Question: Find the magnetic field at point Pfor each of the steady current configurations shown in Fig. 5.23.

(a) A quarter circular ring of inner radius a and outer radius b and carrying current l .

(b) A semi circular wire of radius R extending to infinity at each end and carrying current l .

The field due to a circle of radius and carrying current at the center is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\mu }_0$is the permeability of free space.

The field due to an infinite straight wire carrying current l at a distance R from it is

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{2}\mathbf{\pi R}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

In the first figure, the straight sections produce no field at P because their extended sections pass through it.

From equation (1), the field from the inner ring is

$$\begin{gathered} B=\frac{1}{4}\times \frac{{\mu }_{0}l}{2a} \\ =\frac{{\mu }_{0}l}{8a} \end{gathered}$$

This field is pointed outward according to right hand rule.

From equation (1), the field from the outer ring is

role="math" localid="1657774429401" $$\begin{gathered} B=\frac{1}{4}\times \frac{{\mu }_{0}l}{2b} \\ =\frac{{\mu }_{0}l}{8b} \end{gathered}$$

This field is pointed inward according to right hand rule.

Thus, the net field at P is

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{l}}{\mathbf{8}}\left(\frac{1}{a}-\frac{1}{b}\right)$

The field is pointed outward.

The net field at P is $\frac{{\mathit{\mu }}_{\mathbf{0}}\mathit{l}}{\mathbf{8}}\mathbf{(}\frac{\mathbf{1}}{\mathit{a}}\mathbf{-}\frac{\mathbf{1}}{\mathit{b}}\mathbf{)}$pointed outward.

The two half infinite sections at the top and bottom of the second figure form an infinite wire with field from equation (2) at P

$B=\frac{{\mu }_{0}l}{2\pi R}$

The field from the semi circular section from equation (1) at P is

$$\begin{gathered} B=\frac{1}{2}\times \frac{{\mu }_{0}l}{2R} \\ =\frac{{\mu }_{0}l}{4R} \end{gathered}$$

Both of these fields are pointed inwards. Thus the net field at P is

$$\begin{gathered} B=\frac{{\mu }_{0}l}{2\pi R}+\frac{{\mu }_{0}l}{4R} \\ =\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\pi }\right) \end{gathered}$$

Thus, the net field at P is $\frac{{\mu }_{0}l}{4R}\left(1+\frac{2}{\pi }\right)$pointed inwards.