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 problem requires calculating the magnetic field produced at the center of a composite loop consisting of two concentric semicircular arcs of different radii ($R_1$ and $R_2$) and two radial straight lines. The total magnetic induction is the vector sum of the fields produced by each segment. A fundamental principle of the Biot-Savart Law is that radial segments do not produce a magnetic field at the center of the arc because the angle between the current direction and the position vector is either 0 or 180 degrees. Therefore, we only need to consider the fields from the two semicircles. Step 2: Key Formula or Approach:
1. Magnetic field at the center of a full circular loop: $B = \frac{\mu_0 i}{2R}$.
2. Magnetic field at the center of a semicircle: $B = \frac{\mu_0 i}{4R}$.
3. Total Field ($B_{net}$): Vector sum of the fields from the inner arc and the outer arc.
4. Direction: Determined by the sense of current (Clockwise vs. Counter-Clockwise) using the right-hand grip rule. Step 3: Detailed Explanation:
Let's evaluate the field from the inner semicircular arc of radius $R_1$. The current $i$ passing through it creates a magnetic field at $O$ with magnitude $B_1 = \frac{\mu_0 i}{4R_1}$.
Now, look at the outer semicircular arc of radius $R_2$. The same current $i$ flows through it, but in the opposite angular direction relative to the inner arc. Its magnitude is $B_2 = \frac{\mu_0 i}{4R_2}$.
Based on the visual layout, the current flows clockwise in one arc and counter-clockwise in the other. This means one field points into the page while the other points out of the page.
Since the magnetic field is inversely proportional to the radius ($B \propto 1/R$), the field from the smaller radius ($R_1$) is stronger than the field from the larger radius ($R_2$).
To find the net field, we subtract the magnitude of the smaller field from the larger field: $B_{net} = B_1 - B_2$.
Substituting the expressions: $B_{net} = \frac{\mu_0 i}{4R_1} - \frac{\mu_0 i}{4R_2}$.
Factoring out the common terms: $B_{net} = \frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
Step 4: Final Answer:
The net magnetic induction at the centre $O$ is $\frac{\mu_0 i}{4} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
