Answer:
a. A = 0.0859 m^2
b. A = 0.0178 m^2
Explanation:
Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of from the horizontal, and a net magnetic flux of 8.4 103 Wb passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.
take note that the question has not specified th angle which the surface is tilted so i assume the angle is at [tex]12^{0}[/tex] to the horizontal
flux = BAcos([tex]\alpha[/tex])
B=magnetic flux in Weber
A=area of the flat surface in m^2
[tex]\alpha[/tex]=the angle to the horizontal
a) 8.4 x10^-3= (.47)Acos(78)
alpha has to be the angle from the normal and not the horizontal so 90-12=78,
8.4 x10^-3
/(.47)cos(78)
A = 0.0859 m^2
b) If flux remains the same then for it to be the smallest possible area it needs to be perpendicular to the magnetic field so alpha would be 0.
8.4 x10^-3 = (.47)Acos(0)
A = 0.0178 m^2
