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    What I tried to do:
    -rotate the field vector 45 degrees anticlockwise around the y axis (using the rotation matrix)
    -then I know dS=dxdy/n̂.k̂ and use the limits of x from 0 to 1 and y from 0 to 1
    -then find ∫F.dS=8rt2

    However, the answer is root2 but I'm not sure what I've done wrong?
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    (Original post by bobbricks)
    What I tried to do:
    -rotate the field vector 45 degrees anticlockwise around the y axis (using the rotation matrix)
    -then I know dS=dxdy/n̂.k̂ and use the limits of x from 0 to 1 and y from 0 to 1
    -then find ∫F.dS=8rt2

    However, the answer is root2 but I'm not sure what I've done wrong?
    Rotating the plate anticlockwise is equivalent to rotating the field vector clockwise.
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    (Original post by ghostwalker)
    Rotating the plate anticlockwise is equivalent to rotating the field vector clockwise.
    Cheers- so I then dot the k component (which is perpendicular to the plate) with dxdy and integrate? That gets me root(2) but just making sure I have the theory correct
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    (Original post by bobbricks)
    Cheers- so I then dot the k component (which is perpendicular to the plate) with dxdy and integrate? That gets me root(2) but just making sure I have the theory correct
    Wouldn't like to comment on the finer points - too rusty.
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    (Original post by bobbricks)
    What I tried to do:
    -rotate the field vector 45 degrees anticlockwise around the y axis (using the rotation matrix)
    -then I know dS=dxdy/n̂.k̂ and use the limits of x from 0 to 1 and y from 0 to 1
    -then find ∫F.dS=8rt2

    However, the answer is root2 but I'm not sure what I've done wrong?
    Since the vector field is constant, you need not integrate. The flux will be:

    \phi = \vec{v} \cdot \vec{A}

    where \vec{v} = 3 \hat{i} + 4\hat{j} + 5\hat{k} and \vec{A} = A\hat{n} with A=1 here, the area of the surface.

    Originally the normal is \hat{n} = \hat{k} but after rotation anticlockwise about the y-axis, we will have \hat{n} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}) so we will get:

    \phi = \frac{1}{\sqrt{2}}(3+5) = \frac{8}{\sqrt{2}} = 4\sqrt{2}

    AFAICS. So I disagree with both results that you quoted. To get a flux of \sqrt{2}, you would have to rotate the surface clockwise about the y-axis.
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    (Original post by atsruser)
    after rotation anticlockwise about the y-axis, we will have \hat{n} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}).
    Surely \hat{n} = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})

    and the desired result follows.
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    (Original post by ghostwalker)
    Surely \hat{n} = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})

    and the desired result follows.
    Not that I can see if we are using right-handed axes. Point your thumb along the +ve y-axis, and your fingers rotate the z-axis anti-clockwise into the x-axis. This is analogous to rotating the usual 2D x-y plane anti-clockwise - x rotates towards y, since the +ve z-axis points upwards from the plane.

 
 
 
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