Physics 2113
Jonathan Dowling
Lecture 33: WED 11 NOV
Review Session B: Midterm 3
EXAM 03: 6PM WED 11 NOV in Cox Auditorium
The exam will cover: Ch.27.4 through Ch.30
The exam will be based on: HW08–11
The formula sheet and practice exams are here:
http://www.phys.lsu.edu/~jdowling/PHYS21133-FA15/lectures/index.html
Note this link also includes tutorial videos on the various right hand rules.
You can see examples of even older exam IIIs here:
http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys2102/Phys2102OldTests/
The Biot-Savart Law
For B-Fields
Jean-Baptiste
Biot (1774-1862)
Felix Savart
(1791-1841)
• Quantitative rule for computing
the magnetic field from any
electric current
• Choose a differential element of
wire of length dL and carrying a
current i
• The field dB from this element at
a point located by the vector r is
given by the Biot-Savart Law
 0 =4×10–7 T•m/A
(permeability constant of
free space)
i
Ä
Field of a Straight Wire
m0 ids (r sin q )
dB =
3
4p
r
p -q
q -p /2
sinq = cos(q - p / 2) = R / r
¥
r = (s 2 + R 2 )1/ 2
¥
Rds
m0i ds (r sin q ) m0i
=
B=
ò
3
ò
2
2 3/ 2
4
p
4p -¥
r
- ¥ (s + R )
¥
m 0i
Rds
=
2p ò0 (s 2 + R 2 )3 / 2
¥
ù
m0iR é
s
=
ê 2 2
1/ 2 ú
2
2p êë R (s + R ) úû
0
m 0i
=
2pR
Biot-Savart Law
• A circular arc of wire of radius R
carries a current i.
• What is the magnetic field at the
center of the loop?
i
Direction of B?? Not another
right hand rule?!
m0 idsR m 0 iRdf
dB =
=
3
2
4p R
4p R
m0 idf m0iF
B=
=
ò
4p R
4p R
TWO right hand rules!:
If your thumb points along the
CURRENT, your fingers will point
in the same direction as the
FIELD.
If you curl our fingers around
direction of CURRENT, your
thumb points along FIELD!
Example, Magnetic field at the center of a circular arc of a circle pg 768:
ICPP: What is the direction of the B
field at point C?
(a) Into the board Ä ? ✔
(b) Out of the board ⊙ ?
(c) To the right ® ?
(d) To the left ¬ ?
Ä
Example, Magnetic field at the center of a circular arc of a circle.:
Right Hand Rule & Biot-Savart: Given i Find B
A length of wire is formed into a closed circuit with radii a and b, as
shown in the Figure, and carries a current i.
(a) What are the magnitude and direction of B at point P?
m 0 if
B=
4pR
(b) Find the magnetic dipole moment of the circuit.
 =NiA
Forces Between Wires
Magnetic field due to wire 1
where the wire 2 is,
L I1
I2
F
m 0 I1
B1 =
2p d
Force on wire 2 due to this field,
F21 = L I 2 B1
d
eHarmony’s Rule for Currents:
Same Currents – Attract!
Opposite Currents – Repel!
m 0 L I1 I 2
=
2p d
29.3: Force Between Two Parallel Wires:
F21 =
m0 Li1 i2
= Force on 2 due to 1.
2p d
Like currents attract & Opposites Repel.
1/r Þ double the distance halve the force.
⊙
⊙
⊙
⊙
⊙
⊙
a
a
a
Fbnet > Fcnet > Fanet
Ä
b
c
b
b
Ä
c
Ä
c
29.3.3. The drawing represents a device called Roget’s Spiral. A coil of wire hangs vertically and
its windings are parallel to one another. One end of the coil is connected by a wire to a
terminal of a battery. The other end of the coil is slightly submerged below the surface of a
cup of mercury. Mercury is a liquid metal at room temperature. The bottom of the cup is also
metallic and connected by a wire to a switch. A wire from the switch to the battery completes
the circuit. What is the behavior of this circuit after the switch is closed?
a) When current flows in the circuit, the coils of
the wire move apart and the wire is extended
further into the mercury.
b) Nothing happens to the coil because there will
not be a current in this circuit.
c) A current passes through the circuit until all of the
mercury is boiled away.
d) When current flows in the circuit, the coils of the
wire move together, causing the circuit to break at the surface of the mercury. The coil then
extends and the process begins again when the circuit is once again complete.
29.3: Force Between Two Parallel Wires, Rail Gun:
Ampere’s law: Closed
Loops
i4
The circulation of B
(the integral of B scalar
ds) along an imaginary
closed loop is
proportional to the net
amount of current
traversing the loop.
i2
i3
i1
ds
Thumb rule for sign; ignore i4
If you have a lot of symmetry, knowing the circulation of B
allows you to know B.
Ampere’s law: Closed
Loops
The circulation of B
(the integral of B scalar
ds) along an imaginary
closed loop is
proportional to the net
amount of current
piercing the loop.
Thumb rule for sign; ignore i3
If you have a lot of symmetry, knowing the circulation of B
allows you to know B.
Calculation of ienc. Wecurl the fingers
of the right hand in the direction in which
theAmperian loop was traversed. We note
the direction of thethumb.
All currents inside the loop parallel to the thumb are counted as positive.
All currents inside the loop antiparallel to the thumb are counted as negative.
All currents outside the loop are not counted.
In this example :
ienc = i1 - i2 .
(a) - i + i + i = i
(b) - i + 0 + i = 0
(c) - i + 0 + 0 = -i
(d) 0 + i + i = 2i
d >a=c>b=0
Ampere’s Law: Example 2
• Infinitely long cylindrical
wire of finite radius R
carries a total current i
with uniform current
density
• Compute the magnetic
field at a distance r from
cylinder axis for:
r < a (inside the wire)
r > a (outside the wire)
Current out
of page,
circular field
lines
i
Ampere’s Law: Example 2
(cont)
Current out of
page, field
tangent to the
closed
amperian loop
B(2pr ) = m0ienclosed
m 0ienclosed
Need Current Density J!
B=
2pr
2
i
r
2
2
ienclosed = J (pr ) = 2 pr = i 2
pR
R
m 0ir For r < R
For r>R, i =i, so
B=
2
B= i/2R = LONG
2pR
WIRE!
enc
0
B-Field In/Out Wire: J is
Constant
B
m0i
2p R
O
R
r
r<R r>R
B µ r B µ1/ r
r<R
m0ir
B=
2
2p R
r>R
m0i
B=
2p r
Outside Long Wire!
3Ä
+5 ⊙
-9 Ä
+4 ⊙
3Ä
+5 ⊙
-9 Ä
+4 ⊙
ra = 1m
B
Ba µ 4 /1 = 4 = 16 / 4
Bb µ 4 - 9 / 2 = 5 / 2 = 10 / 4
Bc µ 4 - 9 + 5 / 3 = 0
Bd µ 4 - 9 + 5 - 3 / 4 = 3 / 4
m0ienclosed
B=
2p r
1/r Law too
Ba > Bb > Bd > Bc = 0
29.4.1. A copper cylinder has an outer radius 2R and an inner radius of R and carries
a current i. Which one of the following statements concerning the magnetic
field in the hollow region of the cylinder is true?
a) The magnetic field within the hollow region may be represented as concentric
circles with the direction of the field being the same as that outside the cylinder.
b) The magnetic field within the hollow region may be represented as concentric
circles with the direction of the field being the opposite as that outside the
cylinder.
c) The magnetic field within the hollow region is parallel to the axis of the cylinder
and is directed in the same direction as the current.
d) The magnetic field within the hollow region is parallel to the axis of the cylinder
and is directed in the opposite direction as the current.
e) The magnetic field within the hollow region is equal to zero tesla.
29.4.3. The drawing shows two long, straight wires that are parallel to each other and carry a
current of magnitude i toward you. The wires are separated by a distance d; and the
centers of the wires are a distance d from the y axis. Which one of the following
expressions correctly gives the magnitude of the total magnetic field at the origin of the
x, y coordinate system?
a)
b)
0i
2d
 0i
2d
c)
0i
2 d
d)
0i
d
e)
zero tesla
29.5: Solenoids:
Fig. 29-19 Application of Ampere’s law to a section of
a long ideal solenoid carrying a current i. The Amperian
loop is the rectangle abcda.
Here n be the number of turns per unit length of the solenoid
29.5.5. A solenoid carries current I as shown in the figure. If the
observer could “see” the magnetic field inside the solenoid, how
would it appear?
29.5.1. The drawing shows a rectangular current carrying wire loop that has one side
passing through the center of a solenoid. Which one of the following statements
describes the force, if any, that acts on the rectangular loop when a current is
passing through the solenoid.
®
a) The magnetic force causes the loop
to move upward.
i
b) The magnetic force causes the loop
to move downward.
c) The magnetic force causes the loop
to move to the right.
d) The magnetic force causes the loop to move to the left.
e) The loop is not affected by the current passing through the solenoid or the
magnetic field resulting from it.
Magnetic Field of a Magnetic Dipole
A circular loop or a coil currying electrical current is a
magnetic dipole, with magnetic dipole moment of
magnitude  =NiA. Since the coil curries a current, it
produces a magnetic field, that can be calculated using
Biot-Savart’s law:
All loops in the figure have radius r or 2r. Which of
these arrangements produce the largest magnetic
field at the point indicated?
29.6: A Current Carrying Coil as a Magnetic Dipole:
For small z << R
B(z) =
m0iR
2R 3 (1+ z 2 / R 2 )
3/2
@
m0 i
2R 2
1 / R 2 Law! Double the R one fourth the field.
29.6: A Current Carrying Coil as a Magnetic Dipole:
3
It's a 1/r law!!!
Lenz’s Law
Induction and Inductance
• Faraday’s law: E = - dF B
dt
or
• Inductance: L=N/i
– For a solenoid: L= 0n2Al= 0N2A/l
• Inductor EMF: EL= -L di/dt
• RL circuits: i(t)=(E/R)(1–e–tR/L) or i(t)=i0e–tR/L
• RL Time Constant:  = L/R
Units: [s]
• Magnetic energy: U=Li2/2 Units: [J]
• Magnetic energy density: u=B2/20
Units:
[J/m3]
i
Summary
Two versions of Faradays’ law:
– A time varying magnetic flux produces an EMF:
dFB
E =dt
–A time varying magnetic flux produces an electric
field:
Changing B-Flux Induces EMF
Flux Up
RL Circuits
Flux Down
tL = L / R
E
iup ( t ) = (1- e-t /t L )
R
1 2
U B ( t ) = Li
2
1 E2
-t /t L 2
= L 2 (1- e
)
2 R
To find E L walk the loop : +E + VR + E L = 0
E L = -E - VR = -E - ( -iR )
éE
ù
= -E + ê (1- e-t /t L ) ú R
ëR
û
E L = -E e-t /t L
E -tR/L
idn ( t ) = e
R
1 2
U B ( t ) = Li
2
2
1 E -2tR/L
= L 2e
2 R
Walk the loop!
E L = -VR = -iR
E L = -E e-t /t L
RL Circuits
E/2
t=?
Make or Break?
At t=0 all L’s make breaks so throw out all
loops with a break and solve circuit.
At t=∞ all L’s make solid wires so replace L’s
with wire and solve circuit.
Checkpoints/Questions
Magnitude/direction of induced current?
Magnitude/direction of magnetic field inducing current?
Magnitude of induced emf/current?
Given |∫E∙ds| , direction of
magnetic field?
Given B, dB/dt, magnitude of
electric field?
Current inducing EL?
Largest L?
Current through the battery?
Time for current to rise 50% of max value?
Largest current?
R,L or
2R,L or
R, 2L or
2R,2L?
dF B
dB
EEMF =
=A
dt
dt
EMF is proportional to the slope dB/dt
Eb > Ed = Ee > Ea = Ec = 0
I a = Ib > I c = 0
30.4.1. Consider the situation shown. A triangular, aluminum loop is
slowly moving to the right. Eventually, it will enter and pass
through the uniform magnetic field region represented by the tails
of arrows directed away from you. Initially, there is no current in
the loop. When the loop is entering the magnetic field, what will
be the direction of any induced current present in the loop?
a) clockwise
b) counterclockwise
c) No current is induced.
30.4.3. A rigid, circular metal loop begins at rest in a uniform
magnetic field directed away from you as shown. The loop is then
pulled through the field toward the right, but does not exit the
field. What is the direction of any induced current within the
loop?
a) clockwise
b) counterclockwise
c) No current is induced.
ICPP
• 3 loops are shown.
B
II
• B = 0 everywhere except in
III
I
the circular region I where B
is uniform, pointing out of
the page and is increasing
at a steady rate.
• III encloses no flux so EMF=0
• Rank the 3 loops in order of • I and II enclose same flux so
EMF same.
increasing induced EMF.
• Are Currents in Loops I & II
– (a) III < II < I ?
Clockwise or Counterclockwise?
– (b) III < II = I ?
– (c) III = II = I ?
d =c>b=a
dF
di
E == -L
dt
dt
If i is constant (in time) then E = 0
a?
b?
c?
e?
d?
f?
The RL circuit
•
•
•
Set up a single loop series circuit
with a battery, a resistor, a solenoid
and a switch.
Describe what happens when the
switch is closed.
Key processes to understand:
– What happens JUST AFTER
the switch is closed?
– What happens a LONG TIME
after switch has been closed?
– What happens in between?
Key insights:
• You cannot change the CURRENT
in an inductor instantaneously!
• If you wait long enough, the current
in an RL circuit stops changing!
At t = 0, a capacitor acts like a solid wire and inductor acts like break in the wire.
At t = ∞ a capacitor acts like a break in the wire and inductor acts like a solid wire.
ICPP
Immediately after the
switch is closed, what is
the potential difference
across the inductor?
(a) 0 V
(b) 9 V
(c) 0.9 V
10 
9V
10 H
At t = 0 when you walk the loop
since i(0) = 0 it's like VR = iR is not even there!
E - VL = 0
VL = E = 9.0V
• Immediately after the switch, current in circuit = 0.
• So, potential difference across the resistor = 0!
• So, the potential difference across the inductor = E = 9 V!
Fluxing Up The Inductor
• How does the current in
the circuit change with
time?
i
di
-iR + E - L = 0
dt
E
-t /t
i = (1- e )
R
i(t)
Fast = Small 
E/R
Slow= Large 
Time constant of RL circuit: LR = L/R
t
Fluxing Down an Inductor
The switch is at a for a long time, until the
inductor is charged. Then, the switch is
closed to b.
i
What is the current in the circuit?
Loop rule around the new circuit walking
counter clockwise:
di
iR + L = 0
dt
E - Rt /L E -t /t
i= e
= e
R
R
i(t)
Exponential defluxing
E/R
t RL = L / R
t
When the switch is closed, the
inductor begins to get fluxed up,
and the current is
i=(E/R)(1-e–t/τ).
When the switch is opened, the
inductors begins to deflux.
The current in
this case is then
i= (E/R) e–t/τ
Example, RL circuit, immediately after switching and after a long time: SP30.05