Question

A wooden toroidal core with a square cross section has an inner radius of 10cm and an outer radius of 12cm. It is wound with one layer of wire (diameter, 0.96mm; the resistance per unit length , 21m ohm/m. Calculate a) the inductance and b) the inductive time constant. Ignore the thickness of the insulation.

Answer 1

radius of the inner core r1= 10cm = 0.1m

radious of the outer core r2 = 12cm =0.12m

width of the wire    a = 0.96 mm = 0.96*10^-3 m

no. of layers N= 1

(a)

the self inductance of the toroid of square cross section(L) is given by,

L = [ ( $\mu$ o*N²*a) / (2*pi*log(r2/r1) )]

L = 4*pi*10^-7*0.96*10^-3 / 2*pi*log(0.12/0.10)

L = 10.54*10^-10 H

L = 1.054*10^-9 H

L = 1.054 nH

(b)

time constant t = L/R = 1.054*10^-9/21

t = 0.050235*10^-9 s

t = 50.23 microsecond