# Question #a6a34

Oct 5, 2016

$B = \textcolor{red}{\frac{7}{16}} \frac{\mu I}{R}$

#### Explanation:

We need only add the components due to the different part circles.

For a full circle of radius R' we have $B = \frac{\mu I}{2 R '}$ as per the attachment.

For BC, which is only a quarter circle of radius $2 R$:

${B}_{B C} = \frac{1}{4} \cdot \frac{\mu I}{2 \left(2 R\right)} = \frac{1}{16} \frac{\mu I}{R}$

For DA, we do likewise

${B}_{D A} = \frac{3}{4} \cdot \frac{\mu I}{2 R} = \frac{3}{8} \frac{\mu I}{R}$

And adding $\implies \frac{7}{16} \frac{\mu I}{R}$

NB
We can do it this simply because of the consat radius and because $d \vec{L} \times \hat{r}$ is the same all along a circle: it is the tangent $d \vec{L}$ vector crossed with unit radial $\hat{r}$ vector. So we are merely adding the components of the $d \vec{B}$'s for each circle

The direction of the $\vec{B}$ field will be into the page as the right hand rule would indicate when applied to this cross product.