Physics - Magnetism Question with Solution | TestHub

From the given figure, the dashed line passing through A, O, and C is straight, indicating that the segments ABC and ADC are semicircles.

Let the resistance of segment ADC be $R_{ADC}=2R$. Given that the resistance of segment ABC is half that of ADC, we have $R_{ABC}=R$.

The total current $I_0$ splits at point A into two parallel branches. Let $I_1$ be the current through ABC and $I_2$ be the current through ADC.

Since the branches are in parallel, the potential difference across them is equal:
$I_1{R}_{ABC}=I_2{R}_{ADC}$
$I_1(R)=I_2(2R)\Rightarrow I_1=2I_2$

Applying Kirchhoff's current law at junction A:
$I_1+I_2=I_0$
$2I_2+I_2=I_0\Rightarrow 3I_2=I_0\Rightarrow I_2=\frac{I_0}{3}$
Consequently, $I_1=\frac{2I_0}{3}$.

The magnetic field at the center of a semicircle carrying current $I$ is given by $B=\frac{{\mu }_{0}I}{4r}$.

For the left semicircle ABC, the current $I_1$ flows in a counter-clockwise direction. By the right-hand rule, its magnetic field at O is directed out of the plane:
$B_1=\frac{{\mu }_0{I}_1}{4r}=\frac{{\mu }_0}{4r}\left(\frac{2I_0}{3}\right)=\frac{{\mu }_0{I}_0}{6r}$

For the right semicircle ADC, the current $I_2$ flows in a clockwise direction. Its magnetic field at O is directed into the plane:
$B_2=\frac{{\mu }_0{I}_2}{4r}=\frac{{\mu }_0}{4r}\left(\frac{I_0}{3}\right)=\frac{{\mu }_0{I}_0}{12r}$

The straight wire segments carrying current $I_0$ into A and out of C lie on the line passing through O, so they do not contribute to the magnetic field at O.

The net magnetic field at the center O is the difference between $B_1$ and $B_2$ since they are in opposite directions:
$B_{net}=B_1-B_2=\frac{{\mu }_0{I}_0}{6r}-\frac{{\mu }_0{I}_0}{12r}=\frac{2{\mu }_0{I}_0-{\mu }_0{I}_0}{12r}=\frac{{\mu }_0{I}_0}{12r}$

Answer: $\frac{{\mu }_0{I}_0}{12r}$