The windings of a circular cylindrical solenoid have 4245 turns per meter and cr
ID: 1616154 • Letter: T
Question
The windings of a circular cylindrical solenoid have 4245 turns per meter and cross-sectional radius r = 3.99 cm. A second coil having 75 loops is wrapped around the solenoid. The current in the solenoid increased from zero to I = 0.45 A in elapsed time delta t = 2.40 times 10^-3 seconds. a) Compute the initial and final magnetic fields produced by the solenoid. b) Compute A, the area of the solenoid cross-section. c) Compute delta phi_B, the change in magnetic flux through the solenoid. d) Compute the induced emf in the coil wrapped around the solenoid. e) Computer M, the mutual inductance between the solenoid and the coil wrapped around it.Explanation / Answer
let
n = 4245 turns/m
r = 3.99 cm = 0.0399 m
N = 75 turns
I = 0.45 A
delta_t = 2.4*10^-3 s
a)
Bi = mue*n*Ii
= 0
Bf = mue*n*If
= 4*pi*10^-7*4245*0.45
= 2.4*10^-3 T
b) A = pi*r^2
= pi*0.0399^2
= 5.00*10^-3 m^2
c) the change in flux through the solenoid = A*(Bf - Bi)
= 5*10^-3*(2.4*10^-3 - 0 )
= 1.2*10^-5 T.m^2
d) induced emf in the coil = N*change in megnctic flux/time taken
= 75*1.2*10^-5/(2.4*10^-3)
= 0.375V
e) use induced emf = M*dI/dt
==> M = induced emf/(dI/dt)
= 0.375/(0.45/(2.4*10^-3))
= 2.0*10^-3 H
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