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Module G485.1 Electric and
Magnetic Fields
student booklet
Lesson 1 questions - Electric Fields
1)
Part of the electric field pattern has been drawn in each of the
diagrams below. Use your knowledge of symmetry to complete the
electric field pattern in each case.
Each of the diagrams below shows a pair of electrodes connected to a voltage
supply.
+
2)
–
For each of the diagrams, add solid lines to illustrate the shape of the
electric field between the electrodes. Draw arrows to indicate the
direction of the field.
3)
Add dotted lines to show the shape of equipotential lines
associated with the field. The spacing of the equipotential lines should
give an indication of the relative strength of the electric field. Practical
advice
These questions are intended to be straightforward, to reinforce
understanding and to build confidence.
Lesson 3 questions – Electric field and potential
1
This question is about electric forces.
A very small negatively-charged conducting sphere is suspended by an insulating
thread from support S. It is placed close to a vertical metal plate carrying a positive
charge. The sphere is attracted towards the plate and hangs with the thread at an angle
of 20 degrees to the vertical as shown in fig 1.1.
fig 1.1
a)
Draw at least five field lines on fig 1.1 to show the pattern of the field
between the plate and the sphere.
(3)
b)
The sphere of weight 1.0 x 10-5N carries a charge of –1.2 x 10-9C.
i)
Show that the magnitude of the attractive force between the
sphere and the plate is about 3.6 x 10-6N.
Resolve forces
F = Wtan20
F = 1.0 x 10-5N tan20
F=3.64 x10-6N
F
W
(3)
ii)
Hence show that the value of the electric field strength at the
sphere, treated as a point charge, is 3.0 x 103 in SI units. State
the unit.
E=F/Q
E=3.64 x10-6N/1.2 x 10-9C
E=3.0 x 103
Unit for electric field strength is ……NC-1………… (3)
c)
The plate is removed. Fig 1.2 shows an identical sphere carrying a
charge of +1.2 x 10-9C, mounted on an insulating stand. It is placed so
that the hanging sphere remains at 20 degrees to the vertical.
fig 1.2
Treating the spheres as point charges, calculate the distance r between their centres.
F=kqQ/r2
r2=kqQ/F
k=1/4πε0
r2=qQ/Fx4πε0
r2=(1.2 x 10-9) 2/3.6 x10-6 x4πx(8.85 x 10-12)
r = ……0.06…………….m (3)
d)
On fig 1.2, sketch the electric field pattern between the two charges.
By comparing this sketch to your answer to (a), suggest why the distance between the
plate and the sphere in fig 1.1 is half of the distance between the two spheres in fig
1.2.
……Looking at the symmetry, the lines on the first diagram are half those on the
second…………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
2
a)
Define electric field strength at a point in space.
………Electric field strength is the force per unit positive charge………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
b)
Fig 2.1 shows two point charges of equal magnitude, 1.6 x 10-19C, and
opposite sign, held a distance 8.0 x 10-10m apart at points A and B. The charge A is
positive.
8.0 x 10-10m
A
B
Fig 2.1
i)
On Fig 2.1, draw the electric field lines to represent the field in
the region around the two charges.
(3)
ii)
Calculate the magnitude of the electric field strength at the mid
point between the charges. Give a suitable unit for your answer.
E=kq/r2
E=q/4πє0r2
Sum of both charges since opposite
E= 2 x 1.6 x 10-19/4x3.14x8.85x10-12 x (4.0 x 10-10)2
Electric field strength = ……1.8x1010………………….. unit ……NC-1………….. (5)
c)
Imagine two equal masses, connected by a light rigid link, carrying
equal but opposite charges. This is a system called a dipole. Fig 2.2 and 2.3 show the
dipole placed in different orientations between two uniformly and oppositly sharged
plates.
fig 2.2
fig 2.3
Any effects of gravity are negligible.
i)
Describe the electric forces acting on the charges by drawing
suitable arrows on the diagrams.
ii)
Explain the motion, if any, of the dipole when it is released
from rest
in fig 2.2
……Nothing – attraction and repulsion balance………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
in fig 2.3
……It will start to rotate clockwise………………………………………………..
……Because a couple is set up there will be no translational motion, just rotational.
…………………………………………………………………………………………..
……………………………………………………………………………………… (5)
Lesson 5 questions – projectile comparisons
1
This question is about changing the motion of electrons using electric fields.
Fig 1.1 shows a horizontal beam of electrons moving in a vacuum. The electrons pass
through a hole in the centre of the metal plate A. At B is a metal grid through which
the electrons can pass. At C is a further metal sheet. The three vertical conductors are
maintained at voltages of +600V at A, 0V at B and +1200V at C. the distance from
plate A to grid B is 40mm.
Fig 1.1
a)
On fig 1.1 draw electric field lines to represent the fields in the regions
between the three plates. (2)
b)
Show that the magnitude of the electric field strength between plate A
and grid B is 1.5 x 104Vm-1.
E=V/d
E=600/40x10-3
E=15000 Vm-1
(2)
c)
hole in A.
Calculate the horizontal force on an electron after passing through the
E=F/q
15000 x 1.6x10-19 = F
Force = …2.4 x 10-15……….. N (2)
d)
Show that the minimum speed that an electron in the beam must have
at the hole in A to reach the grid at B is about 1.5x107 ms-1.
½ mv2 = Fd
2.4 x 10-15 40 x 10-3 x 2 /9.11 x 10-31 = v2
v= 1.45 x 10 7 ms-1(2)
e)
Calculate the speed of these electrons when they collide with sheet C.
The potential is twice as much so the energy difference is twice therefore:
½ mv12 =mv22
v2 =2v2
√2v1= v2
Speed = …2.05 x 107…………….. ms-1 (1)
f)
Describe and explain the effect on the current detected at C when the
voltage of the grid B is increased negatively.
……fewer electrons will reach grid B or C
……therefore current will fall
………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………………
…………………………………………………………………………………… (2)
2
a)
Calculate the (i) gravitational force (ii) the electrical force between two
protons that are 2.0 x 10-10 m apart. Take the mass of a proton to be 1.7 x 10-27kg.
i)
F=GmM/r2
F=6.67x10-11x(1.7 x 10-27 ) 2/(2.0 x 10-10)2
Force = ……4.8x10-45……….N
ii)
F=qQ/4πε0r2
F=(1.6x10-19) 2/4π8.85x10-12x(2.0 x 10-10)2
b)
Force = …5x10-9………….N (4)
What is the ratio of the electrical to gravitational force?
Ratio = 5x10-9/4.8x10-45
Ratio = ……1.2x1036……………….. (2)
Lesson 6 questions – Magnetic field patterns
1)
Figs 1.1 and 1.2 illustrate magnetic field patterns caused by current-carrying
conductors in a plane at right angles to the conductors.
fig 1.1
fig 1.2
What shapes of conductor would produce these field patterns above?
Fig. 1.1 ………A current carrying wire………………………………………………..
Fig. 1.2 ………A coil or parallel wires………………………………………………
(2)
Total [2]
2)a) Explain why a compass needle placed very close to a wire may deflect when
the current in the wire is switched on.
…………The wire is surrounded by a magnetic
field……………………………………
…………………………………………………………………………………………
(1)
b)
Fig. 2.1 shows a cross section of a current carrying conductor.
current-carrying
conductor (current out
of plane of page)
fig 2.1
On fig 2.1, draw the magnetic field pattern.
(3)
Total [4]
3)
Fig. 3.1 shows a cell connected to two terminals on a plastic box. When switch
S is closed the magnetic field pattern shown is detected around the box.
fig 3.1
State what is inside the box that produces the field pattern.
………………A coil or solenoid……………………………………………………
(1)
Total [1]
Sketching flux patterns
1
The flux from a long magnet looks like this:
N
S
2.
The flux from a flat thin magnet looks like this:
S
3.
N
The flux from a horseshoe magnet looks like this:
4.
The flux pattern is symmetrical about the line dividing the poles and
about the line joining their centres:
N
S
5.
The flux pattern is symmetrical about the line dividing the coils and
about the line through their centres:
6.
The poles are as shown:
–
7.
+
The poles are as shown:
S
N
8
Because the coil is long, the current turns at places along its length
surround each place in much the same way. A fly at each point would
just see coils around it in all directions. So the field is the same along
all the central part of the length (not near the ends). Thus it must be
straight and uniform.
9.
The flux just runs in circles round inside the coil:
10.
The flux crosses the air gaps, goes through the rotor, and joins up by
going round the
casing of the stator:
flux
flux
Lesson 7 questions – F=BIL and Flemings left hand
rule
1)
Fig. 1.1 shows an arrangement for measuring the magnetic flux density B
between poles of a magnet.
fig 1.1
fig 1.2
The coil shown in figs 4.1 and 4.2 has 50 turns. Its lower side XY is horizontal and as
a mean length of 30mm. Before the current is switched on, the balance reads 0.850N
and with the current of 2.0A switched on the balance reads 0.815N.
a)
State the rule that can be used to determine the direction of the force acting on
the magnet.
…………Flemings Left Hand Rule……………………………………………………
(1)
b)
Determine the magnitude and direction of the electromagnetic force acting
on the magnet.
……………F=0.8500.815........................................………………………………………
……………=0.035N……………………………………………………………………
…
………………Direction is up…………………………………………………………
(2)
c)
Calculate the magnetic flux density B between the poles of the magnet.
……………B=F/Il……………………………………………………………………
……
………………=0.035/(2x0.03x50)……………………………………………………
……
………………=0.0117T………………………………………………………………
……
…………………………………………………………………………………………
……
…………………………………………………………………………………………
(3)
Total [6]
2)
Fig 2.1 shows a wire placed at righ angles to a magnetic field. The wire rests
on two metal supports.
The length of the wire between the supports is 4.5 cm, and this length has weight 6.0
x 10-2N. The current in the wire is slowly increased from zero until the wire starts to
lift off the metal supports.
a)
Calculate the current I in the wire. The magnetic flux density is 0.36T.
………………F=BIL…………………………………………………………………
…...
………………I=(6.0x102
)/(0.36x0.045)………………………………………………...
………………=3.7A……………………………………………………………………
...
…………………………………………………………………………………………
…...
…………………………………………………………………………………………
(3)
b)
Suggest why the overhead cables for the National Grid cannot be freely
supported by the Earths magnetic field.
…………Earths magnetic field is very small
(50μT)………..……………………………
…………Cables have large weight
It is an a.c current and so av force=0…………………………………………………...
(2)
Total [5]
3)
Fig. 3.1 shows a current-carrying metal rod that can roll freely on two parallel
metal rails. The rod is at right angles to the magnetic field lines.
fig 3.1
a)
Determine the direction of the force experienced by the rod. Explain how you
determined this direction.
………………Right…………………………………………………………………….
..
………………Flemings left hand rule…………………………………………………
(2)
b)
The current in the metal rod is 2.0A and it has a length 5.0cm between the two
metal rails. Calculate the force experienced by the metal rod given the magnetic flux
density is 1.8 x 10-3T.
……………F=BIL……………………………………………………………………
…....
………………=1.8 x 10-3 x 2.0 x 0.05…………………………………………………
………………=1.8 x 10-4
N……………………………………………………………....
…………………………………………………………………………………………
(3)
Total [5]
Lesson 9 questions – Moving charge in a magnetic
field
(
1
2
3
/39)…….%…….
Lesson 14 questions – Faraday and Lenz
Name ……………………
(
/36)……..%……..
Class…………………
1
2
3
Lesson 16 questions – AC Generators
Name ……………………
(
/27)……..%……..
Class…………………
1
2
Lesson 17 questions - Transformers
1