A 62.5 m length of insulated copper wire is wound to form a solenoid of radius 1.7 cm. The copper wire has a radius of 0.49 mm. (Assume the resistivity of copper is ? = 1.7 ? 10?8 ? · m.)
(a) What is the resistance of the wire? ?
(b) Treating each turn of the solenoid as a circle, how many turns
can be made with the wire?
turns
(c) How long is the resulting solenoid? m
(d) What is the selfinductance of the solenoid? mH
(e) If the solenoid is attached to a battery with an emf of 6.0 V
and internal resistance of 350 m?, compute the time constant of the
circuit. ms
(f) What is the maximum current attained? A
(g) How long would it take to reach 99.9% of its maximum current?
ms
(h) What maximum energy is stored in the
inductor? mJ
(a) Given that resistivity of copper, ? = 1.7 x 10^8 ?.m
with r = 1.7 cm = 0.017 m
So, resistance, R = ?L/A = (1.7 x 10^8 x 62.5)/?(0.00049²) =
1.41 ?
(b) solenoid radius: 0.017 m, then,
its circumference will be: l = 2?r = (6.28)0.017 = 0.107 m
The total wire length is 62.5 m, thence the # of turns is;
N = 62.5 / 0.107 = 585 turns
(c) Length of the solenoid = 2(0.00049)(585) = 0.57 m
(d) Selfinductance of the solenoid, L = ?oN²A/l =
(4?x10^7)585²(?0.017²)/62.5 = 0.0062 mH
(f) Maximum current = i = E/R[1 – e^(R/L)t] =
Imax = E/R = 6/0.350 = 17.0 A
(g) Time takken to reach 99.9% of its maximum current

i = 6[1 – e^(R/L)t]
=> 0.99 = 1 – e^(0.350/0.0000062)t
=> (56451)t = ln{0.01} = 4.6
t = 4.6/56451 = 0.008 ms
(h) Maximum energy stored in the conductor 
E = ½ LI² = ½ (0.0000062) 17² = 1.79 mJ
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