How to solve it for a scalar output?

조회 수: 6 (최근 30일)
Rahul
Rahul 2024년 9월 6일
댓글: Torsten 2024년 9월 6일
Hi,
I'm trying to find the electron current using the relation J=sigma[E+uxB]
for which my code gives an error
Unable to perform assignment because the left and right sides have a different number of elements.
Error in pde2fshear_fisher_new>pdex2pde (line 738)
je(STEP_x)=sig*(DuDx(1)+cross(ue,Bt));
which I'm unable to debug it
my code is as below
function [c,f,s] = pdex2pde(x,t,u,DuDx)
c = [1;1;1];STEP_x=2; zone=1;B0=4617;a=.9;R0=6; ue = [1.0;0;0];
xmin = 0;xmax = 1;xstep = 100; sig=1;
X = linspace(xmin,xmax,xstep-1);
if STEP_x==1
Btx=B0/(R0/a+zone*X(STEP_x));
Bt=[Btx;0;0];
je(STEP_x)= sig*(DuDx(1)+cross(ue,Bt));
elseif STEP_x > 1 && STEP_x <= 5
Btx=B0/(R0/a+zone*X(STEP_x));
disp(Btx);
Bt=[Btx;0;0];
je(STEP_x)=sig*(DuDx(1)+cross(ue,Bt));
else
Btx=B0/(R0/a+zone*X(STEP_x));
Bt=[Btx;0;0];
je(STEP_x)=sig*(DuDx(1)+cross(ue,Bt));
end
end

답변 (1개)

Torsten
Torsten 2024년 9월 6일
이동: Torsten 2024년 9월 6일
cross(...) usually gives a vector as result, but je(STEP_x) is a scalar. You can't assign a vector to a scalar.
a(1) = cross([1 2 3],[4 5 6])
Unable to perform assignment because the indices on the left side are not compatible with the size of the right side.
  댓글 수: 3
Walter Roberson
Walter Roberson 2024년 9월 6일
je(STEP_x) = sig*(DuDx(1) + norm(cross(ue,Bt)));
Question:
What is the difference between your three branches? The only difference I can see is that you have an extra disp(Btx); in one of the branches.
Torsten
Torsten 2024년 9월 6일
But I wonder if it make sense or not?
I wonder how I could tell.

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Programming에 대해 자세히 알아보기

태그

제품


릴리스

R2024a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by