The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Stuck on finding flux through square

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?
(edited 7 years ago)
Capture.PNG12.2KB
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.
Reply 2
Avatar for bobbricks
bobbricks
OP
19
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 :smile:
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 :smile:


Wouldn't like to comment on the finer points - too rusty.
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:

ϕ=v⃗⋅A⃗\phi = \vec{v} \cdot \vec{A}

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

Originally the normal is n^=k^\hat{n} = \hat{k} but after rotation anticlockwise about the y-axis, we will have n^=12(i^+k^)\hat{n} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}) so we will get:

ϕ=12(3+5)=82=42\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 2\sqrt{2}, you would have to rotate the surface clockwise about the y-axis.
Original post by atsruser

after rotation anticlockwise about the y-axis, we will have n^=12(i^+k^)\hat{n} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{k}).


Surely n^=12(−i^+k^)\hat{n} = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})

and the desired result follows.
(edited 7 years ago)
Original post by ghostwalker
Surely n^=12(−i^+k^)\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.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you