The wire shown in Fig. $31-7$ carries a current of $40 \mathrm{~A}$. Find the field at point- $P$.
Since $P$ lies on the lines of the straight wires, those wires contribute no field at $P$. A circular loop of radius $r$ gives a field of $B=\mu_{0} I / 2 r$ at its center point. Here we have only three-fourths of a loop, and so we can assume that
$$\begin{aligned}
B \text { at point- } P &=\left(\frac{3}{4}\right)\left(\frac{\mu_{0} I}{2 r}\right)=\frac{(3)\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)(40 \mathrm{~A})}{(4)(2)(0.020 \mathrm{~m})} \\
&=9.4 \times 10^{-4} \mathrm{~T}=0.94 \mathrm{mT}
\end{aligned}$$
The field is out of the page.