Question

A solenoid of radius r = 1.25 cm and length ℓ = 27.0 cm has 280 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. _____µWb

(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. _______µWb

Answer 1

(a)

Magnetic flux is given by:

phi = B*A

Given that B = magnetic field in solenoid = u0*n*i = uo*N*i/L

N = number of turns = 280 turns

i = current in solenoid = 12.0 A

L = length of solenoid = 27.0 cm = 0.27 m

So

B = 4*pi*10^-7*280*12.0/0.27

A = Area of solenoid = pi*r^2

r = radius of solenoid = 1.25 cm = 0.0125 m

A = pi*(0.0125)^2

So,

phi = B*A

phi = (4*pi*10^-7*280*12.0/0.27)*(pi*0.0125^2)

phi = 7.68*10^-6 Wb

phi = 7.68 $\mu$Wb

(b).

A = Area of annulus = pi*(b^2 - a^2)

A = pi*[(0.800*10^-2)^2 - (0.400*10^-2)^2]

So,

phi = B*A

phi = (4*pi*10^-7*280*12.0/0.27)*pi*[(0.800*10^-2)^2 - (0.400*10^-2)^2]

phi = 2.36*10^-6 Wb

phi = 2.36 $\mu$Wb

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