Question

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through the inductor and increase the current at a steady rate. What is the current through the inductor at the end of 2.60 s

Answer 1

The current through the inductor at the end of 2.60s is 9.7 mA.

Step-by-step explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

`V = L(di)/(dt) \\\\V = (L(I_1-I_o))/(dt) \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = (V*dt)/(L)\\\\I_1 = (V*dt)/(L) + I_o\\\\I_1 = (41*10^(-3)*2.6)/(13) +1.5*10^(-3)\\\\I_1 = 8.2*10^(-3) + 1.5*10^(-3)\\\\I_1 = 9.7 *10^(-3) \ A\\\\ I_1 = 9.7 \ mA`

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.