Answer:
a) the plastic tube need to be 24.7 m long
b) the kilometer of copper wire required is 15.7
c) the resistance of this solenoid is 5538.8 2 ohms
Explanation:
Given the data in the question;
we determine the length of the plastic tube. assuming the solenoid is long.
the self inductance of a long solenoid is;
L = μ₀n²πr²l
μ₀ = 4π × 10⁻⁷ T-m/A
where
n = number of turns per unit length
r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)
l = length of the solenoid or the length of the plastic tube
we find n = number of turns per unit length
given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter
number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter
So from our previous formula, we find l
L = μ₀n²πr²l
we substitute
1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)
1 = 0.04048 × l
l = 1 / 0.04048
l = 24.7 m
Therefore, the plastic tube need to be 24.7 m long
b)
n = number of turns per unit length = 1265.8 turns/metre
so, the length of the plastic tube over which the copper wire is to be wound,
number of turns of copper wire required = n × l
= 1265.8 turns/meter × 24.7 m
= 31,265.26 turns
Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.
so for each turn length of copper wire required = 2π × r
= 2π × 0.08 m
= 0.5026548 m
So copper wire required for 31,265.26 turns will be;
⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km
Therefore, the kilometer of copper wire required is 15.7
c)
p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m
Resistance = pl/a
where l is length of copper wire, a is cross sectional area;
diameter of copper wire is 0.79-mm
radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m
area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²
Resistance = pl/a
we substitute
Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]
Resistance = 5538.8 2 ohms
Therefore, the resistance of this solenoid is 5538.8 2 ohms
