A current I_0 flows through a metallic circul | vedclass

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(A) Let $R_1$ be the resistance of segment $ABC$ and $R_2$ be the resistance of segment $ADC$. Given $R_1 = \frac{1}{2} R_2$, or $R_2 = 2R_1$.Since th

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$\frac{\mu_0 I_0}{12r}$

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(A) Let $R_1$ be the resistance of segment $ABC$ and $R_2$ be the resistance of segment $ADC$. Given $R_1 = \frac{1}{2} R_2$, or $R_2 = 2R_1$.Since the segments are in parallel, the potential difference across them is the same, so $I_1 R_1 = I_2 R_2$, where $I_1$ and $I_2$ are currents in segments $ABC$ and $ADC$ respectively.Substituting $R_2 = 2R_1$, we get $I_1 R_1 = I_2 (2R_1)$, which implies $I_1 = 2I_2$.Also, $I_1 + I_2 = I_0$. Substituting $I_1 = 2I_2$, we get $3I_2 = I_0$, so $I_2 = \frac{I_0}{3}$ and $I_1 = \frac{2I_0}{3}$.Assuming the loop is a semicircle, the magnetic field at the center due to a circular arc of angle $	heta$ is $B = \frac{\mu_0 I 	heta}{4\pi r}$.For segment $ABC$ (semicircle, $	heta = \pi$), $B_1 = \frac{\mu_0 I_1 \pi}{4\pi r} = \frac{\mu_0 I_1}{4r} = \frac{\mu_0 (2I_0/3)}{4r} = \frac{\mu_0 I_0}{6r}$ (directed into the page).For segment $ADC$ (semicircle, $	heta = \pi$), $B_2 = \frac{\mu_0 I_2 \pi}{4\pi r} = \frac{\mu_0 I_2}{4r} = \frac{\mu_0 (I_0/3)}{4r} = \frac{\mu_0 I_0}{12r}$ (directed out of the page).The net magnetic field $B_{net} = |B_1 - B_2| = |\frac{\mu_0 I_0}{6r} - \frac{\mu_0 I_0}{12r}| = \frac{\mu_0 I_0}{12r}$.

$A$ current $I_0$ flows through a metallic circular loop of radius $r$. The resistance of the segment $ABC$ is half that of the segment $ADC$. The magnitude of the magnetic field at the centre $O$ of the loop is:

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