Different forms of Biot-Savart Law

[guyvsdcsniper]

Homework Statement

What expressions give the magnetic field at the point r due to the moving charge?

Relevant Equations

Biot-Savart Law

My book never references the Biot-Savart law with in terms of 1/r^3, only 1/r^2.

How is it that it can be expressed as 1/r^3 as well

 

[ergospherical]

$\hat{\boldsymbol{r}}/r^2 = \boldsymbol{r}/r^3$

 

[guyvsdcsniper]

$\hat{\boldsymbol{r}}/r^2 = \boldsymbol{r}/r^3$

So after look at some problems I think I kinda get why.

Originally the equation is V x r^hat and that is being used when we just want to know the magnetic field at a point. So you cross product those two values and you will get the direction.

In this problem, at the point (x1,0,z1) we use the actual r vector which is in the direction of k^hat and i^hat. If we cross product those two values with k^hat, the direction the the particle is moving, we are left with j^hat.

So if the point we want to evaluate lies at a point in that has more than one direction vector, we use r/r^3 ?

 

[ergospherical]

You can use whichever form you want / is more convenient.

 

[guyvsdcsniper]

You can use whichever form you want / is more convenient.

So my homework used the formula with r/r^3 which I guess I can see is more convenient.

But when I use the r^hat/r^2 I get a different answer

Just focusing on the part of the equation that deals with the cross product and the distance, my distance isn't raised to the 3/2 power. Am I missing a step?

I know this may seem trivial or unnecessary but I just want to make sure how to properly use both equations.

 

[ergospherical]

$\hat{\boldsymbol{r}} = \boldsymbol{r}/r = (x \boldsymbol{i} + z \boldsymbol{k})/\sqrt{x^2 + z^2}$, as opposed to just $x \boldsymbol{i} + z \boldsymbol{k}$ as you wrote...

 

[guyvsdcsniper]

$\hat{\boldsymbol{r}} = \boldsymbol{r}/r = (x \boldsymbol{i} + z \boldsymbol{k})/\sqrt{x^2 + z^2}$, as opposed to just $x \boldsymbol{i} + z \boldsymbol{k}$ as you wrote...

Oh mannnn I just needed to think a little bit harder. That make so much sense. A unit vector is a vector divided by its magnitude. If I used my approach the problem above, my distance would be x^2+z^2 but then when I use the unit vector the denominator (x^2+z^2)^1 gets multiplied by (x^2+z^2)^1/2 and I ultimately end up with ^3/2.

That makes so much sense. Thank you so much.