Download Topic 5 , 10-11 New Selected Problems 2 - Solutions

Topic 5 , 10-11 New Selected Problems 2 - Solutions
1. a) If the distance from a point charge is doubled the magnitude of the electric
field will _______________________________.
b) The electric field of a point charge at distance r is 1.0 x 10-4 NC-1 . If the distance
r is reduced to one- third , what is the magnitude of the new electric field?
2) An isolated electron is acted on by an electric force of 3.2 x 10-14 N. What is the
magnitude of the electric field at the electron’s location?
Answers :
1. a) If the distance from a point charge is doubled the magnitude of the electric
field will
decrease by one fourth
E=k Q
r2
 1
(2)2
=1
4
b) The electric field of a point charge at distance r is 1.0 x 10-4 NC-1 . If the
distance r is reduced to one- third , what is the magnitude of the new electric
field?
E=
1 =
r2
1
. =
(1/3)2
9
so it is 9 times original field = 9.0 x 10-4 NC-1
1
2. An isolated electron is acted on by an electric force of 3.2 x 10-14 N. What is the
magnitude of the electric field at the electron’s location?
Answer :
no distance given so use E = F
q
electron = 1.6 x 10-19 C
E = 3.2 x 10-14 N
1.6 x 10-19 C
= 2.0 x 105 NC-1
( 2 sig. dig.)
3. There are two separate series circuits using a 12 V battery and one resistor. Which
circuit consumes more power : a circuit with a 5.0 -Ω resistor or a 10 -Ω resistor ?
Explain.
P is proportional to current. The 5.0 -Ω resistor circuit will have less resistance and
therefore more current and consume more power.
Can Calculate using both formulas :
P = RI 2 ( need to calculate I from V = RI )
P = I2 R ( 5 ohm resistor)
calculate current first :
P = V2
R
I = V/R = 12V/ 5 = 2.4 A
P = ( 2.4)2 x 5 = 28.8 W
P = I2 R ( 10 ohm resistor)
calculate current first :
I = V/R = 12V/ 10 = 1.2 A
P = ( 1.2)2 x 5 = 14.4 W
P = V2
R
= ( 12) 2 = 28.8 W
5
P = V2
R
= ( 12) 2 = 14.4W
10
power is inversely proportional to R
2
4.
In the circuit below, the voltmeter has a resistance 100 k. The battery has
negligible internal resistance and emf 6 V.
The reading on the voltmeter is B
A.
0 V.
B.
2 V.
C.
3 V.
D.
4 V.
3
5. OMITTED
4
6.
5
6Answer C
Direction is from Y  X. Y is + and X is of + charge.
. Direction follows movement
In the case of parallel plates the electric field is determined by E = V/ d where d is
the distance between the plates. The electric field is uniform and has the same
value at all points between the plates. Its direction is from high potential to low
potential.
fig. 2.15 p. 294
Work done to move a charge
is not uniform and depends
on distance the charge
moves : W = Eqd = qV
7. OMITTED
6
8.
7
8.
8
Ohm’s law Review
Normal Ohm’s law – ohmic resistor
Filament = NON OHMIC resistor that heats up and this heat will increase the resistance
exponentially NOT linearly as in normal ohm’s law. This increase in resistance will
cause a DECREASE in current ( I ) as V increases.
V
___________________ I
Note if the graph is inverted to I vs V it will look like this
I
V
9
Ohm’s Law Lab Review
Variable resistor also called a potentiometer. It varies the resistance and thus controls the
current. Make sure you can relate the lab equipment to the circuit diagram below it. Be
able to label different parts of the circuit.
Be able to label the position of the potentiometer to give you full resistance and full
voltage drop or zero resistance and zero voltage drop. The more of the potentiometer you
use the more the resistance and the more the voltage drop.
If the arrow coming from the ammeter is placed closest to the positive pole of the
potentiometer then you will get zero resistance and zero voltage drop.
If the arrow coming from the ammeter is placed furthest from the positive pole of the
potentiometer then you will get full resistance and full voltage drop.
10
X
11
9-12 HL only
9. A circular coil of wire of radius r is placed in a uniform magnetic
field of flux density B. The angle between the plane of the coil and
the magnetic field is θ.
B
r
The magnetic flux linking the coil is B
A.
πr2B.
B.
πr2B sin θ.
C.
πr2B θ.
D.
πr2
NOTES REVIEW
Ф = BA
Area ( A) = πr2
Magnetic Flux ( Ф) , Faraday´s Law
Ф = BA
Since the induced emf or voltage depends on magnetic flux or field lines, magnetic flux
(Ф) is defined as the number of field lines in a magnetic field ( B) that pass through a
loop of area (A) :
Ф = BA
units:
magnetic field B = T ( telsa
Area (A )
= m2
Magnetic flux (Ф ) = BA = T m2 ( weber Wb)
12
MAGNETIC FLUX AT DIFFERENT ANGLES
Figs. 20. 4 b, c d and e below show how the loop angle ( θ) with respect to the magnetic
field (B) can change the value of the magnetic flux ( Ф) .
Note: Magnetic flux is MAX at θ = 0 or 180…perpendicular formation Fig.b,c
Magnetic flux is ZERO at θ = 90 …..parallel formation
13
10. The magnetic flux Φ through a coil having 500 turns varies with
time t as shown below.
2.5
2.0
1.5
–3
/ 10 Wb
1.0
0.5
0.0
0
1
3
2
–3
t / 10 s
4
5
The magnitude of the emf induced in the coil is C
A.
0.25 V.
B.
0.50 V.
C.
250 V.
D.
1 000 V.
FARADAY`S LAW and N :
ε=NΔФ
t
ε=NΔФ
t
= 500 x 2 x 10-3 Wb
4 x 10-3 s
induced emf or voltage (V) with
# of loops N
= 250V
note can choose any 2 values for Ф and t off straight line
14
Notes Review
FARADAY’S LAW
ε= ΔФ
t
induced emf or voltage (V)
With these definitions for flux and flux linkage, we can now write the results of
Faraday’s investigations.
  N
 B
t
(Faraday's Law)
FARADAY`S LAW and N :
ε=NΔФ
t
induced emf or voltage (V) with
# of loops N
States that the induced emf also depends on the number of loops (N) in a coil along with
the time rate change in magnetic flux .
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
15
11. This question is about electrical energy and associated phenomena.
electromagnet
The current in the circuit is switched on.
(i)
a) State Faraday’s law of electromagnetic induction and
b) use the law to explain why an emf is induced in the coil of the electromagnet.
Answer a) : (induced) e.m.f. proportional to rate of change of magnetic
flux (linkage);
FARADAY’S LAW
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.
Answer b) : as current increases, magnetic field in coil increases and thus change in
flux and e.m.f. is induced
NOTES REVIEW
States that the induced emf also depends on the number of loops (N) in a coil along with
the time rate change in magnetic flux .
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
16
(ii)
State Lenz’s law and use the law to predict the direction of the induced emf in
(i).
Answer: induced e.m.f. must oppose e.m.f. of battery
LENZ´S LAW : Induced field Opposes the External-Original
field to resist any change in magnetic flux
Due to conservation of energy and action – reaction forces :
LENZ´S LAW states that when a magnetic field induces an emf in a loop (called the
external - original field) ; the loop in turn, generates a secondary magnetic field (called
the induced field )that is OPPOSITE in direction of the original magnetic field that
caused the induction:
Newton´s 3rd Law:
Axn - Rxn
Same as Rt. Hand rule except 2 magnetic fields are involved and you have to find
the INDUCED Field fig. B above NOT THE EXTERNAL FIELD fig. a above.
17
12.
18
12. Answer :
(i)
Calculate the rms value of the emf of the generator.
From graph : peak reading = 1400 ± 50 V
 peak reading 
  990 V
2


Vrms = 
(ii)
The speed of rotation of the generator is halved with no other changes
being made.
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
19
13.
Two long, vertical wires X and Y carry currents in the same direction and
pass through a horizontal sheet of card.
X
Y
Iron filings are scattered on the card. Which one of the following diagrams
best shows the pattern formed by the iron filings? (The dots show where the
wires X and Y enter the card.)
A.
B.
C.
D.
Answer = A C is close but does not show circular magnetic field produced by
straight wire.
20