The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 ×
×
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss =
×
×
× 
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.
We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as
B = μ₀ . I / 2r
where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.
From the question above, we know that:
r = 4 cm = 0.04 m
I = 1.7 A
By substituting the parameter, we get
B = μ₀ . I / 2r
B = 4π×10¯⁷ . 1.7 / (2.0.04)
B = 2.67 x 10¯⁵ T
Due to the perpendicular plane of loops, the total magnetic field at center will be
Btotal = √(2(B²))
Btotal = √(2(2.67 x 10¯⁵²))
Btotal = 3.78 x 10¯⁵ T
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