Question

In recent years it has been possible to buy a 1.0 F capacitor. This is an enormously large amount of capacitance. Suppose you want to build a 4.0 Hz oscillator with a 1.0 F capacitor. You have a spool of 0.20-mm -diameter wire and a 4.2-cm -diameter plastic cylinder. How long must your inductor be if you wrap it with 2 layers of closely spaced turns?

Answer 1

frequency of L-C oscillator=1/(2*pi*sqrt(L*C))

given frequency=4 Hz and C=1 F

==>4=1/(2*pi*sqrt(L*1))

==>L=1.58314 mH

inductance is given by wheelers approximation:

L=(31.6*r1^2*n^2)/(6*r1+9*l+10*(r2-r1))

where L=inductance in mH=1583.14

r1=inside coil radius in meters=2.1 cm=0.021 m

r2=outside coil radius in meters=(2.1 cm+2*(0.2 mm/2))=0.0212 m

l=coil length in meters

n=number of turns totals

here 1 turn takes distance=diameter os the wire

=0.2 mm=2*10^(-4) m

then number of turns=n=l/(2*10^(-4)) =5000*l

then 1583.14=31.6*0.021^2*25*10^6*l^2/(6*0.021+9*l+10*2*10^(-4))

==>220.06*l^2=0.128+9*l

==>l=0.05207 m

hence length of the cylinder required is 0.05207 m