A solenoid is 65 cm long and has a cross sectional area of 30cm^2. The 23.2 mH i
ID: 1491935 • Letter: A
Question
A solenoid is 65 cm long and has a cross sectional area of 30cm^2. The 23.2 mH inductor is wound with N loops of wire.
A)How many loops are required to make this inductor?
B) The resistance of the solenoid wire is 2 ohms. If the inductor is connected across a 12 volt battery, how long will it take the current to reach 95% of its final value?
C) Now consider the RL circuit with R1= 180 ohm, L= 2.5H and emf = 45V
What is the current through the resistor R1 immediately after the switch is closed? What is the current through resistor R1 a long time after the switch is closed?
Explanation / Answer
A) Let us find diameter(d) of the wire,
A=r2
30 = 3.14*r^2 => r= 3.09
d= 2r= 2*3.09 = 6.18cm
No of loops = N = L/d = 65/6.18 = 10.52 = 11 loops
B)
L= 0N2A/L = (4*10^-7*11^2*30*10^-4)/0.65 = 7.0*10^-7 H
IL = (V/R)[1-e-t/(L/R) ]
0.95 = (12/2) [1-e-t/[(7.0*10^-7)/2]] => t= 6.0*10^-8 s
C) IL = (V/R)[1-e-t/(L/R) ]
At t= 0s
IL = (45/180)[1-e-0/(L/R) ] = 45/100 = 0 A
At t= infinity
IL = (45/180)[1-e-infinity/(L/R)] = 0.45 A
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