Calculating Magnetic Field at Point P from a Short Current Element

[mortymoose]

Homework Statement

[/B]
A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.20 A in the same direction as dl⃗ . Point P is located at r⃗ = ( -0.730 m)i^+ (0.390m)k^.
Find the magnetic field at P produced by this current.
So basically what they are asking me for is the x, y, and z components of dB.

Homework Equations

dB= mu0/4pi * (I*(dl x r^)/r^2)
B=Bxi^ +Byj^ +Bzk^

The Attempt at a Solution

dB= 10^-7 * 5.2A *(1.95*10^-4m i^ +3.65*10^-4m k^ )/r^2

So, I cross multiplied dl and r^, now I am completely lost. I don't understand how I am supposed to find the x,y, and z components.
The idea i have is:
use the r^ as r and square it and finish the above equation and solve for dB, and then whatever i^ and k^ component is in the answer is equivalent to my x and z?

So dB=(-2.67*10^-4)i^ +(4.86*10^-10)k^
i^= x k^=z and y=0=j^But i get this answer wrong. Can anyone direct me into the right direction or tell me what I did wrong?

 

[haruspex]

dB= 10^-7 * 5.2A *(1.95*10^-4m i^ +3.65*10^-4m k^ )/r^2

Looks right.

use the r^ as r and square it

Yes. $\vec r^2=\vec r.\vec r=|\vec r|^2$.

dB=(-2.67*10^-4)i^ +(4.86*10^-10)k^

I don't understand how you get these numbers (nor the minus sign). Please post the intermediate steps.

 

[rude man]

They don't tell the location of the current element. You have assumed it is at (0,0,0). OK. Gotta assume something.

But your answer of dB=(-2.67*10^-4)i^ +(4.86*10^-10)k^ has a sign error. (I didn't check your numbers).

 

[mortymoose]

Looks right.

Yes. $\vec r^2=\vec r.\vec r=|\vec r|^2$.

I don't understand how you get these numbers (nor the minus sign). Please post the intermediate steps.

dB= 10^-7 * 5.2A * (1.95*10^-4 i^ + 3.65*10^-4 k^)/r^2

r^2= (-0.73i^ + 0.39k^)^2 = (0.5329 i^ + 0.1521 k^)

Then i plug r^ in...
dB= 1.9027*10^-10 i^ + 1.248*10^-9 k^

Does this look better? I think i must of multiplied something wrong the first time maybe?

 

[rude man]

r^2= (-0.73i^ + 0.39k^)^2 = (0.5329 i^ + 0.1521 k^)

This line makes no sense. r^2 is a scalar, not a vector. But the final result might be OK.
Your previous mistake was to say j x -i = -k.

 

[haruspex]

r^2= (-0.73i^ + 0.39k^)^2 = (0.5329 i^ + 0.1521 k^)

As rude man points out, this is wrong.
$(x \vec i+y\vec j)^2=x^2\vec i.\vec i+2xy\vec i.\vec j+y^2\vec j.\vec j=x^2+y^2$
$\vec i.\vec i=\vec j.\vec j=1$ (dot product is a scalar, and these are unit vectors).
$\vec i.\vec j=0$ ( these vectors are perpendicular).