Module 3 : MAGNETIC FIELD
Lecture 15 : Biot- Savarts' Law

The field due to each straight line section is obtained by putting $\phi_1 = 90^\circ$and $\phi_2 = 0^\circ$in the expression obtained in Example 5 above. The field due to each wire is $\mu_0I/4\pi a$.
For the semi-circular arc, each length element on the circumference is perpendicular to $\vec r$, the vector from the length element to the point P. Thus

\begin{eqnarray*} B_{arc} &=& \frac{\mu_0I}{4\pi}\int \frac{\vec{dl}\times\hat r... ...\int dl = \frac{\mu_0I}{4\pi a^2}\cdot \pi a = \frac{\mu_0I}{4a} \end{eqnarray*}

The net field due to the current in the hairpin bend at P is

\begin{displaymath}B = \frac{\mu_0I}{2\pi a} + \frac{\mu_0 I}{4a}\end{displaymath}

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