Question

Calculate the self-inductance (in mH) of a 49.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 22.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s †

Answer 1

given that

length l = 0.49 m

radius r = 0.05 m

number of loops N = 1000

a) self inductance due to a solenoid

L = 0*N^2*A/l

L = 4*3.14*10^-7*1000^2*(3.14*0.05^2)/0.49

L = 20.12 *10^-3 H

b) energy U = 1/2*Li^2

U = 1/2*20.12*10^-3*22^2

U = 4.87 J

c) induced emf V = Bvl

B = 4*3.14*10^-7*1000*22/0.49 = 0.0564 T

speed v = 3/0.0564*0.49

v = 108.6 m/s

emf = L*di/dt

time dt = 20.12*10^-3*22/3 = 0.148 s