20 -12 points SerPSE9 30.P048.W My Notes Ask Your Teacher A solenoid of radius r
ID: 1562606 • Letter: 2
Question
20 -12 points SerPSE9 30.P048.W My Notes Ask Your Teacher A solenoid of radius r 1.25 cm and length t 32.0 cm has 295 turns and carries 12.0 A. (a) Calculate the flux through the surface of a disk-shaped area of radius R 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid as in the figure (a) above. HWb (b) Figure (b) above shows an enlarged end view of the same solenoid. Calculate the flux through the tan area, which is an annulus with an inner radius of a 0.400 cm and outer radius of b 0.800 cm. HWb Need Help? L Read it L Watch It Submit Answer Save Progress Practice Another VersionExplanation / Answer
The flux through the surface of the disk is given by B*N*A where B is the magnetic field, N is the number of turns, and A is the area. The radius is 1.25 cm or 0.0125 m.
(a) MagFlux = [ (4* x 10^-7 T*m/A) x 295 x 12.0 A ] x [ x (0.0125)^2 ]
= 2.18 x 10^-6 T*m^2 or 2.18 x 10^-6 Wb = 2.18 micro Wb
(b) The area in this situation is
A = [ (0.008 m)^2 - (0.004 m)^2 ] = 1.51 x 10^-4 m^2
MagFlux = (4.45 x 10^-3 T) x (1.51 x 10^-4 m^2)
= 6.72 x 10^-7 T*m^2 or 6.72 x 10^-7 Wb
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