00:01
All right, so here you have a flux, a cube, excuse me, with side length l.
00:13
Right.
00:13
And so you have to find the magnetic flux through this shaded region of the cube.
00:19
Please excuse my terrible drawing here, but basically you have to find the flux.
00:23
And so flux, as we know, is equal to b a, v times a times cosine theta.
00:31
And so the magnetic field, for the magnetic field, we're given i -hat, j -hat, x, y, hat, x, y, z components.
00:42
And in this case, as we know, area will be l squared.
00:45
That's the length of a cube squared.
00:51
The field component that is perpendicular, perpendicular to the cube face, is the one we're calculating.
01:00
So for this cube face, the phi sub b will be b sub x times l squared.
01:08
So b sub x is the x component, right, of the magnetic field.
01:15
Of the magnetic field, so this will be five tesla times l, which is 0 .025 meters.
01:25
That's 2 .5 centimeters squared.
01:27
And this gives you phi sub b of flux coming out of that phase of 3 .125 times 10 to the negative 3 weber's.
01:42
And so that's part a.
01:45
For part b, you want the net flux...