Exam (elaborations)
ELE2601 Exam pack 2601 (Questions and answers)
- Course
- ELE2601
- Institution
- University Of South Africa
ELE2601 Exam pack 2601 (Questions and answers) With accurate answers and assurance that they are in exam
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ELE2601 EXAM PACK2024
QUESTIONS AND
ANSWERS
FOR ASSISTANCE CONTACT
EMAIL:
, lOMoARcPSD|31863004
Supplementary Examples #5
Magnetism
Supplementary Example 6.1
A loudspeaker coil is placed in a radial magnetic field of flux density 0.7T. The coil is circular in
shape, with a diameter of 30mm, and consists of 110 turns. Calculate the force acting upon the
coil when the current is 25mA.
= πd × N
= π× 30 × 10 −3 × 110
= 10.37 m
F = B I
= 0.7 × 10.37 × 25 × 10 −3
= 0.181N
Supplementary Example 6.2
A current of 10A in a 4000 turn coil produces a flux of 6mWb. Assuming the current and flux are
proportional calculate the emf induced in the coil if:
6.2.1 The current is reduced to 1A in 0.13 seconds.
6.2.2 The current is increased from 10A to 14A in 0.4 seconds.
6.2.1
10A 6mWb
1A 0.6mWb
0.13s
− N (Φ 1 − Φ 0 )
E1 =
t1 −t0
− 4000 × (0.6 × 10 −3 −6 × 10 −3 )
=
0.13
= 166V
, lOMoARcPSD|31863004
6.2.2
14A 8.4mWb
10A 6mWb
0.4s
− N (Φ 1 − Φ 0 )
E2 =
t1 −t0
− 4000 × (8.4 × 10 −3 −6 × 10 −3 )
=
0.4
= −24V
Supplementary Example 6.3
A 600mm long conductor is moved at a uniform rate at right angle to its length and to the
magnetic field having a density of 0.5T. If the emf generated in the conductor is 3V and the
conductor forms part of a close circuit having a resistance of 0.6Ω calculate the following;
6.3.1 The velocity of the conductor.
6.3.2 The force acting on the conductor.
6.3.3 The work done when the conductor has moved 700mm.
(v = 10m.s-1; F = 1.5N; W = 1.05J)
6.3.1
e = Bv
e
v =
B
3
=
0.5 × 600 × 10 −3
= 10m.s −1
6.3.2
e
F = B I = B
R
3
= 0.5 × 600 × 10 −3 ×
0 .6
= 1 .5 N
6.3.3
W = F× s
= 1.5 × 700 × 10 −3
= 1.05 J
, lOMoARcPSD|31863004
Supplementary Example 6.4
Two coils A and B are wound on the same ferromagnetic core. There are 500 turns on A and
4000 turns on B. A current of 6A through the A coil produces a flux of 800µWb in the core. If
the current is reversed in 35ms, calculate the average emf induced in coils A and B.
(EA = 22.86V, EB = 182.9V)
− N A (Φ 2 −Φ1 )
EA =
∆t
(
−800 × 10 −6 −800 × 10 −6
= −500 ×
)
35 × 10 −3
= 22.86V
− N B (Φ 2 −Φ1 )
EB =
∆t
(
−800 × 10 −6 −800 × 10 −6
= −4000 ×
)
35 × 10 −3
= 182.9V
Supplementary Example 6.5
A DC motor consists of 800 armature conductors. The effective length of the conductors in
the field is 160mm and the conductors are situated at a radius of 100mm from the centre of
the motor shaft. The magnetic flux density is 0.6T and the current of 25A flows through the
windings. Calculate the torque available in the motor shaft. (T = 192Nm)
F = B I
( )
= 0.6 × 1.6 × 10 −3 × 800 × 25
= 19.2 N
T =Fr
= 19.2 × 100 × 10 −3
= 1.92 Nm
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