waecE-LEARNING
Applied Electricity Paper 2 , Nov/Dec. 2010  
Questions:   1 2 3 4 5 6   Main
General Comments
Weakness/Remedies
Strength

Question 3

(a) A coil of 300 turns is wound uniformly over a wooden ring having a mean
circumference of 900 mm and a uniform cross-sectional area of 400 mrrr'.
If the current through the coil of 6 A, calculate the:
(i) magnetic field strength;
(H) flux density;
(Hi) total flux.
[Take µr of wood = 1; µo = x10-7H/m]

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Observation

The expected answers were:
Number of turns, n = 300
Mean length, 1= 900mm = 0.9m
Cross-sectional area, A = 400mm7² = 400 x 10–6
Current through coil, I = 6A
(a)The magnetic field strength
H = NI/I
= 300 x 6/α
= 2000A/m
(b)The flux density, B = µoµoH
= 4Π X 10-7x 1x 2000
= 2.513mT/0.00251T/2.513 x 10–3T
(c)Total flux, φ = BA
= 2.513 X 10–3X 400 X 0–6 = 1.005µb
The question was on Electromagnetic fields. The question was unpopular with the candidates. Very few candidates attempted this question with fair performance.

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