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Transcript
```SOLUTIONS TOPIC 5 , 10 -11 (New) Part 2
Electricity and Magnetism
2. Magnetic Fields Created by Current Carrying Wires
1
2
3. Magnetic force ( Fb) , Magnetic Field ( B) ,Current ( I) velocity of charge (
v) and charge ( q)
PROBLEMS
1. A proton moves vertically upward perpendicularly to a uniform magnetic field
and deflects to the right as you watch it. What is the magnetic field direction?
2. If a negatively charged particle were moving downward along the right edge of
this page , which way should a magnetic field ( known to be perpendicular to
the plane of the paper) be oriented so that the particle would initially be
deflected to the left?
3
AP EXAMPLE
4
PROBLEMS ( cont.)
4. A proton and an electron are moving perpendicularly to a constant magnetic
field that produces a constant magnetic force. Which particle will move faster?
Why ?
F = qvB
5–6:
5
6
7
8
Tsokos # 2 p. 344
NOTE in ( C ) - Correct B is OUT OF PAGE
9
IB examples:
1. A charged particle of mass m and charge q is travelling in a uniform magnetic
field with speed v such that the magnetic force on the particle is F. The
magnetic force on a particle of mass 2m, charge q and speed 2v travelling in
the same direction in the magnetic field is C
A.
4F.
B.
2F.
C.
F.
D.
1
F.
2
(1)
F = qvB. Double v = double in F. Magnetic force independent of mass.
Remember Other Formulas for F:
F = E q ( E – electric field)
F= k q1 q2
r2
2. Note: “experienced BY wire z”
10
SOLUTIONS TOPIC 11 HL ONLY (New)
Electromagnetic Induction
3. Magnetic Flux ( Ф) , Faraday´s Law
PROBLEMS
1. List the unit of magnetic flux :
Tm2 or Wb ( weber)
2. The magnetic flux through a loop can change due to a change in :
a)
b)
c)
d)
the area of the coil
the strength of the magnetic field
the orientation of the loop
all of the preceding
SUMMARY of conditions needed to induce an emf ( voltage and subsequent
current) . Only one of the following needs to be met :
1. The strength of the magnetic field ( B) changes
2. The loop area ( A) changes
3. The angle between the loop and the field changes
Key word: change = fluctuates = flux
11
3. Magnetic Flux ( Ф) , Faraday´s Law
PROBLEMS (cont. )
3. A circular loop with an area of 0.015m2 is in a uniform magnetic field of
0.30 T. What is the flux through the loop’s plane if it is :
a) parallel to the field
ZERO : Cos 90 = 0 or Sin 180 = 0
b) at an angle of 370 to the field
12
Ф = BA cos θ
= (0.30 ) (0.015m2 ) ( cos 37 or sin 53) = 3.6 x 10-3 Tm2 or Wb
c) perpendicular to the field
Ф = BA cos θ
(0.30 ) (0.015m2 ) ( cos 0 or Sin 90 = 1) = 4.5 x 10-3 Tm2 or Wb
IB Example
13
ε= ΔФ
induced emf or voltage (V)
t
States that the induced emf or voltage depends on the change in the magnetic flux
over time.
4. For an induced current to appear in a wire loop:
a) there must be a large magnetic flux through the loop
b) the loop’s plane must be parallel to the magnetic field
c) the loop’s plane must be perpendicular to the magnetic field
d) the magnetic flux through the loop must vary with time
  N
 B
t
key words : with the time rate change in magnetic flux .
14
5. What are the factors that affect the magnitude of the induced emf in a
solenoid?
SUMMARY of conditions needed to induce an emf ( voltage and subsequent
current) . Only one of the following needs to be met :
1. The strength of the magnetic field ( B) changes
2. The loop area ( A) changes
3. The angle between the loop and the field changes
Key word: change = fluctuates = flux
6. The plane of a conductive loop with an area of 0.020m2 is perpendicular to a
uniform magnetic field of 0.30 T. If the field drops to zero in 0.0045s, what is the
magnitude of the average emf induced in the loop? ASSUME N = 1
Ф = BA = (0.30) (0.02m2) = 6 x 10-3 Tm2
assume N = 1
ε = 1 (6 x 10-3 Tm2) = 1.3 V
0.0045s
7. Does the induced emf in a closed loop depend on the value of the
magnetic flux in the loop? Explain
YES!
  N
 B
t
15
MORE PROBLEMS
a) Ф = BA
A = π r2
r = 3cm = 0.03m
Ф = (0.2) (π 0.032) = 5.7 x 10-4 Tm2 = 6 x 10-4 Tm2 ( 1 sig. dig.)
b) Ф = BA cos 45 = 4.0 x 10-4 Tm2
16
IB Example
A square loop of wire of side 0.20 m has a total resistance o 0.60Ω. The loop is
positioned in a uniform magnetic field B of 0 .030 T. The field is directed into the
page, perpendicular to the plane of the loop.
A) Calculate the magnetic flux Φ through the loop.
A = 0.20 x 0.20 = 0.04 m2
Ф = BA
= ( 0.03 T ) ( 0.04m2 )
= 0.0012 T * m2 or Wb ( weber)
The field strength now increases uniformly to 0.20 T in 0.50 s :
B) Calculate the emf (ε) induced in the loop during this period
Be careful: The filed strength increased B increased from 0.030 T to 0.20 T so
you have to calculate the change in magnetic flux Ф = BA then you can
calculate the emf ε :
Δ Ф = Ф2 – Ф1
Ф2 = B2 A = (0.20 T) ( 0.04m2 ) = 0.008 Tm2
Ф1 = B1 A = ( 0.03 T ) ( 0.04m2 ) = 0.0012 T m2
ε= ΔФ
t
= Ф2 – Ф1
t
= 0.008 – 0.0012 = 0.014 V
0.5s
C) Calculate the magnitude of the current ( I)
I = V/R = 0.014V/ 0.60 Ω = 0.023A
17
a) Ф = BA cos 45 = 4.0 x 10-4 Tm2
( From example 14.1 before)
ε= ΔФ
t
ε= ΔФ
t
=
4.0 x 10-4 Tm2
0.5s
=
8 x 10-4 V
b) Counterclockwise looking down from loop
18
4. LENZ´S LAW : Induced field Opposes the External-Original field
IB Examples :
Example 1
Explain which way the current is directed and why:
I is clockwise because it needs to generate a downward induced magnetic field.
This is to oppose any change in magnetic flux due to upward external magnetic field.
19
Example 2
Explain which way the current is directed and why:
I is counter clockwise because it needs to generate an
upward induced magnetic field. This is to oppose any
change in magnetic flux due to downward external
magnetic field.
Example 3
I is counter clockwise because it needs to generate an upward induced magnetic
field. This is to oppose any change in magnetic flux due to downward external
magnetic field
20
Example 4
a) The south end of a bar magnet is pulled away from a tiny wire coil see fig. 1
below. Looking from behind the coil, which way is the induced current?
counter clockwise.
b) Suppose that the magnetic field across the area of the coil is initially 0.040T, the
coil’s radius is 2.0 mm, and there are 100 loops in the coil. Determine the
magnitude of the average induced emf in the coil if the bar magnet is removed in
0.75 s. Assume that the magnetic field is constant over the coil area.
Fig.1 Looking from behind the coil.
The magnetic field is INTO the page.
To answer a) above show the direction of the current in the loop using arrows and
state whether the direction is clockwise or counter clockwise.
ε=NΔФ
t
=
(100) (0.040 T) ( π x 0.0022)m2
0.75s
=
6.7 x 10-5 V
21
5. ε = Bvl
The induced emf can also be calculated knowing the magnetic field (B) , width of the
loop (l) OR length of rod (l) , and the velocity at which the loop OR ROD changes
motion ( v) .
ε = BLv
= 0.40 x 0.20 x 0.60 = 0.048 V
22
23
IB Example
24
f)
ε = BLv
Bv = ε = 2.4 x 10-3V = 0.015 Vm-1
L
0.16 m
F = qvB = ( 1.6 x 10-19C) ( 0.015Vm-1) = 2.4 x 10-21 N
Example ( Wilson- Buffa)
A uniform magnetic field of 0.25 T is established in the figure below. If the bar is 20
cm long and is pulled at a steady speed of 10 cms-1 , what would be the induced
current if the resistor has a value of 5.0 Ω ?
ε = BLv
= 0.25 T x 0.20m x 0.10 ms-1 = 5 x 10-3 V ( 1 sig. dig.)
Ohms’ law V= RI
ε=V
I = ε / R = 5 x 10-3 V = 1 x 10-3 A ( 1 sig. dig.)
5.0 Ω
25
TOPIC 11.2 HL ONLY: Alternating Current
Root –mean –square (rms) values for AC
Example - Find the rms values of current and voltage from fig. 8.3 and 8.4 below.
Assume the resistance in the circuits is 16 Ω.
Fig. 8.3
Answers : Given R = 16 Ω
From graph : V0 = Peak voltage = 40.0 V
I0 = V0 / R = 40 / 16 = 2.5 A
Irms = I0 = 1.8 A
√2
V0 = Peak voltage = 40.0 V .
Vms = V0
√2
also called εrms
Vms = 28 V
26
Fig. 8.4
Answers : Given R = 16 Ω
From graph : I0 = current = 2.5 A
Irms = I0 = 1.8 A
√2
V0 = Peak voltage = = RI = 16 x 2.5 = 40.0 V
Vms = V0
√2
also called εrms
Vms = 28 V
27
Example
Each channel of a stereo receiver is capable of a power output of 100.0 W into an
8 .0 Ω resistor.
A) Calculate the peak voltage and current.
P = V2
R
V0 = 28 V ( 2 sig. dig. )
P = I2 R
I0 = 3.5 A ( 2 sig. dig. )
B) Calculate the Vrms and Irms .
Vms = V0
√2
also called εrms
Vms = 19. 8 V
Irms = I0 = 2.5 A
√2
28
a) Peak power = 20.0 W
P = I2 R
I0 = √20 =
2.5
Irms = I0
√2
=
2.828
2.0 A
b) Peak power = 20.0 W
P = V2
R
V0 = √20.0 x 2.5 = 7.07
V rms = V0
√2
= 5.0 V
29
Transformers
MISCELLANEOUS EXAMPLES:
1.This question is about an ideal transformer.
(a)
State Faraday’s law of electromagnetic induction. (2)
(b)
The diagram below shows an ideal transformer.
laminated core
primary coil
secondary coil
(i)
Use Faraday’s law to explain why, for normal operation of the
transformer, the current in the primary coil must vary
continuously. (2)
(ii)
Outline why the core is laminated. (2)
(iii) The primary coil of an ideal transformer is connected to an
alternating supply rated at 230V. The transformer is designed to
provide power for a lamp rated as 12V, 42W and has 450 turns of
wire on its secondary coil. Determine the number of turns of wire
on the primary coil and the current from the supply for the lamp
to operate at normal brightness. (3)
Solution
(a) e.m.f. induced proportional to/equal to;
rate of change of flux (linkage) / rate of flux cutting;
(b)
(i)
(ii)
2
for e.m.f./current to be induced in secondary, flux must be
changing in the core;
changing flux is caused by varying current in primary;
2
induced currents in core are kept small; (do not allow reduced/
prevented) to reduce heating/energy losses;
(do not allow mere “eddy current losses”)
2
(iii) use of
N S VS

N P VP
;
to give NP = 8600 turns;
and
 42 
IP 

 230 
= 180 mA;
3
30
2.
3.
31
4.
This question is about the emf induced in a coil.
(a)
Define magnetic flux.
the product of the magnetic field strength and area through which it passes
 = BAcosθ
(b)
A coil is rotated at constant speed in a region of uniform magnetic field.
The graph shows the variation with time t of the emf ε induced in the coil for
one cycle of rotation.
32
(i)
On the graph label, with the letter T, a time at which the flux linkage in
the coil is a maximum.
(1)
(ii)
Use the graph to determine the rate of change of flux at t = 4.0 ms.
From graph emf at 4.0ms = 2.0 V /
emf equals rate o change of flux =
2
ε= ΔФ
t
(iii)
Calculate the root mean square value of the induced emf.
From graph V0 = 6.0 V
V rms = V0
√2
= 4.2 V
33
5.
This question is about a generator.
(a)
Define electromotive force.
electromotive force = voltage = W/q = work done per unit charge
(b)
The graph shows the variation with time of electromotive force (emf)
for a generator.
(i)
Calculate the rms value of the emf of the generator.
From graph : peak reading = 1400 ± 50 V
  990 V
2


Vrms = 
34
(ii)
The speed of rotation of the generator is halved with no other changes
On the graph, sketch the variation of emf with time.
Voltage is directly proportional to speed of rotation. Speed is halved so
voltage is halved giving peak of 700V: 1400v/ 2 = 700 V
(iii) Explain why the graph you drew in (ii) is different from the original
graph.
Voltage is directly proportional to speed of rotation. Speed is halved so
voltage is halved giving peak of 700V: 1400v/ 2 = 700 V
35
TOPIC 11.3 HL ONLY: Capacitance
36
37
3. Capacitance of a Parallel Plate Capacitor, Permittivity ε
Example Questions:
1. A dielectric material is inserted between the plates of a fully charged capacitor.
State the effect, if any, on:
a) The potential difference across the capacitor
It increases the overall potential of the capacitor by increasing the potential across
the negative end of the plate and decreasing the potential across the positive end of
the plate
b) The charge on one of the capacitor plates
The charge on the negative end decreases and the charge on th positive end
increases
38
2. A capacitor is connected to a cell of emf 12 V. The capacitor has a capacitance of
4 F. The permittivity of a dielectric material that is inserted between the plates of
the capacitor is twice that of a vacuum.
The capacitor is fully charged.
a) Calculate the capacitance of the capacitor with the dielectric material in it.
C doubles so C = 8 F = 8x 10 -6 F
b) Calculate the energy stored in the capacitor with the dielectric material in it.
E = ½ CV2
E = 6.5 x 10-4 J
4.Combining Capacitors in Parallel and Series
Worked Example: parallel
Worked Example: series
39
2.
40
6. Discharging a Capacitor – Details: The Math and Equations
41
42
Example 3 :
43
7. Charging a Capacitor
IB Example 1
44
IB Example 2
45
46
```