A 57.0 m length of insulated copperwire is wound to form a solenoid of radius 2.

Question

A 57.0 m length of insulated copperwire is wound to form a solenoid of radius 2.2 cm. The copper wire has a radius of0.53 mm. (a) What is the resistance of the wire?
_______________

(b) Treating each turn of the solenoid as a circle, how many turnscan be made with the wire?
_____________ turns

(c) How long is the resulting solenoid?
___________ m

(d) What is the self-inductance of the solenoid?
_________________ mH

(e) If the solenoid is attached to a battery with an emf of 6.0 Vand internal resistance of 350 m, compute the time constantof the circuit.
______________ms

(f) What is the maximum current attained?
_______________ A

(g) How long would it take to reach 99.9% of its maximumcurrent?
____________ ms

(h) What maximum energy is stored in the inductor?
________________ mJ neep really bad................... (a) What is the resistance of the wire?
_______________

(b) Treating each turn of the solenoid as a circle, how many turnscan be made with the wire?
_____________ turns

(c) How long is the resulting solenoid?
___________ m

(d) What is the self-inductance of the solenoid?
_________________ mH

(e) If the solenoid is attached to a battery with an emf of 6.0 Vand internal resistance of 350 m, compute the time constantof the circuit.
______________ms

(f) What is the maximum current attained?
_______________ A

(g) How long would it take to reach 99.9% of its maximumcurrent?
____________ ms

(h) What maximum energy is stored in the inductor?
________________ mJ neep really bad...................

Explanation / Answer

Given that         l = 57 m         r = 0.53 mm = 0.53 x10-3 m         = 1.7 x10-8 m         r ' = 2.2 cm= 2.2 x 10-2 m (a) we know that              R = l / A = L / r2 = 1.7x 10-8 * 57 / 3.14 * (0.53 x 10-3)2 = 1.098 (b) l = N (2r' )       ==> N = l / 2r' =57 / 2*3.14 * 2.2 x 10-2 = 415 turns (c) l' = N * 2r = 415 * 2 * 0.53 x10-3 = 0.4410 m (d) self - inductance L = o N2A/ l' = 4 x 10-7 * 415*415* 3.14 * (2.2 x 10-2)2 / 0.4410 = 7.45 x 10-1 mH (e) Ri + R = 0.35 + 1.098 = 1.448     ==> = L / ( Ri + R) =7.45 x 10-1 mH / (1448 m) = 0.516 ms (f) Imax = / (Ri + R) = 6.0 / 1.448 =4.143 A (g) Given that I = 0.99 Imax        Therefore we have                   t = 6.9 = 6.9 * 0.516 ms = 3.5604ms (h) U = 0.5LImax2 = 0.5 * 7.45 x10-4 * 4.143 * 4.143 = 6.393mJ

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