose velocities are such that the magnetic and
electric forces exactly cancel will pass through both slits. We want qE = qvB, so v =
E/B.
Mar 24, 2014
PHYS 1442-004
COULD I ADD GRAVITY TO THIS PROBLEM?
Torque on a Current Loop
• So what would be the magnitude of this
torque?
– What is the magnitude of the force on the
section of the wire with length a?
• Fa=IaB
• The moment arm of the coil is b/2
– So the total torque is the sum of the torques by each of the forces
  IaB b  IaB b  IabB  IAB
2
2
• Where A=ab is the area of the coil
– What is the total net torque if the coil consists of N loops of wire?
  NIAB
– If the coil makes an angle  w/ the field   NIAB sin 
Mar 24, 2014
PHYS 1442-004
Review Chapter 20
m0 I
B
2 r
Magnetic field from long straight wire
F m0 I1I 2

l 2 d
Magnetic force for two parallel wires
 B  dl  m I
0 encl
B  m0 nI
Mar 24, 2014
Ampére’s Law
solenoid
PHYS 1442-004
Ampère’s Law
Example : Field inside and outside a wire.
A long straight cylindrical wire conductor
of radius R carries a current I of uniform
current density in the conductor.
Determine the magnetic field due to this
current at (a) points outside the conductor
(r > R) and (b) points inside the conductor
(r < R). Assume that r, the radial distance
from the axis, is much less than the length
of the wire. (c) If R = 2.0 mm and I = 60
A, what is B at r = 1.0 mm, r = 2.0 mm,
and r = 3.0 mm?
Mar 24, 2014
PHYS 1442-004
Solution: We choose a circular path around
the wire; if the wire is very long the field
will be tangent to the path.
a. The enclosed current is the total current;
this is the same as a thin wire. B = μ0I/2πr.
b. Now only a fraction of the current is
enclosed within the path; if the current
density is uniform the fraction of the current
enclosed is the fraction of area enclosed:
Iencl = Ir2/R2. Substituting and integrating
gives B = μ0Ir/2πR2.
c. 1 mm is inside the wire and 3 mm is
outside; 2 mm is at the surface (so the two
results should be the same). Substitution
gives B = 3.0 x 10-3 T at 1.0 mm, 6.0 x 10-3
T atMar
2.0
mm, and 4.0 x 10-3 T atPHYS
3.01442-004
mm.
24, 2014
Example
Suspending a wire with current. A horizontal wire carries
a current I1=80A DC. A second parallel wire 20cm below it
must carry how much current I2 so that it doesn’t fall due
to the gravity? The lower has a mass of 0.12g per meter
of length.
Which direction is the gravitational force? Downward
This force must be balanced by the magnetic force exerted on the wire by
the first wire. Fg mg FM m0 I1 I 2



l
l
l
2 d
Solving for I2
mg 2 d

I2 
l m0 I1



2 9.8 m s 2  0.12  103 kg   0.20m 
Mar 24, 2014
 4  10
7
PHYS 1442-004

T  m A   80 A 
 15 A
Solenoid Magnetic Field
• Now let’s use Ampere’s law to determine the magnetic field
inside a very long, densely packed solenoid
• Let’s choose the path abcd, far away from the ends
– We can consider four segments of the loop for integral
– B l  m0 Iencl  B lab  B lbc  B lcd  B lda
– The field outside the solenoid is negligible, and the internal field is
perpendicular to the end paths, so these terms also are 0
– Thus Ampere’s law gives us Bl  m NI
0
Mar 24, 2014
PHYS 1442-004
B  m0 nI
Example
Coaxial cable. A coaxial cable is a single wire surrounded by a
cylindrical metallic braid, as shown in the figure. The two
conductors are separated by an insulator. The central wire
carries current to the other end of the cable, and the outer braid
carries the return current and is usually considered ground.
Describe the magnetic field (a) in the space between the
conductors and (b) outside the cable.
(a) The magnetic field between the conductors is the same
as the long, straight wire case since the current in the outer
conductor does not impact the enclosed current.
m0 I
B
2 r
(b) Outside the cable, we can draw a similar circular path, since we
expect the field to have a circular symmetry. What is the sum of the total
current inside the closed path? I encl  I  I  0.
So there is no magnetic field outside a coaxial cable. In other words, the
coaxial cable is self-shielding. The outer conductor also shields against
external electric fields, which could cause noise.
Mar 24, 2014
PHYS 1442-004
Faraday’s Law of Induction
• In terms of magnetic flux
 B  B A  BA cos   B  A
– The emf induced in a circuit is equal to the rate of change
of magnetic flux through the circuit
d B
  N
Faraday’s Law of Induction
dt
• For a single loop of wire N=1, for closely wrapped loops, N
is the number of loops
• The negative sign has to do with the direction of the
induced emf (Lenz’s Law)
Mar 24, 2014
PHYS 1442-004
Lenz’s Law
• It is experimentally found that
– An induced emf gives rise to a current whose magnetic field
opposes the original change in flux  This is known as Lenz’s
Law
– We can use Lenz’s law to explain the following cases :
• When the magnet is moving into the coil
– Since the external flux increases, the field inside the
coil takes the opposite direction to minimize the
change and causes the current to flow clockwise
• When the magnet is moving out
– Since the external flux decreases, the field inside the
coil takes the opposite direction to compensate the
loss, causing the current to flow counter-clockwise
Mar 24, 2014
PHYS 1442-004
How does a transformer work?
• When an AC voltage is applied to the primary, the
changing B it produces will induce voltage of the
same frequency in the secondary wire
• So how would we make the voltage different?
– By varying the number of loops in each coil
– From Faraday’s law, the induced emf in the secondary is
d B
– VS  N S
dt
–The input primary voltage is
d B
– VP  N P
dt
–Since dB/dt is the same, we obtain
– VS N S
Transformer

Mar 24, 2014
Equation
VP N P
PHYS 1442-004