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NCEA Level 2 Physics (90257) 2011 — page 1 of 4
Assessment Schedule – 2011
Physics: Demonstrate understanding of electricity and magnetism (90257)
Evidence Statement
Q
Evidence
Achievement
The left hand plate, A. Field lines go from positive to
negative.
OR
Field lines show the direction a positive test charge
would move.
1
ONE part correct.
(b)
The field lines are equally spaced.
1
Correct answer.
(c)
V
d
V = Ed
2
2
ONE
(a)
E=
Merit
1
BOTH parts correct.
Correct working except for ONE
error.
2
Correct answer.
Correct calculation of energy
change.
2
Correct working with ONE
error.
Excellence
V = 3.33 ´ 106 ´ 0.12
V = 399 600 = 4.0 ´ 105 V = 400 kV
(d)
V=
DE
q
DE = Vq
DE = 4 ´ 10 ´ 1.5 ´ 10
5
-10
DE = 6 ´ 10-5 J
EK =
v=
1
2 mv
2
= 6 ´ 10-5 J
2 ´ 6 ´ 10 -5
2.5 ´ 10-2
= 4.8 ´ 10 -3 = 0.069 m s -1
OR F = 4.995  10–4
a = 0.01998
2
Correct working and answer.
NCEA Level 2 Physics (90257) 2011 — page 2 of 4
(e)
TWO
(a)
(b)
When the ball touches the negative plate, it will gain
electrons until it has an overall negative charge.
It then experiences a force in the opposite direction to
the field (OR is attracted to the positive plate OR is
repelled from the negative plate).
When the ball touches the positive plate, it loses
electrons until it has an overall positive charge.
It then experiences a force in the same direction as the
field (OR is attracted to the negative plate OR is
repelled from the positive plate).
1
Electrons flow from the negative terminal in the
direction X→Y
These electrons are cutting across a magnetic field that
is towards the bottom of the page. Each electron
experiences a force in the direction “A”
This causes the wire to experience a force and to
swing in direction “A”
OR can explain in terms of conventional current.
1
V = IR
1
ONE correct idea.
Eg moves towards positive plate
The loop swings in direction “A”.
1
Moves towards + and
attraction/force/repulsion
1
Direction of charge flow and
loop movement correct.
1
Full explanation linking the
charging process and the force due
to the field.
ONE correct calculation.
2
All correct except for ONE
error.
Full explanation.
Loop movement + Current flow +
perpendicular / cutting / across
(not in).
OR
Current direction and wire
perpendicular/crossing/cutting
(not in) field.
V
R
6.0
I=
= 3.33 A
1.8
F = BIl
2
Correct answer.
Must have = 3.8 cm.
Stops at 0.038 m.
F
0.25
=
BI 2.0 ´ 3.33
l = 0.038 m
l=
Electron – direction “C” (or left).
TWO correct ideas.
M+ Electron movement and
repetition
I=
(c)
1
1
Correct answer.
NCEA Level 2 Physics (90257) 2011 — page 3 of 4
(d)
V = Bvl
1
Correct except for one error.
1
Correct answer.
(Must link power of 5W to 12V)
2
Correct answer.
1
Correct LED current AND ONE
reason.
V = 12 - 3.40 = 8.6 V
2
Correct answer.
Must state 2 Sig Fig.
1
Correct answer.
2
Correct answer.
1
Correct LED current AND
BOTH reasons.
V
Bl
11´ 10-3
v=
2 ´ 0.0375
v=
v = 0.1467 = 0.15 m s-1
THREE
(a)
(b)
When the bulb has 12 V across it, the power output is
5 W.
P = VI
P
V
5
I=
= 0.42 A
12
I=
(c)
LED “B” is forward biased, so that the resistance is
small (as long as the voltage is greater than the knee
voltage) so current will flow through it.
LED “A” is reverse biased so that the resistance is
infinite (very large) (as long as the voltage is less than
the break down voltage), so no current will flow
through it.
OR
LED “B” has low resistance, so the current through it
is large.
LED “A” has infinite/large resistance, so the current
through it is zero.
(d)
NCEA Level 2 Physics (90257) 2011 — page 4 of 4
(e)
If a third LED is added in parallel, this will decrease
the total resistance (of the circuit).
The supply voltage doesn’t change, so the total current
(and therefore the resistor current) increases.
V = IR, so the voltage across the resistor will increase.
1
ONE correct idea.
Voltage increases.
Voltage +1 idea
(f)
V 3.5

 5.0  (Must be 5 – no other option).
I 0.7
OR if tangent drawn then R=0.6
2
Correct answer.
(g)
As the current increases, resistance decreases.
1
Correct answer.
R
1
TWO correct ideas.
1
Full explanation including all
THREE ideas linked.
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
2 A1 + 2 A2+ 4 A
1 M1 + 1 M2 + 11 A
1 E1 + 1 E2 + 1 E + 10 A
OR
4E+8A
Note: where the criterion is not specified, the required grade(s) can be from either.