Figure here shows three cases : in all cases the circular path has radius r and straight ones are infinitely long. For same current the magnetic field at the centre P in cases 1, 2, 3 have the ratio

 

For first figure magnetic field at O due to straight part (a),

$B_a=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}$ 

Magnetic field at O due to circular part (b)

$B_b=\frac{{\mu }_0\pi i}{4\pi }$ 

Magnetic field at O due to straight part (c)

$B_c=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}$

So, net magnetic field at the centre of first figure,

$B_1=B_b+B_c-B_a=\frac{{\mu }_0}{4\pi }\cdot \frac{\pi i}{r}$                  (i)

For second figure, magnetic field at O.due to two straight and one circular part is

$B_2=\frac{{\mu }_0}{4\pi }\frac{\pi i}{r}$                                          (ii)

For third figure magnetic field at O due to straight part (a)

$B_a=0$

Magnetic field at O due to circular part (b)

$B_b=\frac{{\mu }_0(2\pi -\pi /2)i}{4\pi r}$

Magnetic field at C due to straight part ( c)

$B_c=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}$

So, net magnetic field at centre of third figure

$B_3=\frac{{\mu }_0}{4\pi }\cdot \frac{i}{r}\left(\frac{3\pi }{2}-1\right)$                         (iii)

From Eqs. (i), (ii) and (iii), we obtain

$$\begin{gathered} B_1:B_2:B_3=\pi :\pi :\left(\frac{3\pi }{2}-1\right) \\ =\left(-\frac{\pi }{2}\right):\left(\frac{\pi }{2}\right):\left(\frac{3\pi }{4}-\frac{1}{2}\right) \end{gathered}$$