1.	(a)	(i)	Additional force
Correct answer [3.9 × 10^–3 N ] (1)	1
Example of answer:
0.4 × 10^–3 kg × 9.81 N kg^–1 = 4 × 10^–3 N
(ii)	Explanation
Quality of written communication (1)
(Current produces) a magnetic field around the rod (1)
[Do not accept in the rod]
	There is an interaction between the two magnetic fields / fields
combine to give catapult field (1)
	Fleming’s Left Hand Rule/ Fleming’s Motor Rule (1)
The rod experiences an upward force (1)
	Using Newton 3 → downward force on magnet	Max 4
(b)	(i)	Diagram
Lower pole labelled North/N and upper pole labelled South/S (1)	1
(ii)	Calculation of current in rod
Use of F = BIl. (Ignore 10^x. F is their force and l is 5cm) (1)
See conversions; mT to T and cm to m (1)
Correct answer [2.6/2.7 A ] (1)	3
	Example of answer:
I = 3.9 × 10^–3 N / (30 × 10^–3 T × 5 × 10^–2 m) = 2.6 A
(iii)	New reading on the balance
Value < 85g [not a negative value] (1)
84.6 g (1)	2
[11]

2.	Diagram
Reasonably concentric circles  [no breaks]    (1)
Circles become more widely spaced [min 3 circles]    (1)
Correct direction/anticlockwise    (1)
[One correctly labelled field line is acceptable.  Mark lost if
contradictory arrows on diagram]	3
	Current in wire
	B = 
	1.4 × 10^–5 T =  ×     (1)

	I = 8.4 A    (1)	2
    (1)	1
	Must be labelled and originate on line
    (1)	1
	F_y: F_z = 1:1    (1)	1
	F_y can be reduced by
	Reduce I/current in Y OR reduce 2I/current in Z OR reduces currents    (1)
Increase separation/r    (1)
Reduce length of wire Y    (1)	Max 2
[10]

3.	Direction of current:  B to A  OR North [Not upwards]	(1)
1
Diagram:

Points to left  [anywhere to left]	(1)
At  ( 45° this way	(1)
2
Explanation:
Field due to current perpendicular to page at (2)	(1)
so sets as Earth’s field
because:
•	cannot pivot up/down	(1)
•	too far from wire
•	field due to wire negligible
OR only in Earth’s field
2
[5]

4.	The magnitude of the force on a current-carrying conductor in a magnetic field is directly proportional to the magnitude of the current in the conductor.  With the aid of a diagram describe how you could demonstrate this in a school laboratory.
A diagram with a wire perpendicular to a magnetic field with the means 
to measure the force on the wire.    (1)
Method of providing and measuring d.c.    (1)
Method of varying and measuring force with details.    (1)
For various values of current, measure F.    (1)
Plot F against I – straight line through origin.    (1)
MAX 3
(4 marks)
	At a certain point on the Earth’s surface the horizontal component of the Earth’s magnetic field is 1.8 × 10^-5 T.  A straight piece of conducting wire 2.0 m long, of mass 1.5 g lies on a horizontal wooden bench in an East-West direction.  When a very large current flows momentarily in the wire it is just sufficient to cause the wire to lift up off the surface of the bench.
State the direction of the current in the wire.
To the East    (1)
Calculate the current.
F = BI l
(1.5 × 10^-3 kg)(9.81 N kg^-1) = (1.8 × 10^-5 T) × I × (2.0 m)    (1)
Current = 410 A     (1)
What other noticeable effect will this current produce?
Wire melts    (1)
(4 marks)
[Total 8 marks]

5.	Explanation:
The wire, when carrying a current, feels a force when in a magnetic field    (1)
The current is at 90^o to the magnetic field    (1)
hence PQ feels an upward force;  RS a downward force; they produce a couple 
which causes rotation    (1)
Addition to diagram of an arrow ↑    (1)
	Three factors which would affect the magnitude of force:
(1)	Strength/magnitude of B field    (1)
(2)	Size of current    (1)
(3)	Number of turns of coil/length of PQ    (1)	7
	Explanation of observation:
Flux cut →  e.m.f. induced    (1)
e.m.f. opposite to applied p.d. → V_TOTAL less, hence I less    (1)	2
[Total 9 marks]

6.	F proportional to I
Quality of written communication (1)
Any two from:
[In words or on diagram]
•	Method of producing and measuring a varying direct current
•	Wire perpendicular to B field
•	Method of measuring/detennining forces, e.g. moments / acceleration (1) (1)
	Graph of F - I straight line through origin for F = added weight (1)	4
	[OR correct straight line if F is total weight OR  constant]
	Calculation of Initial acceleration
	F = BIl
= 0.20 T × 4.5 A × 5.0 × 10^–2 m
= 4.5 ×10^–2 N
a = F/m
= 
= 0.90 m s^–2
	Recall/state/use F – ma and F = BIl (1)
Use 5.0 × (10^–2 m) for length (1)
Conversion of g to kg 50 × 10^–3 (1)
a = 0.90 m s^–2 (1)	4
[8]

7.	Meaning of uniform magnetic field
	Magnetic flux density constant / magnetic field lines parallel / (1)
magnetic field strength is constant/ does not vary	1
	Sizes and directions of forces on LM and NO
	Force on LM: 2.4 × 10^–4 N/ 0.24 mN (1)
Direction:
Downwards/into (paper) (1)
	Force on NO:	)
2.4 × 10^–4 N / 0.24 mN [No unit penalty]	)  Must have both (1)	3
	Direction:	)
Upwards / out of (paper)	)

	Why no forces on MN and OL
	Wires/current and B field directions are parallel [allow ‘same
direction’] / field due to current and B field of magnet are
perpendicular to each other (1)	1
	The effect on the square
	A (turning) moment will be applied / it will (begin) to turn / spin / rotate (1)	1
	Moving the pole pieces further apart
	Reduces the size of the forces, (1)
Because the flux density is reduced/ magnetic field (strength)
reduced / B (field) reduced (1)	2
[8]

St Mary's College