Physics - Magnetism Question with Solution | TestHub

The conductor can be divided into three parts: the semi-infinite straight wire PQ, the circular arc QRS, and the semi-infinite straight wire ST.

1. Magnetic field due to the straight wire PQ at O:
The extended line of the wire PQ passes through the center O. Therefore, the angle between the current element $d\vec{l}$ and the position vector $\vec{r}$ is zero. Thus, the magnetic field produced by this segment at O is zero.
${\vec{B}}_{PQ}=0$

2. Magnetic field due to the circular arc QRS at O:
The arc QRS subtends an angle of $360^{\circ }-90^{\circ }=270^{\circ }=\frac{3\pi }{2}$ radians at the center O. The current flows in a counterclockwise direction. According to the right-hand thumb rule, the magnetic field at O is directed outwards, i.e., in the $+\hat{k}$ direction.
${\vec{B}}_{QRS}=\frac{{\mu }_{0}i}{4\pi r}\theta \hat{k}=\frac{{\mu }_{0}i}{4\pi r}\left(\frac{3\pi }{2}\right)\hat{k}$

3. Magnetic field due to the semi-infinite straight wire ST at O:
The wire ST is a semi-infinite straight wire starting from S and extending to infinity. The perpendicular distance from O to the wire is $r$. The point O lies exactly above the starting point S of the wire. The angles subtended by the ends of the wire at O are ${\theta }_1=90^{\circ }$ (for the end at infinity) and ${\theta }_2=0^{\circ }$ (for the end at S). The current flows to the right, so by the right-hand thumb rule, the magnetic field at O is directed outwards ($+\hat{k}$).
${\vec{B}}_{ST}=\frac{{\mu }_{0}i}{4\pi r}(\sin 90^{\circ }+\sin 0^{\circ })\hat{k}=\frac{{\mu }_{0}i}{4\pi r}(1+0)\hat{k}=\frac{{\mu }_{0}i}{4\pi r}\hat{k}$

Total magnetic field at O is the vector sum of the magnetic fields due to all three parts:
${\vec{B}}_{net}={\vec{B}}_{PQ}+{\vec{B}}_{QRS}+{\vec{B}}_{ST}$

${\vec{B}}_{net}=0+\frac{{\mu }_{0}i}{4\pi r}\left(\frac{3\pi }{2}\right)\hat{k}+\frac{{\mu }_{0}i}{4\pi r}\hat{k}$

${\vec{B}}_{net}=\frac{{\mu }_{0}i}{4\pi r}\left(\frac{3\pi }{2}+1\right)\hat{k}$

Answer: $\frac{{\mu }_0\mathrm{i}}{4\pi \mathrm{r}}\left(\frac{3\pi }{2}+1\right)\hat{\mathrm{k}}$