Question

Chapter 30, Problem 046 Your answer is partially correct. Try again. The current i through a 6.5 H inductor varies with time t as shown by the graph of the figure. The inductor has a resistance of 14 2. Find the magnitude of the induced emf during the time intervals (a) t = 0 to 2 ms; (b) 2 ms to 5 ms, and (c) 5 ms to 6 ms. (Ignore the behavior at the ends of the intervals.) 6 2 0 2 4 6 1 (ms) Units (a) Number 22750 (b) Number 4333.33 Units (c) Number 1 26000 Units

Answer 1

From self inductance concept we have induced emf

e = -L di/dt

e = L (slope)

e(0 to 2 ms) = 6.5((7-0)/(2*10^-3-0*10^-3)) = 22750 V

e(2 to 5 ms) = 6.5((5-7)/(5*10^-3-2*10^-3)) V = -4333.3 V

e(5 to 6 ms) = 6.5((0-5)/(6*10^-3-5*10^-3)) V = -32500 V

Negative sign says about the direction only

--

Given

solenoid of length l = 79.3 cm = 0.793 m
cross section area A = 25.3 cm^2 = 25.3*10^-4 m^2
with number of turns is N = 807

carying current I = 6.47 A

energy density , u, and total energy stored is U = u*Volume of the solenoid ( called cylinder)

we have energy density is u = 27.24 J/m^3

so total energy stored is U = u*volume

   U = 27.24*A*l

   U = 27.24*25.3*10^-4*0.793 J

   U = 0.0546 J