Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]
From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)
(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)
(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)
Response:
(b) 10 Wb
Clarification:
Given;
angle of the magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
The equation for magnetic flux is;
Φ = BACosθ
where;
B denotes the magnetic field strength
A represents the area of the plane
θ is the inclination angle
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Next, calculate the magnetic flux through a 2.0 m² section of the same plane:
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
<pHence, the magnetic flux through a 2.0 m² area of the same plane is
10 Wb.Option "b"
