Please answer on file paper.
1.
A straight wire of length 0.25 m is carrying a constant current of 4.5 A. It is placed in a magnetic field of magnetic field strength 85 mT. a.
What is 85 mT in T?
O.085 T (  )
(1) b.
What is the force acting on the wire if the angle between the wire and the field is: i.
90 o
F = 0.085 × 4.5 × 0.25 × sin 90
(  )
= 0.096 N
(  ) ii.
0 o
F = 0.085 × 4.5 × 0.25 × sin 0 = 0 N
(  ) iii.
60 o
F = 0.085 × 4.5 × 0.25 × sin 60
(  )
= 0.083 N
(  )
(4)
2.
Calculate the magnitude and direction of the missing quantities on each line, using the information given on the line: a b c d
Current
2.5 A
Vertically up
3.6 A
Due south (  )
50 A (  )
Horizontal due
South
60 mA
Vertically down
Magnetic Field Strength
30 mT
Horizontal due North
120 mT
Vertically down
200 mT
Horizontal due West
75 mT
30 o
east of north
Force per unit length*
0.075 N
(  )
Due West (  )
0.432 N (  )
Due East
10 N
Vertically downwards
(  )
4.5 × 10 -3 N
(  )
120 o east of North (  )
* means force on 1 metre of wire
(8)

3.
The Earth’s magnetic field strength at a certain point
P has a horizontal component of
18 mT due north, and a downwards vertical component of 55 mT. Calculate: a.
The magnitude and angle of the Earth’s magnetic field strength at
P . (3)
18 mT
55 mT
(  )
By Pythagoras:
Br 2 = 55 2 + 18 2 = 3349
Br = 57.9 mT
(  )
Angle = tan -1 (18 ÷ 55) = tan -1 (0.327) = 18.1 o from the vertical (or
71.9 o to the horizontal)
(  ) b.
The force and its direction on a 0.5 m length of wire carrying a steady current of 4.0 A when the wire is: i.
Vertical with the current going downwards;
F = 55 × 10 -3 × 4.0 × 0.5 = 0.11 N (  )
Force is due East
(  ) ii.
Horizontal so that the current passes from east to west.
F = 18 × 10 -3 × 4 × 0.5 = 0.036 N
(  )
Vertically downwards (  )
(4)

4.
A current balance consists of a rectangular wire frame ABCD balanced on pivots midway between AB and CD .
Current in
B Rider A
C
X
D
Y
Current out
A rider of mass m = 4.5 × 10
-4
kg is positioned at a distance of 40 mm from end B to balance the frame horizontally. End AD is positioned at right angles to a uniform magnetic field as shown. When a current of 3.2 A is passed through AD , the rider needs to be moved to a new position 95 mm from end B to regain the balance.
AB = CD = 240 mm, and AD = BC = 100 mm. a.
Why is there a gap between C and D ?
So that the current passes through BD
(  )
(1) b.
Calculate the magnitude of the force on the end AD due to the magnetic field.
What is the direction?
Weight of rider = 4.5 × 10 -4 × 9.8 = 4.41 × 10 -3 N
(  )
(5)
Distance of rider from the pivot = 120 – 40 = 80 mm = 0.08 m
With current flowing, new distance = 120 – 95 = 25 mm from pivot.
Change in distance = 0.08 – 0.025 = 0.055 m
(  )
Change in moment = 0.055 × 4.41 × 10 -3 N = 2.43 × 10 -4 Nm (  )
Force = 2.43 × 10 -4 Nm ÷ 0.12m = 2.02 × 10 -3 N
(  )
Direction is downwards (  ) c.
Calculate the strength and direction of the magnetic field.
B = F/IL = 2.02 × 10 -3 N ÷ (3.2 × 0.1)= 6.31 × 10 -3 T (  )
Field is from Y to X
(  )
(2)

d.
Calculate the position of the rider is there were NO gap between A and C . (4)
Current would be 1.6 A, as there are two ways. (  )
Change in force =6.31 × 10 -3 × 1.6 × 0.1 = 1.01 × 10 -3 N
Change in moment = 1.01 × 10 -3 × 0.12 = 1.32 × 10 -4 Nm (  )
Change in distance = 1.32 × 10 -4 ÷ 4.41 × 10 -3 N = 0.030 m (  )
Position will be 0.12 – 0.03 = 0.09 m = 90 mm from B
(  )
32 marks