SAMPLE ASSESSMENT SCHEDULE Physics 90937 (1.3): Demonstrate understanding of aspects of electricity and magnetism

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NCEA Level 1 Physics 90937 (1.3)


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1

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SAMPLE ASSESSMENT SCHEDULE

Physics 90937 (1.3): Demonstrate understanding of aspects of electricity and magnetism

Assessment Criteria

Achievement

Merit

Excellence

Demonstrate understanding

r
e-
quires the candidate to provide ev
i-
dence that will typically s
how an
awareness of how simple facets of
phenomena, concepts or principles
relate to a given situations. This
may include using methods for
sol
v
ing problems involving aspects
of electricity and magnetism.


Demonstrate in
-
depth understan
d-
ing

requires the ca
ndidate to pr
o-
vide evidence that will typically
show how or why phenomena, co
n-
cepts or principles relate to given
situ
a
tions
.


Demonstrate comprehensive u
n-
derstanding
requires the candidate
to provide evidence that will typ
i
ca
l-
ly show insight into how or w
hy
ph
e
nomena, concepts or principles
are connected in the context of gi
v-
en situations.

Statements will
demonstrate understanding of co
n-
nections between concepts.



Evidence Statement

One

Expected Coverage

Achievement

Merit

Excellence

(a)

Description of

a magnetic field
:

A magnetic field is a r
e
gion in which
a force is felt from a magnet or an
electric cu
r
rent.

Explanation of how field varies
:

The magnetic field strength of a bar
can be shown by the proximity of
lines of force
.

Where the lines are close
t
o
gether,
the magnetic field is strong. Where
the lines are far apart, the magnetic
field is weak.

Explanation of effect on compass
needle
:

The magnetic fields of the two bar
magnets are of equal strength
where the compass is located. This
results in equal

forces on the co
m-
pass needle. The forces are also in
opposite dire
c
tions so the net force
on the needle is zero, and the ne
e-
dle points northward as drawn b
e-
cause of Earth’s magnetic field.

TWO of:



in (a)
,

describes
what
a magnetic
field



in (a)
,

describes
or draws

the
strength of the
magnetic field
around a bar
magnet
(see
Appendix A)



in (b),
selects the
appropriate fo
r-
mula for finding
the current, mo
d-
ifies it, and su
b-
stitutes va
l
ues
but does not
substitute the
correct unit for
power.

TWO of:



in (a)
,

explai
ns
how the strength
of a magnetic
field varies, with
the field stron
g-
est
near

the
poles and wea
k-
er t
o
wards the
centre of the bar
ma
g
net



in (a)
,

states that
the needle mai
n-
tains its shown
position
because

the mag
netic
fo
r
ces are equal
and opposite



in (b)
,

s
hows an
understanding
of
how the relevant
formulae relate
to the situation in
the

attempt to
calculat
e

the cu
r-
rent
,
then the
magnetic field
strength
.

BOTH of:



in (a) connects
the equal
strengths and
opposite dire
c-
tions of
the two
magnetic field
s

with a zer
o net
force on the
ne
e
dle
, and
hence the co
m-
pass needle
points nort
h-
wards b
e
cause
of Earth’s ma
g-
netic field



in (b)
,

shows i
n-
sight into how
the relevant fo
r-
mulae relate to
the situ
a
tion by
using the correct
units, rearran
g-
ing a formula,
and correctly
ca
l
cul
ating the
ma
g
netic field
strength.

(b)

Calculation of current
:

P = VI

so I = P/V = 50 000/20 000

= 2.5 A

Calculation of magnetic field
strength:

B = kI/d = 2.0 x 10
-
7

x 2.5/12

= 4.2 x 10
-
8

T

NCEA Level 1 Physics 90937 (1.3)


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4



Two

Expected Coverage

Achievement

Merit

Excellence

(a)

Ca
lculation of resistance of siren
:

R
T

= R
S
+ R
FL

+ R
H

+ R
H

so R
S

= R
T

-

R
FL

-

R
H

-

R
H

R
S

= 22.5


8.0
-

5.8


5.8 = 2.9 Ω

Calculation of current in circuit
:

V= IR

so I = V/R = 9/22.5 = 0.40 A

Calculation of individual voltages
:

V
FL

= IR = 0.40 x 8.0 = 3.2 V

V
S

= IR = 0.40 x 2.9 = 1.16 V

Calculation of total voltage
:

Total voltage = V
FL

+ V
S

= 3.2 + 1.16 = 4.36 V

TWO of:



in (a)
,

makes a
stat
e
ment about
the total r
e-
si
s
t
ance in the
ci
r
cuit being the
sum of the ind
i-
vidual resi
s
tan
c-
es, or other
equivalent
stat
e
men
t for
a
n
other aspect of
the ca
l
culation



in (b)
,

states that
the
eye
bulb
s are

not as bright as
the bulb in the
body

because
the eye bulbs are
in series



in (b),

finds

the
t
o
tal resistance

and selects the
appropriate fo
r-
mula for finding
the current, but
does

not calc
u-
late the current
correctly.

TWO of:



in (a), shows an
understanding
of
how relevant
fo
r
mulae relate
to the situation
by calculating
the
current

and a
t-
tempting to ca
l-
culate
the total
voltage



in (b), shows an
understanding
of
how principles
r
e
late t
o the co
n-
text
by explaining
that
bulbs in s
e-
ries share the
vol
t
age, which is
why the bulbs in
the eyes are
dimmer




in (b),
shows an
u
n
derstanding
of
how the relevant
fo
r
mulae relate
to the situation,
by

calculat
ing

the
current flowing
through the ba
t-
tery
,
then a
t-
tempting to ca
l-
culate the power
output of the ba
t-
tery
.

TWO of:



in (a), connects
the calculated
current

with vo
l
t-
ages across the
flas
h
ing light and
siren to ca
l
culate
the total voltage
across the flas
h-
ing light and s
i-
ren



in (b),
connects
the brightne
ss of
the bulbs to
power, and that
power is a fun
c-
tion of both vo
l
t-
age and current,
and shows that
the power
through each
‘eye’ bulb is ¼ of
that of the ‘body’
light



in (
b
)
,

shows i
n-
sight into how
the relevant fo
r-
mulae relate to
the situ
a
tion by
using the
correct
units, rearran
g-
ing formulae as
needed, and co
r-
rectly calcula
t
ing
the power.

(b) (i)



(b)
(ii)

Comparison of brightness
:

The bulbs in each eye are di
m
mer
than the bulb in the body.

Explanation on comparison
:

Each bulb requires 6.0 V to light up
t
o its full brightness. The bulb in the
body has 6.0V across it b
e
cause it
is connected in parallel with the
battery, so it lights up to full brigh
t-
ness.

The bulbs in the eyes are co
n
nec
t-
ed in series with the battery. Each
bulb has only 3.0 V across it (i.
e.
their equal share of the 6V), and
will light up to half the brigh
t
ness of
the bulb in the body

(a
s
suming their
resistance is co
n
stant).


Further, each ‘eye’ bulb receives
half the current to that of the body
light. Brightness is determined by
power. As
P=V.I and V = ½ and I =
½, for each eye light then P = ¼ of
the power of the body light.

(b
)

(iii)

Calculation of current from battery
:

Total resistance of bulbs in the
eyes = 4



+ 4




= 8




V =IR, so I

=
V
/
R

= 6/8

= 0.75 A

Calculation of power
:

P

=
V x I

= 6.0 x 0.75

= 4.5 W

NCEA Level 1 Physics 90937 (1.3)


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4



Three

Expected Coverage

Achievement

Merit

Excellence

(a)

Drawing of charge distribution
:

Ball shows positive charges on the
left side (close to the neg
a
tively
charged rod) and negative charges
on the opp
o
site side (away
from the
charged rod).

Explanation of attraction
:

Positive charges are induced at the
end of the ball closer to the rod.
Since opposite charges attract, the
ball now moves towards the rod.

TWO of:



in (a)
,

illustrates
(or describes)
the di
s
tribution of
char
ge on the
ball



in (b)
,

illu
s
trates
(or describes)
the charge distr
i-
bution on the ball



in (c)
,

illu
s
trates
(or describes)
the charge distr
i-
bution on th
e

ball.


TWO of:



in (a)
,

shows an
understanding
of
how principles
r
e
late to the sit
u-
ation
by e
x
plai
n-
ing th
e a
t
traction
of the ball to the
rod



in (b
),

shows an
unde
r
standing
of
how principles
r
e
late to the sit
u-
ation by

e
x
plai
n-
ing the transfer
of charge to the
ball



in (c)
,

shows an
understanding
of
how principles
r
e
late to the sit
u-
ation
by e
x
plai
n-
ing by giving a

partial expla
n
a-
tion of the m
o
tion
of the ball.


TWO of:



in (a)
,

connects
the greater a
t-
tractive force
with the smaller
distance of the
opposite charges
and so there is a
net a
t
tractive
force on the ball
to explain the a
t-
tra
c
tion of the
ball to the rod



in
(b)
,

connects
the transfer of
charge to the ball
with the then like
charges on the
rod and ball to
explain the r
e-
pulsion of the
ball from the rod



in (c)
,

e
x
plains
that earthing r
e-
moves the e
x-
cess charges
from the ball
making the ball
u
n
charged, and
then co
nsiders
the
charge distr
i-
butions, the di
s-
tances i
n
volved
,

and the forces
acting on the ball
to explain the
motion of the
ball.

(b)

Drawing of charge distribution
:

Ball

show
s

negative charge
.

Explanation of repulsion
:

W
hen the ball touches the rod,
some ne
gative charges are tran
s-
ferred to the ball.

This action means that the ball now
has excess negative charges.

Since like charges repel, the ball
is
repel
led

away from the rod.

(c)

Drawing of charge distribution
:

Ball now has no excess charge, OR
the sa
me as in Diagram 5 (showing
separation).

Description and explanation of m
o-
tion
:

The negative charges are tran
s-
ferred from the ball to the girl when
she touches it. The ball is then u
n-
charged and no force of repulsion
acts on it, so it is now attracted
back

to the neg
a
tively charged rod.


Judgement Statement

Achievement

Achievement with Merit

Achievement with Excellence

Minimum of:

2A

Minimum of:

2M

Minimum of:

2E

NCEA Level 1 Physics 90937 (1.3)


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Appendix A