An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
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Always use the Right Hand Thumb Rule carefully to check if fields from different segments reinforce each other or cancel out.
To find the magnetic field at the center \( O \) of the circular loop due to the current-carrying wire, we need to consider the contributions from both the circular and straight portions of the wire.
Let's denote the magnetic field due to the straight parts as \( \mathbf{B}_{\text{straight}} \) and due to the circular part as \( \mathbf{B}_{\text{circle}} \).
According to the Biot-Savart law, the magnetic field due to a circular arc of current at the center of the arc is given by:
\[\mathbf{B}_{\text{circle}} = \frac{\mu_0 I}{4 \pi r} \times 2\pi \left(\frac{\theta}{2\pi}\right)\]
Here, the angle \(\theta = \pi\) since it's a semicircle.
\[\mathbf{B}_{\text{circle}} = \frac{\mu_0 I}{4 \pi r} \times \pi = \frac{\mu_0 I}{4 r} \hat{i}\]
The contribution from the infinite straight wire part can be derived from:
\[\mathbf{B}_{\text{straight}} = \frac{\mu_0 I}{2 \pi r} \arctan\left(\frac{y}{r}\right)\]
For an infinite wire, this becomes:
\[\mathbf{B}_{\text{straight}} = \frac{\mu_0 I}{2 \pi r} \hat{i}\]
Therefore, the net magnetic field at \( O \) is:
\[\mathbf{B} = \mathbf{B}_{\text{circle}} + \mathbf{B}_{\text{straight}}\]\[\mathbf{B} = \frac{\mu_0 I}{4 r} \hat{i} + \frac{\mu_0 I}{2 \pi r} \hat{i}\]
