Calculate the flux that passes through the floor

[figs]

A house has a floor area of 200 m^2 and an outside wall that has an area of 50.0 m^2. The Earth's magnetic field here has a horizontal component of 2.58×10^-5 T that points due north and a vertical component of
4.14×10^-5 T that points straight down, toward the earth. a)Determine the magnetic flux through the wall if the wall faces north. b)calculate the flux that passes through the floor.

a)
flux=BAcos(theta)
=(4.14*10^-5)(50 m^2)(cos 90 deg)
=0 Wb

b)
flux=BAcos(theta)
=(2.58*10^-5)(200 m^2)(cos 0 deg)
=5.16*10^-3 Wb

I'm not sure what I did wrong.

 

[dextercioby]

At point a) it's the horizontal component which gives the nonzero flux...

As for point b),it's viceversa,only the vertical one does.

Daniel.

 

[Astronuc]

A house has a floor area of 200 m^2 - normal points vertical - a vertical component of 4.14×10^-5 T that points straight down,

An outside wall has an area of 50.0 m^2. The wall's normal is horizontal, The Earth's magnetic field here has a horizontal component of 2.58×10^-5 T that points due north.

 

[figs]

im still in the dark about this vertical horizontal component
do i need to find some magnitude before i find the flux?

 

[Astronuc]

See if this helps - http://scienceworld.wolfram.com/physics/MagneticFlux.html