A current \(I_0\) flows through a metallic circular loop of radius \(r\) as shown.
The resistance of arc \(ABC\) is half that of arc \(ADC\).
Find the magnetic field at the centre \(O\).
Show Hint
For parallel branches, current divides inversely proportional to resistance.
Magnetic field due to a semicircle:
\[
B=\frac{\mu_0 I}{4r}
\]
Always check whether the fields add or subtract.
Step 1: Set up the split. The current $I_0$ enters the loop and divides between arc $ABC$ and arc $ADC$. We are told $R_{ABC} = \tfrac12 R_{ADC}$. Let $R_{ADC} = R$, so $R_{ABC} = R/2$. Step 2: Share the current. Parallel branches share current inversely with resistance. The lower-resistance arc $ABC$ carries the larger share. \[ I_{ABC} = I_0\frac{R}{R + R/2} = \frac{2I_0}{3}, \qquad I_{ADC} = I_0\frac{R/2}{R + R/2} = \frac{I_0}{3} \] Step 3: Field from a semicircular arc. Each arc behaves like a semicircle at the centre, giving $B = \dfrac{\mu_0 I}{4r}$. Step 4: Field from arc $ABC$. \[ B_1 = \frac{\mu_0}{4r}\cdot\frac{2I_0}{3} = \frac{\mu_0 I_0}{6r} \] Step 5: Field from arc $ADC$. \[ B_2 = \frac{\mu_0}{4r}\cdot\frac{I_0}{3} = \frac{\mu_0 I_0}{12r} \] Step 6: Combine the opposing fields. The two arcs carry current in opposite senses around the centre, so the fields subtract. \[ B = B_1 - B_2 = \frac{\mu_0 I_0}{6r} - \frac{\mu_0 I_0}{12r} = \frac{\mu_0 I_0}{12r} \] \[ \boxed{\dfrac{\mu_0 I_0}{12r}} \]
