Magnetic Field Equation in Spherical Coordinates to Cartesian Coordinates

[jhosamelly]

Homework Statement

The magnetic field around a long, straight wire carrying a steady current I is given in spherical coordinates by the expression

$\vec{B} = \frac{\mu_{o} I }{2∏ R} \hat{\phi}$ ,

where $\mu_{o}$ is a constant and R is the perpendicular distance from the wire to the observation points. Find the expression for $\vec{B}$ in cartesian coordinates.

Homework Equations

The Attempt at a Solution

I know I need to get the partial derivative of this with respect to some variable.. but I don't know what that variable is. can someone help me please?

 

[Mindscrape]

Jacobians are pretty helpful moving between coordinates. You don't necessarily need to go that in depth though. The R is easy. How will the unit vector transform?

 

[jhosamelly]

Jacobians are pretty helpful moving between coordinates. You don't necessarily need to go that in depth though. The R is easy. How will the unit vector transform?

I didn't really get your point.. i mean, is that formula a unit vector?? I think not.

In terms of R. do u mean I should make it

R = $\frac{\mu_{o} I}{2∏ \vec{B}}$ $\hat{\phi}$

 

[paris1244bc]

He means the unit vector in your basis.

 

[jhosamelly]

He means the unit vector in your basis.

basis meaning the observation points..

its

R=Rx i + Ry j + Rz k

 

[vela]

He means $\hat{\phi}$.

 

[jhosamelly]

He means $\hat{\phi}$.

hmmm.. now I'm really confused. Can you please tell me the steps on how to do this, then i'll try. I'll show you what I did then you can tell me if I'm wrong or right. Thanks.

 

[jhosamelly]

Can someone please help me with this one?? This is the last problem I wasn't able to solve in our problem set. Help will be much appreciated. Thanks.

 

[jhosamelly]

Spherical coordinates to cartesian coordinates

Homework Statement

$\vec{B} = \frac{\mu_{o} I }{2∏ R} \hat{\phi}$ , is the equation of Magnetic Field in spherical coordinates. where $\mu_{o}$ is a constant and R is the perpendicular distance from the wire to the observation points. Find the expression for $\vec{B}$ in cartesian coordinates.

Homework Equations

The Attempt at a Solution

I tried to equate this equation in terms of $\hat{\phi}$ but after that I'm stuck.. I know also that R being the perpendicular distance from the wire to the observation points will also do the trick but I don't know how it will help. Can someone please help me figure this thing out? help will be much appreciated

 

[tiny-tim]

hi jhosamelly!

hint: what shape are the field-lines?

 

[jhosamelly]

hi jhosamelly!

hint: what shape are the field-lines?

circular.. do you mean I should use equations for circle?

 

[tiny-tim]

seems a good idea!

what do you get? ​

 

[jhosamelly]

seems a good idea!

what do you get? ​

$(x-a)^2 + (y-b)^2 = r^2$

or if the center is at the origin


$x^2 + y^2 = r^2$


how does this help? should I equate this with the R in the equation?

 

[tiny-tim]

…or if the center is at the origin


$x^2 + y^2 = r^2$

the centre isn't at the origin, is it?

it's anywhere along the z-axis

and it isn't r, it's R …

ok, now that you know what shape everything is, write the original formula for B in spherical coordinates

 

[jhosamelly]

the centre isn't at the origin, is it?

it's anywhere along the z-axis

and it isn't r, it's R …

ok, now that you know what shape everything is, write the original formula for B in spherical coordinates

do you mean this??

$\vec{B} = \frac{\mu_{o} I }{2∏ R} \hat{\phi}$

I think this is already in spherical coordinates.

should i change R now to $x^2 + y^2$

$\vec{B} = \frac{\mu_{o} I }{2∏ (x^2 + y^2)} \hat{\phi}$

 

[tiny-tim]

oops! i meant cartesian coordinates!

should i change R now to $x^2 + y^2$

$\vec{B} = \frac{\mu_{o} I }{2∏ (x^2 + y^2)} \hat{\phi}$

yes, and finally you need to change phi to x and y (or i and j)

 

[jhosamelly]

oops! i meant cartesian coordinates! yes, and finally you need to change phi to x and y (or i and j)

$\vec{B} = \frac{\mu_{o} I }{2∏ (x^2 + y^2)} \hat{\phi}$

$\hat{\phi} = - sin \phi \hat{i} + cos \phi \hat{j}$

$\vec{B} = \frac{\mu_{o} I }{2∏ (x^2 + y^2)} (-sin \phi \hat{i} + cos \phi \hat{j})$

is this it? Thanks for your help )

 

[tiny-tim]

(type \pi for π in latex )

$\vec{B} = \frac{\mu_{o} I }{2∏ (x^2 + y^2)} \hat{\phi}$

$\hat{\phi} = - sin \phi \hat{i} + cos \phi \hat{j}$

$\vec{B} = \frac{\mu_{o} I }{2\pi (x^2 + y^2)} (-sin \phi \hat{i} + cos \phi \hat{j})$

is this it? Thanks for your help )

that's it!

(except for the missing square-root )

(btw, now that you've got the idea, you don't need to find the shape of the field-lines …

that was just to help you visualise everything)

 

[jhosamelly]

(type \pi for π in latex )


that's it!

(except for the missing square-root )

(btw, now that you've got the idea, you don't need to find the shape of the field-lines …

that was just to help you visualise everything)

ow yah.. because its R^2.. hehe.. thanks for the reminder..

$\vec{B} = \frac{\mu_{o} I }{2\pi (\sqrt{(x^2 + y^2)})} (-sin \phi \hat{i} + cos \phi \hat{j})$

is this ok now or should i also change sin to y/r and cos to x/r ??

 

[tiny-tim]

… should i also change sin to y/r and cos to x/r ??

yes!

 

[jhosamelly]

yes!

so the final answer is

$\vec{B} = \frac{\mu_{o} I }{2\pi (\sqrt{(x^2 + y^2)})} (-\frac{y}{r} \hat{i} + \frac{x}{r} \hat{j})$

Thank you so much for your help.. God Bless

 

[tiny-tim]

hold on!

you need to change those r's to x and y also!

 

[jhosamelly]

hold on!

you need to change those r's to x and y also!

ow.. ok. So, the final answer should be

$\vec{B} = \frac{\mu_{o} I }{2\pi (\sqrt{(x^2 + y^2)})} (-\frac{y}{\sqrt{(x^2 + y^2)}} \hat{i} + \frac{x}{\sqrt{(x^2 + y^2)}} \hat{j})$

if i distribute $\frac{\mu_{o} I }{2\pi (\sqrt{(x^2 + y^2)})}$$\vec{B} = - \frac{\mu_{o} I y }{2 \pi (x^2 + y^2)} \hat{i} + \frac{\mu_{o} I x}{2 \pi (x^2 + y^2)} \hat{j}$
Thanks ))) Because you helped me I'll return the favor by helping others here as well as much as I can ;))) Thank you so much

 

[tiny-tim]

and how about multiplying those √s together?

Thanks ))) Because you helped me I'll return the favor by helping others here as well as much as I can ;))) Thank you so much

that's the way it works!

 

[jhosamelly]

and how about multiplying those √s together?

i think I already did this.. can you please look at my final answer to clarify. Thanks )

 

[tiny-tim]

(ohh, you edited!)

yes, that's perfect!

 

[jhosamelly]

(ohh, you edited!)

yes, that's perfect!

maybe I'm not yet done posting my answer when you saw it (I need to save it first to see if all the "codes" for the equations are correct.) Big thanks! God Bless )) See you around.