A coil with an inductance of 0.03 H is connected in series to a dc battery and a

Question

A coil with an inductance of 0.03 H is connected in series to a dc battery and a switch. The switch is closed at t = 0. The current reaches a final steady value of 30mA. (a) if it takes millisecond for the current to reach 40 % of its final value, find the time constant of the circuit. (b) Sketch the behavior of the current for times t > 0 on the axes. (c) Find the resistance of the coil if the battery voltage if 6V. (d) How many microjoules of energy does the magnetic field of the coil store once the current has stabilized? (e) The coil has a radius of r and a length of l. it is wrapped around air. It can be approximated as a long solenoid. Write an expression below that will give the number of turns per length (N/l) in the coil. The expression can only use the symbols r, t, the inductance I, and the permeability Mu0. (f) After the current has stabilized, the switch is then opened. Complete the following sentences by circling the correct choices. ?The energy stored by the inductor will ( increase / decrease / remain the same)? ? the inductance of the inductor will ( increase / decrease / remain the same )?

Explanation / Answer

a)

time constant T=L/R

i=io(1-e^-t*R/L)

40*io/100=io(1-e^-t*R/L)

2/5=1-e^-t*R/L

e^-t*R/L=1-2/5

e^-t*R/L=3/5

e^t*R/L=5/3

t*R/L=ln(1.666)

R/L=0.51/t

R/L=0.51/1*10^-3

=510.825

time constant T=L/R=0.00196

=1.96 msec

b)

current increases exponentially

c)

L=0.03 H, io=30 mA ,V=6v
now

time constant T=L/R=1.96 msec

R=L/1.96

=0.03/1.96*10^-3

=15.306 ohms

d)

U=1/2*Lio^2

=0.5*0.03*(30*10^-3)^2

=13.5 microJoules

e)

flux=B.A

=uo*(N/l)*A

emf=-Nd(flux)/dt

emf=-N*uo*(N/l)*A*di/dt

but

emf=-Ldi/dt

hence

L=N*uo*(N/l)*A (n=N/l)

L=uo*(N*N/l)*A

N/l=L/uo*N*A

(N/l)=(L/uo*nl*A)

f)

the energy stored by the inductor will decrease

the inductance of the inductor will remain the same

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