As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\mathrm{\pi }}{10}$ is carrying current I=3A. The magnitude of the magnetic field at the centre O of the arc is: (The permeability of the vacuum $=4\mathrm{\pi }\times {10}^{-7}{\mathrm{NA}}^{-2}$)
 

The effective magnetic field at ‘0’ due to total wire = 
magnetic field at 0 due to I+ 
magnetic field at 0 due to II+ 
magnetic field at 0 due to III
${\left({\mathrm{B}}_{\mathrm{I}}\right)}_{\mathrm{at}0}=0,(\mathrm{B}{)}_{\mathrm{II}}=\frac{{\mathrm{\mu }}_0\mathrm{i\theta }}{4\mathrm{\pi r}}=\frac{{\mathrm{\mu }}_0\mathrm{i\pi }}{4\mathrm{\pi r}}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 3\times \mathrm{\pi }}{4\mathrm{\pi }\times \frac{\mathrm{\pi }}{10}}=3\times {10}^{-6}$
$\begin{array}{l}(\mathrm{B}{)}_{\text{IIlato }}=0\\ \mathrm{B}={\mathrm{B}}_{\mathrm{I}}+{\mathrm{B}}_{\mathrm{II}}+{\mathrm{B}}_{\mathrm{III}}=3\mathrm{\mu T}\end{array}$