TOPIC 5, 10-11 New ELECTROMAGNETISM - SELECTED PROBLEMS 1 SOLUTIONS
Answer C
Know Your Electric Fields
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Example 1: Electric field of a negative charge
All vectors point toward the center of the negative charge ( - q) because a positive test charge
(+q ) would move in this direction due to attraction.
Example 2 : Electric field of a positive charge
All vectors point away from the center because a positive test charge will move in this direction
due to repulsion.
Example 3: Electric Field between opposite charges.
2
Example 4 : Electric field between like charges.
Example 5 : Opposite and UNEQUAL Charges
Strength of an Electric Filed based on field line:
The relative strength of an electric field is determined by the spacing or density of the filed lines :
Stronger Field
Weaker Field
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2. The graph below shows the current/voltage characteristics of a filament
lamp. A
24
16
–3
Current / × 10 A
8
0
0
2
6
4
Voltage / V
8
10
The resistance of the filament at 4.0 V is
A.
250 Ω.
B.
4 000 Ω.
C.
8 000 Ω.
D.
64 000 Ω.
SOLUTION : R = V/ I = 4 / 16 x 10-3 = 250
4
3. This question is about forces on charged particles in electric and magnetic
fields.
The diagram shows two parallel plates situated in a vacuum. One plate is at a positive
potential with respect to the other.
+
Path of positively charged particle
–
A positively charged particle passes into the region between the plates. Initially, the particle
is travelling parallel to the plates.
(a)
On the diagram,
(i)
draw lines to represent the electric field between the plates.
(ii)
show the path of the charged particle as it passes between, and beyond, the
plates.
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(b)
An electron is accelerated from rest in a vacuum through a potential difference of 750
V.
(i)
Determine the change in electric potential energy of the electron.
W = ∆ Ue = q V
q of electron = 1.6 ×10–19
= 1.6 ×10–19 × 750
= 1.2 × 10–16 J
(ii)
Deduce that the final speed of the electron is 1.6 × 107 m s–1.
Kinetic Energy = Electric Potential Energy = 1.2 × 10–16
mass of electron = 9.1 × 10–31
½mv2 = 1.2 × 10–16
Solve for v:
½ × 9.1 × 10–31 × v2 = 1.2 ×10–16
v2 = 2.64 × 1014
to give v = 1.67 × 107 m s–1;
6
4. A cell of emf E and internal resistance r is connected to a variable resistor.
A voltmeter is connected so as to measure the potential difference across
the terminals of the cell. Which one of the following is the correct circuit
diagram of the arrangement? B
A.
B.
E
r
E
r
V
V
C.
r
E
V
D.
r
E
V
(1)
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8
5. This question is about electrical energy and associated phenomena.
Current electricity
A cell of electromotive force (emf) E and internal resistance r is connected in series with a
resistor R, as shown below.
E
r
R
The cell supplies 8.1 × 103 J of energy when 5.8 × 103 C of charge moves completely round
the circuit. The current in the circuit is constant.
(i)
Calculate the emf E of the cell.
Δ E/ q = energy / charge
V = e.m.f. =
=
(8.1  10 3 )
(5.8  10 3 )
= 1.4 V
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(ii)
The resistor R has resistance 6.0 Ω. The potential difference between its terminals is
1.2 V. Determine the internal resistance r of the cell.
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(iii)
Calculate the total energy transfer in the resistor R.
3
Total energy transfer = 8.1 × 10 J given and is proportional to resistance. Total resistance of circuit is 7 Ω.
R = 6 Ω so this resistor gets
3
3
X = 6/7 x 8.1 × 10 J = 6.9 x 10 J
(iv) Describe, in terms of a simple model of electrical conduction, the mechanism by
which the energy transfer in the resistor R takes place.
Current = electrons moving = kinetic energy which is transferred into the resistor and due to
frictioin is converted to thermal energy = resistor gets hot and temp increases.
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6.
(ii) Calculate the power dissipated in the external circuit.
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(iii) Determine the resistance of the resistor R.
13
7. A battery is connected in series with a resistor R. The battery transfers 2
000 C of charge completely round the circuit. During this process, 2 500
J of energy is dissipated in the resistor R and 1 500 J is expended in the
battery. The emf of the battery is A
A.
2.00 V.
B.
1.25 V.
C.
0.75 V.
D.
0.50 V.
Emf = V = Δ E/q = W/q = total energy/ charge = 40000J/ 2000C = 2.00 V
8. In the circuit below, which meter is not correctly connected? D
A
1
3
V
2
A
4
V
A.
1
B.
2
C.
3
D.
4
Voltmeter 4 must be placed ACROSS the resistor NOT IN SERIES.
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15
9. 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 B
A.
4F.
B.
2F.
C.
F.
D.
1
F.
2
Fb = qvB
F = ma
(1)
Always remember formulas for force , in this case, the two above. So force is directly proportional to
change in velocity ( acceleration) and mass. If the mass doubles so the force also doubles to get an
increase in velocity or acceleration to 2v : answer B : 2F
Don’t forget other formulas relating to F
F = ma
F = Eq
F = kqq
r2
W = Fd 
F = W/d
SEE OTHER NOTES…..
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17
A magnetic field will exert a force ( Fb) on the moving electric charge ( q) and cause it to move in a
circular arc much like the centripetal force acts on a car in a curve or roller coaster. The magnetic force
acts perpendicular to the motion of the electric charge and is directly proportional to q and v:
Fb = qvB
It is important to note that the charge is moving perpendicular or at an angle to the field:
Perpendicular
At an angle
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10. This question is about a filament lamp.
(a)
On the axes below, draw a sketch-graph to show the variation with potential difference
V of the current I in a typical filament lamp (the I–V characteristic). (Note: this is a
sketch-graph; you do not need to add any values to the axes).
V
0
I
0
(1)
V
0
I
0
(b)
(i)
Explain how the resistance of the filament is determined from the graph.
Answer : from the value of V / I at any point on the curve
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(ii)
Explain whether the graph you have sketched indicates ohmic behaviour or
non-ohmic behaviour.
Answer : non-ohmic because the resistance (V / I at each point)
is not constant
A filament lamp operates at maximum brightness when connected to a 6.0 V supply. At
maximum brightness, the current in the filament is 120 mA.
(c)
(i)
Calculate the resistance of the filament when it is operating at maximum
brightness.
R = V/I = 6.0V/ .12A = 50
(ii)
You have available a 24 V supply and a collection of resistors of a suitable
power rating and with different values of resistance. Calculate the resistance of
the resistor that is required to be connected in series with the supply such that
the voltage across the filament lamp will be 6.0 V. Draw circuit.
Solution:
Given :
24 V ( from battery) - 6V drop ( from filament) leaves 18V left over for unknown
resistor. Also given 50 resistance of filament.
So need to find unknown resistor needed for the 18 V drop.
R= V/I = 18V / 0.12A = 150 
Drawing
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11. This question is about the electric field due to a charged sphere and the
motion of electrons in that field.
The diagram below shows an isolated, metal sphere in a vacuum that carries a negative
electric charge of 9.0 nC.
–
(a)
On the diagram draw arrows to represent the electric field pattern due to the charged
sphere.
–
(b)
The electric field strength at the surface of the sphere and at points outside the sphere
can be determined by assuming that the sphere acts as though a point charge of
magnitude 9.0 nC is situated at its centre. The radius of the sphere is 4.5 × 10–2 m.
Deduce that the magnitude of the field strength at the surface of the sphere is 4.0 × 104
Vm–1.
E= kq
r2
E = ( 8.99 x 109) ( 9 x10-9)
= 4.0 × 104 V m–1
(4.5 × 10–2 m.)2
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An electron is initially at rest on the surface of the sphere.
(c)
(i)
Describe the path followed by the electron as it leaves the surface of the sphere.
Answer : The electron will move along one of the field lines shown below:
Remember that field line follow where a + charge would go. Sufficient arrows to show
decreasing radial field and direction.
No field in the center of sphere
(ii)
Calculate the initial acceleration of the electron.
GIVEN :
m electron = 9.11 x 10-31kg
E from part b above =4.0 × 104 V m–1
q of electron = 1.60 x 10-19C
F = ma = qE
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a=
q
E
m
a = 7.2 × 1015 m s–2
(iii) State and explain whether the acceleration of the electron remains constant,
increases or decreases as it moves away from the sphere.
Answer :
Decreasing: electric field strength is decreasing so force on electron is decreasing
(iv)
At a certain point P, the speed of the electron is 6.0 × 106 ms–1. Determine the
potential difference between the point P and the surface of the sphere.
V = ΔE/q = W/ q
ΔE = ½ mv2
q = 1.60 x 10-19C
m of electron = 9.11 x 10-31kg
v = 6.0 × 106 ms–1
ΔE = ½ (9.11 x 10-31kg ) (6.0 × 106 ms-1) 2 = 1.60 x 10-17 J
V = Δ E/ q = 1.60 x 10-17 J / 1.60 x 10-19C = 100V
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12. Need to understand below that the voltmeter contributes its own resistance ( 20 k to overall
resistance as a PARALLEL circuit. So you must reduce this to series circuit giving overall
resistance of 10 kfor this part of circuit.
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OHMS LAW LAB REVIEW
Set up circuit as shown:
Remember that the variable resistor is used to control the amount of current running through the
resistor. As you vary the current you will measure the voltage across the resistor and try to verify Ohm’s
law and hopefully get a straight line .
Use the variable resistor to give you 5 distinct ammeter readings that you can clearly measure: for
example 0.6A, 0.7A., 0.8A and 0.9A. As you adjust the variable resistor for each ammeter reading record
the corresponding voltmeter reading.
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Circuit Diagram – Know for IB Exam
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13. This question is about electric circuits.
Susan sets up the circuit below in order to measure the current-voltage (I-V) characteristic of
a small filament lamp.
A
3.0 V
V
S
The supply is a battery that has an emf of 3.0 V and the ammeter and voltmeter are
considered to be ideal. The lamp is labelled by the manufacturer as “3 Volts, 0.6 Watts”.
(a)
(i)
Explain what information this labelling provides about the normal operation of
the lamp.
Answer :
when connected to a 3 V supply, the lamp will be at normal brightness;
and energy is produced in the filament at the rate of 0.60 W;
Look for the idea that 3 V is the operating voltage and the idea
of energy transformation.
when connected to a 3 V supply, the lamp will be at normal brightness;
and the resistance of the filament is 15 Ω / the current in the filament is
0.20 A;
13. ( cont. ) (a) (ii) Calculate the current in the filament of the lamp when it is
operating at normal
I=
P
V
to give I = 0.20 A
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13.(cont.)
She then sets the variable resistor to its zero value of resistance and records the
following readings.
Ammeter reading = 0.20 A
(b)
(c)
Voltmeter reading = 2.6 V
Determine the internal resistance of the battery.
Calculate the resistance of the filament when the reading on the voltmeter is
(i)
0.60 V.
answer : 3.3 
(ii)
2.6 V
answer : 13 
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(d)
Explain why there is a difference between your answers to (c)(i) and (c)(ii).
Answer : at the higher potential difference ( voltage) , greater current and therefore
hotter so the resistance of a metal increases with increasing temperature
13. ( cont. )
(e)
Using the axes below, draw a sketch-graph of the I-V characteristic of the
filament of the lamp. (Note: this is a sketch-graph; you do not need to add any
values to the axes.)
I
0
V
0
I
0
0
V
correct approximate shape (ie showing decreasing gradient with increasing V)
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13. ( cont. ) The diagram below shows an alternative circuit for varying the potential difference
across the lamp.
X
Y
3.0 V
Z
The potential divider XZ has a potential of 3.0 V across it. When the contact is at the
position Y, the resistance of XY equals the resistance of YZ which equals 12 Ω. The
resistance of the lamp is 4 Ω.
(f)
Calculate the potential difference across the lamp.
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14. In the circuit below, n charge carriers pass the point P in a time t. Each
charge carrier has charge q.
P
The current in the circuit is given by the expression B
A.
q
.
t
B.
nq
.
t
C.
qt
.
n
D.
nqt.
(1)
Answer ; Current = TOTAL charge
time
= nq
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15. Find the current in , and the potential difference across, each resistor in the circuit shown below:
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FINAL ANSWER:
16.
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