Recent content by sittingOnTheFloor
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Cylindrical Vector Field Equation Convsersion to Cartesian
I guess I misunderstood what you meant by expressing x and y in terms of cylinder coordinates then. Could you explain more what you meant by that? And also how do the trig functions make it not completely general?- sittingOnTheFloor
- Post #17
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
Ah right well here's where the context of the problems comes in cause this is actually for a program I have to write so as long as it is right it doesn't matter, and it is necessitated that I do keep it in terms of x, y, and z. But if that is right then I would like to thank you a whole whole...- sittingOnTheFloor
- Post #15
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
okay, that makes more sense. I can just replace ##\theta## with ##\arctan{\frac {y} {x}}## right, leaving me with \begin{equation} B(x,y,z) = (- \frac {B_1} {2} {\sqrt{x^2 + y ^ 2}}) ({\cos({{\arctan{\frac {y}{x}}}}) \hat x + \sin({{\arctan{\frac {y}{x}}}}) \hat y + 0\hat z}) + ({B_0 +...- sittingOnTheFloor
- Post #13
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
By second term do you mean the arctan? That is what I am most unsure about, because you had said that I do need to take theta into account but from the original equation it is 0, so instead of the arctan should it be 0?- sittingOnTheFloor
- Post #11
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
Well I am pretty sure the matrix multiplication for this looks like \begin{bmatrix} \cos{\theta} \hat x + \sin{\theta} \hat y + 0\hat z \\ -\sin{\theta} \hat x+ \cos{\theta} \hat y + 0 \hat z \\ 0 \hat x + 0 \hat y + 1 \hat z \end{bmatrix} so does that mean that each row of that matrix is what...- sittingOnTheFloor
- Post #9
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
I am considering ##r## to be equal to ##\sqrt{x^2 + y^2}## and ##\hat r## to be ##\cos{\theta}\sin{\theta}\hat x## which doesn't sit well with me but that is my (knowingly wrong) reading of the attached pictures- sittingOnTheFloor
- Post #7
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
shoot didn't properly copy what I had written out \begin{equation} B(x,y,z) = - \frac {B_1} {2} {\sqrt{x^2 + y ^ 2}} {\cos{\theta}} {\sin{\theta}}\hat{x} - {\sin{\theta}} {\cos{\theta}} \hat{y} + (B_0 + B_1z)\hat{z} \end{equation} You said the r term was wrong though too did you just mean...- sittingOnTheFloor
- Post #5
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
okay this is maybe stemming from my misunderstanding of the ##\theta## term. Are you saying it would be \begin{equation} B(x,y,z) = - \frac {B_1} {2} {\sqrt{(x^2 + y ^ 2}} \hat{x} - {\sin{\theta}} {\cos{\theta}} \hat{y} + (B_0 + B_1z)\hat{z} \end{equation} because this doesn't feel right to...- sittingOnTheFloor
- Post #3
- Forum: Introductory Physics Homework Help
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Cylindrical Vector Field Equation Convsersion to Cartesian
Homework Statement I have been given a changing magnetic field in cylindrical coordinates. The equation is: \begin{equation} B(r,\phi,z) = - \frac {B_1} {2} r \hat{r} + (B_0 + B_1z)\hat{z} \end{equation} I need to be able to find the magnetic field as a function of x, y, and z. Homework...- sittingOnTheFloor
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- Cartesian Cylindrical Field Vector Vector field
- Replies: 17
- Forum: Introductory Physics Homework Help