An inductor is made from a copper ribbon that has been bent to form a tube with
ID: 2100057 • Letter: A
Question
An inductor is made from a copper ribbon that has been bent to form a tube with two flat extensions, as shown. The width of the copper ribbon is a=20 cm, the radius of the tube is r=10 cm, and the space between the flat extensions is negligible.
(a) What is L, the inductance of the inductor? (Hint: What is the B field inside the tube if there is a current flowing through the ribbon?
(b) The inductor is now hooked up to an ideal battery (emf = 12V) and a resistor (R = 10 ohms) to make the circuit shown. How long does it take for the current in the circuit to reach one half its maximum value imax?
Explanation / Answer
a) L = uN^2A/L..........here L = 20 cm........A = 3.14*10^2
PartB) current at any time, i(t) = I_max(1-exp(-R*t/L)),,,,,,,,,
,to reach one half pu
t i(t)= 1/2 * I_max, R= 10 and L=1.97x10-7 H, on solving you get
t= 1.36 X 10-8 sec