(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid h
ID: 1376329 • Letter: #
Question
(a) Calculate the self-inductance of a 50.0 cm long, 10.0 cm diameter solenoid having 1000 loops.
(No Response) mH
(b) How much energy is stored in this inductor when 21.5 A of current flows through it?
(No Response) J
(c) How fast can it be turned off if the induced emf cannot exceed 3.00 V?
(No Response) s
A very large, superconducting solenoid such as one used in MRI scans, stores 1.00 MJ of energy in its magnetic field when 115 A flows.
(a) Find its self-inductance.
(No Response) H
(b) If the coils "go normal," they gain resistance and start to dissipate thermal energy. What temperature increase is produced if all the stored energy goes into heating the 1040 kg magnet, given its average specific heat is 200 J/kg
Explanation / Answer
1)
a)
No of loops, N = 1000
length, l = 50 cm = 0.5 m
diameter, d = 10 cm = 0.1
A = pi*d^2/4 = pi*0.1^2/4 = 7.85*10^-3 m^2
self inductance, L = mue*N^2*A/l
= 4*pi*10^-7*1000^2*7.85*10^-3/0.5
= 1.973*10^-2 H
b) Energy stored ina inductor, U = 0.5*L*I^2
= 0.5*1.973*10^-2*21.5^2
= 4.56 J
c) induced emf = L*dI/dt
==> dI/dt = induced emf/L
= 3/(1.973*10^-2)
= 152.02 A/s
2)
a) Apply, U = 0.5*L*I^2
==> L = 2*U/I^2
= 2*1*10^6/115^2
= 151.23 H
b) U = m*c*dT
==> dT = U/(m*c)
= 1*10^6/(1040*200)
= 4.8 degrees celsius
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