Consider the following current carrying structure. Find the magnetic field at the centre. Given that r₁ = 2π units and r₂= 4π units. Assume current divides equally

The correct answer is 5 × 10^-8 T.

1. Image Analysis & Given Data:
Based on the provided image, we observe the following details:
• A total current of $I=4A$ enters from the left and splits into two parallel paths.
• The upper path contains a semicircle of radius $r_2=4\pi$ units.
• The lower path contains a smaller semicircle of radius $r_1=2\pi$ units, joined by straight horizontal segments.
• The incoming and outgoing straight wires, as well as the horizontal segments connecting to the lower semicircle, lie along the line passing through the center point.

2. Current Distribution:
Since the current divides equally between the two branches:
$I_1=I_2=\frac{I}{2}=\frac{4A}{2}=2A$

3. Magnetic Field Contribution of Straight Segments:
All straight wire segments (the incoming wire, the outgoing wire, and the horizontal segments leading to and from the inner semicircle) are collinear with the center point. According to the Biot-Savart law, the magnetic field produced by any straight current element along its axis is zero:
$B_{\text{straight}}=0$

4. Magnetic Field due to Semicircular Arcs:
The magnetic field at the center of a semicircular arc of radius $r$ carrying current $i$ is given by:
$B=\frac{{\mu }_{0}i}{4r}$
where ${\mu }_0=4\pi \times {10}^{-7}T·\text{m/A}$.

Using the right-hand rule, we determine the directions of the fields:
• For the lower semicircle of radius $r_1$, the current flows in a clockwise direction, producing a magnetic field $B_1$ pointing into the page (represented as negative):
$B_1=\frac{{\mu }_0{I}_1}{4r_1}=\frac{(4\pi \times {10}^{-7})\times 2}{4\times 2\pi }={10}^{-7}T$

• For the upper semicircle of radius $r_2$, the current flows in a counter-clockwise direction, producing a magnetic field $B_2$ pointing out of the page (represented as positive):
$B_2=\frac{{\mu }_0{I}_2}{4r_2}=\frac{(4\pi \times {10}^{-7})\times 2}{4\times 4\pi }=0.5\times {10}^{-7}T=5\times {10}^{-8}T$

5. Net Magnetic Field at the Centre:
Since the two magnetic fields oppose each other, the net magnetic field magnitude is:
$B_{\text{net}}=B_1-B_2$
$B_{\text{net}}={10}^{-7}T-0.5\times {10}^{-7}T=0.5\times {10}^{-7}T=5\times {10}^{-8}T$