Magnetic Field of an infinite current-carrying wire at a point

[BuggyWungos]

Homework Statement

Find the magnetic field strength at point P (illustration below)

Relevant Equations

$$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}\times\hat{r}}{r^2}$$

My attempt:

$$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}r\sin{\theta}}{r^2}$$

$$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}\sin{\theta}}{r}$$
$$ \sin{\theta} = \dfrac{y}{(x^2+y^2)^{1/2}}$$

$$ d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}}{r}\dfrac{y}{(x^2+y^2)^{1/2}}$$
$$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}y}{r^2}$$


$$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Iyd\vec{l}}{r^2}$$



This is what I determined to be the magnetic force at P due to $d\vec{I}$


$$B(r) = \int_{-\infty}^{+\infty} d\vec{B}(r)$$


$$B(r) =\int_{-\infty}^{+\infty} \dfrac{\mu_0}{4\pi}\dfrac{Iyd\vec{l}}{r^2}$$


$$B(r) =\dfrac{\mu_0}{4\pi}\dfrac{Iy\vec{l}}{r^2} \Big|_{-\infty}^{+\infty}$$

I don't think my solution is solvable, it just becomes positive infinity minus negative infinity

My professor had a different solution to $d\vec{B}(r)$

I don't understand where $r$ went and how $dx$ was brought forth in the second step

 

[kuruman]

Please fix your LaTeX and make it legible.

My professor had a different solution to $d\vec{B}(r)$

View attachment 349372

I don't understand where $r$ went and how $dx$ was brought forth in the second step

$r$ didn't go anywhere. It is still there as $\sqrt{x^2+y^2}.$
$d\vec l$ is an element in the direction of the current. Its magnitude in this case is $dx$.

 

[BuggyWungos]

Please fix you LaTeX and make it legible.

$r$ didn't go anywhere. It is still there as $\sqrt{x^2+y^2}.$
$d\vec l$ is an element in the direction of the current. Its magnitude in this case is $dx$.

I'm surprised you were able to understand all that before I fixed it.

I understand the second part of your comment, but where is $\sqrt{x^2+y^2}$ in the second step of my prof's solution?

Edit: I would have expected it to end up like $d\vec{B}(r) = \dfrac{\mu_0}{4\pi} \dfrac{Idx\sin{\theta}}{r}$

 

[Steve4Physics]

@BuggyWungos, note that in the expression
$~~~~\dfrac {\mu_0}{4 \pi} \dfrac {I \vec {dl} \times \hat r}{r^2}$
$\hat r$ is the unit vector in the r-direction; it’s magnitude is $1$. Don't confuse it with $\vec r$.

 

[BuggyWungos]

@BuggyWungos, note that in the expression
$~~~~\dfrac {\mu_0}{4 \pi} \dfrac {I \vec {dl} \times \hat r}{r^2}$
$\hat r$ is the unit vector in the r-direction; it’s magnitude is $1$. Don't confuse it with $\vec r$.

Thank you that's such a silly oversight, I even wrote it as r hat in latex and missed it lol

 

[Orodruin]

Not the way you have solved it, but some symmetry often goes a long way.

https://www.physicsforums.com/insights/symmetry-arguments-and-the-infinite-wire-with-a-current/