A current of $10 A$ flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. The circuit consists of eight alternating arcs of radii $r_1=0.08 m$ and $r_2=0.12 m$. Each subtends the same angle at the centre.

An infinitely long straight wire carrying a current of $10 A$ is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. What is the force acting  on the wire at the centre due to the current in the circuit? What is the force acting on the arc $AC$ and the straight segment $CD$ due to the current at the centre?

Force on $AC$

Force on circular portions of the circuit, i.e., $AC$ etc, due to the wire at the centre will be zero because magnetic field due to the central wire at these arcs will be tangential $\left(\theta =180°\right)$.

Force on $CD$

Current in central wire is also $i=10 A$. Magnetic field at distance $x$ due to central wire $B=\frac{{\mu }_0}{2\mathrm{\pi }}.\frac{i}{x}$

$\therefore$ Magnetic force on element $dx$ due to this magnetic field 

$dF=\left(i\right)\left(\frac{{\mu }_0}{2\mathrm{\pi }}.\frac{i}{x}\right).dx=\left(\frac{{\mu }_0}{2\mathrm{\pi }}\right)i^2\frac{dx}{x}$     $\left[F=ilB \sin 90°\right]$

Therefore,net force on $CD$ is 

$F={\int }_{x=r_1}^{x=r_2}dF=\frac{{\mu }_0{i}^2}{2\mathrm{\pi }}{\int }_{0.08}^{0.12}\frac{dx}{x}=\frac{{\mu }_0}{2\mathrm{\pi }}{i}^2\ln \left(\frac{3}{2}\right)$

Substituting the values,

$F=\left(2\times {10}^{-7}\right)\left({10}^{-2}\right)\ln \left(1.5\right)$

   $=8.1\times {10}^{-6 }N$ (inwards)

Force on wire at the centre

Net magnetic field at the centre due to the circuit is in vertical direction and current in the wire in centre is also in vertical direction. Therefore, net force on the wire at the centre will be zero. $\left(\theta =180°\right)$ Hence, 

(i) Force acting on the wire at the centre is zero.

(ii) Force on arc $AC=0$

(iii) Force on segment $CD =8.1\times {10}^{-6} N$ (inwards)