Physics: Biot_Savort Law Bent wire

Question

An infinitely long wire carrying a current I is bent at right angle: \begin{align*} dB&=\frac{\mu_0}{4\pi}I\frac{d\mathbf{s} \times \mathbf{\hat{r}}}{r^2} \\ \int dB&= \frac{\mu_0}{4\pi}I\int\frac{d\mathbf{s}}{x^2} \end{align*} Sorry but I can't provide a picture, its like this. Am I on the right track?

Answer 1

I assume you are asked to find the magnetic field at $P$ and that $P$ is at coordinates $(x_0,0)$ in the frame where the wire crner is at the origin and the wire lies along the negative $X$ axis and the negative $Y$ axis.

Since $ds$ is taken as a length along the wire, you will need to split your integral into two integrals:

$$ B = \frac{\mu_0}{4\pi} I \int_{y=-\infty}^0 \frac{\hat{y}\, dy \times \hat{r}(y)}{|r(y)|^2} + \frac{\mu_0}{4\pi} I \int_{x=0}^{-\infty} \frac{\hat{x}\, dx \times \hat{r}(x)}{|r(x)|^2} $$ In the first integral $ds$ is a vector in the $+y$ direction. In the second integral, $ds$ is a vector in the $-x$ direction.

It is easy to see that $\hat{r}(x) = \hat{x}$, amd that along this path $|r(x)|= x_0-x$. So the $x$-travelling part of the path contributes zero to the integral because $\hat{x}\times\hat{x} = 0$.

By Pythagoras, $|r(y)|^2 = y^2+x_0^2$. A bit of work with the triangle $yOP$ tells you that $$ \hat{y}\, dy \times \hat{r}(y) = 1\cdot 1 \cdot \sin(\angle OyP) = \frac{x_0}{|r(y)|} $$ So the integral becomes $$ B = \frac{\mu_0}{4\pi} I \int_{y=-\infty}^0 \frac{\frac{x_0}{|r(y)|}} {|r(y)|^2}dy= \frac{\mu_0}{4\pi} I \int_{y=-\infty}^0 \hat{z} x_0 \left( y^2+x_0^2 \right)^{-3/2}dy $$ That integral is just a little bit tricky because the lower limit is infinite, but the indefinite integral will look like $\frac{y}{x_0^2\sqrt{y_2+x_0^2}}$ and that can be re-rewritten as $$ B = -\hat{z}\frac{\mu_0}{4\pi} I x_0\lim_{y\to-\infty} \frac{1}{x_0^2\sqrt{1+x_0^2/y^2}} =-\hat{z}\frac{\mu_0}{4\pi} \frac{I}{x_0} $$