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 :
Show Hint
Always use the Right Hand Thumb Rule carefully to check if fields from different segments reinforce each other or cancel out.
Step 1: Understanding the Concept:
The total magnetic field at O is the vector sum of the fields produced by the circular loop and the infinitely long straight wire.
Step 2: Key Formula or Approach:
1. Field due to a circular loop at its center: \( B_{loop} = \frac{\mu_0 I}{2r} \)
2. Field due to an infinitely long straight wire at distance $r$: \( B_{wire} = \frac{\mu_0 I}{2 \pi r} \)
Step 3: Detailed Explanation:
Based on the current direction in the diagram:
- The current in the loop flows in a counter-clockwise direction. By the right-hand thumb rule, the magnetic field at the center O is pointing out of the page (which corresponds to the positive x-axis in this specific coordinate setup, \( \hat{i} \)).
- The current in the straight wire also produces a magnetic field at distance $r$ (center O) that points in the same direction (\( \hat{i} \)).
The net field is:
\[ \vec{B}_{net} = \vec{B}_{loop} + \vec{B}_{wire} \]
\[ \vec{B}_{net} = \left( \frac{\mu_0 I}{2r} + \frac{\mu_0 I}{2 \pi r} \right) \hat{i} \]
\[ \vec{B}_{net} = \frac{\mu_0 I}{2 \pi r} (\pi + 1) \hat{i} \]
Step 4: Final Answer:
The magnetic field is \(\frac{\mu_0 I}{2 \pi r} (\pi + 1) \hat{i}\).
