The magnetic induction at the centre $O$ in the figure shown is
Show Hint
Whenever straight wire leads are aligned perfectly with the test point, you can cross them off your scratchpad immediately since their field contribution is exactly zero. From there, if the two arcs flow in opposite rotational directions (clockwise vs. counter-clockwise), their fields will subtract, instantly telling you that the correct formula must feature a minus sign.
\nobreak
Concept:
By applying the Biot-Savart Law and the principle of superposition, the net magnetic field ($\vec{B}_{\text{net}}$) at the center point $O$ is calculated as the vector sum of the magnetic fields generated by each individual segment of the current-carrying loop.
We can split this closed loop configuration into four distinct wire segments:
• Segment 1: Inner semicircular arc of radius $R_1$.
• Segment 2: Outer semicircular arc of radius $R_2$.
• Segments 3 & 4: The two small, straight horizontal wire pieces connecting the inner and outer arcs.
Step 1: Evaluate the magnetic field contribution from the straight wire segments.
According to the Biot-Savart law, the magnetic field $B$ at a point due to a straight current element is given by an integral involving $\text{d}\vec{l} \times \hat{r}$. For both of the small horizontal straight wire connections at the base, the line of action passes directly through the center point $O$.
Because the current position vector and the displacement vector $\vec{r}$ are collinear ($\theta = 0^\circ$ or $180^\circ$), the cross product $\text{d}\vec{l} \times \hat{r} = 0$. Therefore:
\[
B_{\text{straight}} = 0
\]
The straight horizontal sections contribute absolutely nothing to the magnetic field at the center.
Step 2: Determine the field due to the inner semicircular arc ($B_1$).
The inner segment is a semicircle subtending an angle of $\theta = \pi$ radians at the center $O$, with a radius of $R_1$. The current flows in a clockwise direction along this inner path (indicated by the pink arrow).
The formula for the magnetic field at the center of a circular arc is:
\[
B_1 = \frac{\mu_0 i}{4\pi R_1}\theta = \frac{\mu_0 i}{4\pi R_1}(\pi) = \frac{\mu_0 i}{4R_1}
\]
Using the Right-Hand Thumb Rule (curling the fingers of your right hand in the clockwise direction of the current), your thumb points straight into the page. Thus, the direction of $\vec{B}_1$ is into the page ($\otimes$).
Step 3: Determine the field due to the outer semicircular arc ($B_2$).
The outer segment is also a semicircle subtending an angle of $\theta = \pi$ radians, but with a larger radius of $R_2$. Following the continuous closed loop path, the current flows in a counter-clockwise direction along this outer track.
Calculating its field magnitude gives:
\[
B_2 = \frac{\mu_0 i}{4\pi R_2}\theta = \frac{\mu_0 i}{4\pi R_2}(\pi) = \frac{\mu_0 i}{4R_2}
\]
Using the Right-Hand Thumb Rule (curling the fingers of your right hand counter-clockwise), your thumb points straight out of the page. Thus, the direction of $\vec{B}_2$ is out of the page ($\odot$).
Step 4: Calculate the net magnetic field ($\vec{B}_{\text{net}}$) through vector addition.
Since the two magnetic field vectors point in exactly opposite directions, they oppose each other. Because the radius is in the denominator ($B \propto \frac{1}{R}$), the inner arc has a smaller radius ($R_1 & Lt; R_2$) and therefore creates a stronger magnetic field ($B_1 > B_2$).
Taking the dominant into the page ($\otimes$) direction as our positive reference framework:
\[
B_{\text{net}} = B_1 - B_2
\]
Substituting our expressions into the subtraction formula:
\[
B_{\text{net}} = \frac{\mu_0 i}{4R_1} - \frac{\mu_0 i}{4R_2}
\]
Factoring out the common terms yields:
\[
B_{\text{net}} = \frac{\mu_0 i}{4}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)
\]
