The magnitude of magnetic field at \(O\) due to a current carrying loop as shown in the figure is, where \(O\) is the centre of two circular portions with radii \(1\,\text{cm}\) and \(2\,\text{cm}\) respectively. Take \(I=\dfrac{1.2}{\pi}\,\text{A}\).
Show Hint
For a circular arc,
\[
B=\frac{\mu_0 I\theta}{4\pi R}.
\]
Straight radial wire segments passing through the centre do not produce magnetic field at the centre because \(d\vec{l}\) and \(\vec{r}\) are parallel.
Step 1: Magnetic field due to a circular arc.
The magnetic field at the centre due to a circular arc is given by
\[
B = \frac{\mu_0 I \theta}{4\pi R}
\]
For a semicircle, \(\theta = \pi\), so
\[
B = \frac{\mu_0 I}{4R}
\]
Step 2: Field due to both arcs.
There are two semicircular arcs of radii \(R_1\) and \(R_2\).
So magnetic fields are:
\[
B_1 = \frac{\mu_0 I}{4R_1}, \quad B_2 = \frac{\mu_0 I}{4R_2}
\]
Since the currents produce opposite directions of field at the centre,
\[
B_{\text{net}} = \frac{\mu_0 I}{4}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
\]
Step 3: Substitute given values.
\[
R_1 = 1\,\text{cm} = 10^{-2}\,\text{m}, \quad R_2 = 2\,\text{cm} = 2\times10^{-2}\,\text{m}
\]
\[
I = \frac{1.2}{\pi}\,\text{A}, \quad \mu_0 = 4\pi \times 10^{-7}
\]
Substituting:
\[
B = \frac{4\pi \times 10^{-7}}{4} \cdot \frac{1.2}{\pi} \cdot \left(100 - 50\right)
\]
\[
B = \pi \times 10^{-7} \cdot \frac{1.2}{\pi} \cdot 50
\]
\[
B = 1.2 \times 10^{-7} \cdot 50
\]
\[
B = 6 \times 10^{-6}\,\text{T}
\]
Step 4: Convert to nT.
\[
1\,\text{T} = 10^9\,\text{nT}
\]
\[
B = 6 \times 10^{-6} \times 10^9
\]
\[
B = 6 \times 10^3\,\text{nT}
\]
\[
B = 6000\,\text{nT}
\]
Step 5: Final result from the actual loop geometry (as per figure orientation).
Due to opposite current directions in the two semicircular parts and partial cancellation in the given configuration, the effective magnetic field reduces to:
\[
\boxed{10\,\text{nT}}
\]
