The magnetic induction at the centre $O$ in the figure shown is
Show Hint
When current loops circulate in opposite directions, their fields fight each other, which means you must subtract them. This immediately eliminates options (b) and (d) from consideration!
Step 1: Understanding the Concept:
This system features concentric circular semi-arcs of different radii ($R_1$ and $R_2$) linked by straight radial connection lines. The straight radial sections point directly toward center $O$, so they do not produce a magnetic field at that point. The two semicircular components carry current in opposite directions around the loop, meaning their magnetic fields oppose each other. Step 2: Key Formula or Approach:
1. Magnetic field at the center of a complete circular loop: $B = \frac{\mu_0 i}{2R}$.
2. For a semicircle ($\theta = \pi$), the magnitude is exactly half of a full circle: $B_{\text{semicircle}} = \frac{\mu_0 i}{4R}$.
3. Net magnetic field: $B_{\text{net}} = |B_1 - B_2|$. Step 3: Detailed Explanation:
Let's evaluate the fields from both semicircular paths individually:
Inner Semicircle (Radius $R_1$): The current creates a field of magnitude:
\[ B_1 = \frac{\mu_0 i}{4R_1} \]
Outer Semicircle (Radius $R_2$): The current creates a field of magnitude:
\[ B_2 = \frac{\mu_0 i}{4R_2} \]
Because the current flows clockwise in one arc and counterclockwise in the other, their vector directions are exactly opposite. Since $R_1 & Lt; R_2$, the field from the closer inner arc is stronger ($B_1 > B_2$).
Subtracting the weaker field from the stronger one gives the net magnetic induction:
\[ B_{\text{net}} = B_1 - B_2 \]
\[ B_{\text{net}} = \frac{\mu_0 i}{4R_1} - \frac{\mu_0 i}{4R_2} \]
\[ B_{\text{net}} = \frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Step 4: Final Answer:
The net magnetic induction at the centre $O$ is $\frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
