EXAMPLE $7\mathrm{.}2$

Find $H$ at $(-3,4,0)$ due to the current filament shown in Figure $7\mathrm{.}7($a$).$

Solution:

Let $H=H_1+H_2,$ where $H_1$ and $H_2$ are the contributions to the magnetic field intensity at $P(-3,4,0)$ due to the portions of the filament along $x$ and $z,$ respectively.

$H_2=\frac{\text{I}}{4}(cos{}_2-cos{}_1)a_{}$

At $P(-3,4,0),=(9+16{)}^{\frac{1}{2}}=5,{}_1=180,{}_2=90,$ and $a_{}$ is obtained as a unit vector along the circular path through $P$ on plane $z=0$ as in Figure $7\mathrm{.}7($b$).$ The direction of $a_{}$

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CHAPTER $7$ MAGNETOSTATIC FIELDS