Physics - Magnetism Question with Solution | TestHub

From the figure, the circular part subtended angle at the centre of the circle$=3\pi /2$

$\therefore$Magnetic field at$O$due to the circular part,

$B_1=\frac{{\mu }_0}{2}\cdot \frac{I}{R}\cdot \frac{\theta }{2\pi }$

$=\frac{{\mu }_0}{2}\cdot \frac{I}{R}\cdot \frac{3\pi /2}{2\pi }=\frac{{\mu }_0}{8}\cdot \frac{3I}{R}$...(i)

$\because$Point$O$at the axial position of wire$CD$

$\therefore$It will not produce magnetic field at$O.$

$\because$Point$O$is at the end of the wire$AE.$

$\therefore$Magnetic field due to wire$AE,$

$B_2=\frac{{\mu }_0}{4\pi }\cdot \frac{I}{R}$...(ii)

Thus,$B_{\mathrm{n}\mathrm{e}\mathrm{t}}=B_1+B_2$

Putting the value of$B_1$and$B_1$from equations (i) and (ii), we get

$=\frac{3{\mu }_{0}I}{8R}+\frac{{\mu }_{0}I}{4\pi R}=\frac{{\mu }_0}{4\pi }\cdot \frac{I}{R}\left[\frac{3\pi }{2}+1\right]$