Questions
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?
(1)
O.085 T ()
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
Current
2.5 A
Vertically up
Magnetic Field Strength
30 mT
Horizontal due North
Force per unit length*
0.075 N ()
3.6 A
Due south ()
120 mT
Vertically down
0.432 N ()
50 A ()
200 mT
Horizontal due West
Horizontal due
South
d
60 mA
Vertically down
Due West ()
Due East
10 N
Vertically
downwards ()
75 mT
30 o east of north
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:
Br2 = 552 + 182 = 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
A
Rider
X
C
Y
D
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?
(5)
-4
-3
Weight of rider = 4.5 × 10 × 9.8 = 4.41 × 10 N ()
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