Q.
A very long wire $ABDMNDC$ is shown in figure carrying current $I$. $AB$ and $BC$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $DMND$ of radius $R$.
$AB, BC$ parts are tangential to circular turn at $N$ and $D$. Magnetic field at the centre of circle is :
Solution:
We say we have 3 parts $(A, B, C)$
$B=B_{A}+B_{B}+B_{C}$
$=\frac{\mu_{0}\,I}{2\pi R}\left(sin\,90^{\circ}-sin\,45^{\circ}\right)\otimes+\frac{\mu_{0}\,I}{2R}⊙\frac{\mu _0\,I}{4\pi R}\left(sin\,45^{\circ}+sin\,90^{\circ}\right)⊙$
$=\frac{\mu_{0}\,I}{2\pi R}\left(sin\,45^{\circ}+\pi\right)$
$=\frac{\mu_{0}\,I}{2\pi R}\left(\pi-\frac{1}{\sqrt{2}}\right)$
