An inductor is made from a copper ribbon that has been bent to form a tube with
ID: 2197689 • 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. Inductance is L=1.97x10-7 H.
a)The inductor is now hooked up to an ideal battery (emf E=12 V) 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 max value i_max?
b)Suppose that there is a magnetic field B=[1(T/s)*t]k_hat piercing the inductor. Does i_max for the circuit increase, decrease, or remain the same as in part (b)? If it changes, by how much does i_max change?
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. Inductance is L=1.97x10-7 H. a)The inductor is now hooked up to an ideal battery (emf E=12 V) 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 max value i_max? b)Suppose that there is a magnetic field B=[1(T/s)*t]k_hat piercing the inductor. Does i_max for the circuit increase, decrease, or remain the same as in part (b)? If it changes, by how much does i_max change?Explanation / Answer
Part A) current at any time, i(t) = I_max(1-exp(-R*t/L)),,,,,,,,,,to reach one half put 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