- #1

BuggyWungos

- 13

- 1

- 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