Electromotive Force Voltage Example

조회 수: 6(최근 30일)
Kutlu Yigitturk 2021년 6월 18일
답변: Mouhamed Niasse 2021년 6월 18일
B = [0 0 1]'; v = [0 1 0]'; length = [1 1 1]'; length = length/norm(length);
eVec = cross(v,B); eMag = dot(eVec,length); az = 50; el = 15;
close all;
grn = [0 .8 0]; brn = [.8 .5 0];
Fig = figure; Fig.Position = Fig.Position + [-200 0 100 0];
subplot(1,2,1)
plot3([0 B(1)],[0 B(2)],[0 B(3)],'b','linewidth',3); hold on, grid on
plot3([0 v(1)],[0 v(2)],[0 v(3)],'color',grn,'linewidth',3);
plot3([0 eVec(1)],[0 eVec(2)],[0 eVec(3)],'r','linewidth',3);
text(B(1),B(2),B(3)*1.2,'B','color','b')
text(v(1),v(2)*1.2,v(3),'v','color',grn)
text(eVec(1)*1.0,eVec(2)-.2,eVec(3)+.2,'eVec','color','r')
text(-1.2,-1.5,-1,num2str(B),'color','b'), text(-1.2,-1.5,-.2,'B','color','b'),
text(-.5,-1.5,-1,num2str(v),'color',grn), text(-.5,-1.5,-.2,'v','color','b'),
text(0.2,-1.5,-1,num2str(eVec),'color','r'),text(0.2,-1.7,-.2,'eVec','color','r'),
text(0.9,-1.5,-1,num2str(length,4),'color',brn), text(0.9,-1.6,-.3,'Rod','color',brn)
title('eVec = cross(v,B) eMag = dot(eVec,length)')
set(gca,'color',[1 1 1]*.9)
axis('equal'), axis([-1 1 -1 1 -1 1]*1.5), view(az,el)
Rod0 = [-length length]*0.75; Rod = Rod0 + v*0;
plot3(Rod(1,:),Rod(2,:),Rod(3,:),'linewidth',6,'color',brn);
text(Rod(1,1),Rod(2,1),Rod(3,1)+.3,'Rod','color',brn)
text(Rod(1,2)-.4,Rod(2,2),Rod(3,2)-.3,['eMag = ',num2str(eMag)],'color',brn)
xlabel('x'), ylabel('y'), zlabel('z')
subplot(1,2,2)
plot3([0 B(1)],[0 B(2)],[0 B(3)],'b','linewidth',1); hold on, grid on
plot3([0 v(1)],[0 v(2)],[0 v(3)],'g','linewidth',1);
plot3([0 eVec(1)],[0 eVec(2)],[0 eVec(3)],'r','linewidth',1);
set(gca,'color',[1 1 1]*.9)
axis('equal'), axis([-1 1 -1 1 -1 1]*2),view(az,el)
xlabel('x'), ylabel('y'), zlabel('z')
BFldx = [0 0 -.02 .02 0 0 0 0]; BFldy = [0 0 0 0 0 -.02 .02 0]; BFldz = [-1 1 0.9 0.9 1 .9 .9 1]*1.5;
for i = -2:0.5:2
for j = -2:0.5:2
plot3(BFldx'+i,BFldy'+j,BFldz','linewidth',0.1,'color',[.5 .5 1]);
end
end
drawnow
DrawRod = plot3(Rod(1,:),Rod(2,:),Rod(3,:),'linewidth',6,'color',brn);
TexteMag = text(Rod(1,2)+.1,Rod(2,2)+.1,Rod(3,2)+.1,['+',num2str(eMag)],'color',brn);
Tit = title(['eVec^t = [',num2str([0 0 0]),'] eMag = ',num2str(0)]);
for n = 1:3
Rod = Rod0;
set(DrawRod,'xdata',Rod(1,:),'ydata',Rod(2,:),'zdata',Rod(3,:));
set(TexteMag,'pos',Rod(:,2)+.1,'string',num2str(0))
set(Tit,'string',['eVec^t = [',num2str([0 0 0]),'] eMag = ',num2str(0)]), pause(1)
for m = 0:0.02:1
eVec = cross(v,B); eMag = dot(eVec,length);
Rod = Rod0 + v*m;
set(DrawRod,'xdata',Rod(1,:),'ydata',Rod(2,:),'zdata',Rod(3,:));
set(TexteMag,'pos',Rod(:,2)+.1,'string',num2str(eMag))
set(Tit,'string',['eVec^t = [',num2str(eVec'),'] eMag = ',num2str(eMag)])
drawnow, pause(0.001)
end
end
OUTPUT:
What exactly do eVec and e in this code? Thanks for your help.

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채택된 답변

Mouhamed Niasse 2021년 6월 18일
Hello,
From the second line of your code, you need to know first what is the physical meaning of each paramaters ('v', 'B', 'length', etc).
eVec = cross(v,B); eMag = dot(eVec,length);
Then you go into MATLAB documentation and read the description of the used functions 'cross' and 'dot'.
Repeat the same process for the rest of the code and you will soon become an expert.

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