To demonstrate the large size of the henry unit, a physics professor wants to wi
ID: 1516938 • Letter: T
Question
To demonstrate the large size of the henry unit, a physics professor wants to wind an air-filled solenoid with self-inductance of 1.0 H on the outside of a 11 cm diameter plastic hollow tube using copper wire with a 0. 79 mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact). How long will the plastic tube need to be? Express your answer to two significant figures and include the appropriate units. How many kilometers of copper wire will be required? Express your answer to two significant figures and include the appropriate units. What will be the resistance of this solenoid? Express your answer to two significant figures and include the appropriate units.Explanation / Answer
Hi,
Part A. In this case, as the diameter would be very small compared to the lenght we can consider that we have an infinitely long solenoid, whose inductance can be calculated as:
L = uo (N2 / l) A ; where L is the inductance of the solenoid, N is the number of turns, A is the cross sectional area of one turn, l is the length of the solenoid (equal to the lenght of the tube) and uo is the vacuum permeability (which value is 4*10-7 H/m).
The area A can be calculated as follows:
A = *(D/2 + d)2 ; where D is the diameter of the tube and d is the diameter of the copper wire. The form of this equation is due to the fact that the copper wire is put outside of the hollow tube.
A = *(11*10-2 / 2 + 0.79*10-3 )2 m2 = 9.78*10-3 m2
If we considered the effect of the copper wire as negligible we find the value of A is:
A = 9.5*10-3 m2 (as the difference between this value and the previous one is less than 3%, we will use this value instead)
The number of turns and the lenght of the tube are related as follows:
l = Nd ::::::::: N = l/d
Therefore, the expression to calculate the inductance can be rewritten as:
L = uo( l/d2) A :::::::: l = Ld2 / (uo A) = (1 H)(0.79*10-3 m)2/[(4*10-7 H/m)(9.5*10-3 m2)] = 52.28 m
Which means that the lenght is 52 meters.
Part B. For each turn we would need d meters of copper, so we need to find the number of turns and then we would know the length of the copper wire that we need.
N = l/d = (52.28 m)/(0.79*10-3 m) = 66177
The total distance will be:
x = N(d) = 66177(*0.79*10-3 m) = 164 m ::::::: x = 0.164 km
Part. C
In this case we must calculate the resistance through the entire copper wire of length x. We know that the resistivity (p) of copper is 1.7*10-8 m, therefore:
R = p (L/A) = p (L/(d/2)2) = 1.7*10-8 m [ 164 m/ ((0.79*10-3 m/2)2) ] = 5.69 = 5.7
I hope it helps.
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