An ideal solenoid with 880 turns is 0.550 m long and has a cross-sectional area

Question

An ideal solenoid with 880 turns is 0.550 m long and has a cross-sectional area of 3.00 times 10^-3 m^2. (a) What is the inductance of this solenoid? H (b) If we connect this solenoid in series with a 12.0-V battery and a 200-ohm resistor, what is the maximum value of the current? A (c) Assuming the current is zero when t = 0 as the circuit elements are connected, at what time will the current reach 50% of its maximum value? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. s

Explanation / Answer

N = no of turns = 880

L = length of solenoid = 0.550m

A = area = 3*10-3m2

L = inductance=?

from L = u0*N2*A / L

= 4*pi*10-7*(880)2*3*10-3 / 0.550

=5.305*10-3 H = 0.0053 H

ii) V = voltage = 12 V

R = resistance = 200 ohm

I0 = maximum current

I0 = V/ R

= 12 / 200

= 0.06 A

iii) t = time = ?

when I = (50/100) *I0

from I = I0 ( 1- e-t/tow) ---------------------1   

( tow = time constant = L / R = 5.305*10-3 / 200 = 2.652*10-5)

put values in eq 1:

(50/100)*I0 = I0 ( 1- e-t/2.652*10 -5)

e-t/2.652*10 -5  = 1- 50/100

  e-t/2.652*10 -5 = 50/100

solve and taking log both sides

- t/ 2.652*10-5 = ln 1 - ln 2

= 0 - 0.693

t = 0.693* 2.652*10-5

t = 1.83*10-5s

  

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