Animated Solution for Physics - Magnetic Effects of Current: The magnitude of the magnetic field ($B$) due to loop $ABCD$ at the origin ($O$) is

Imagine you are looking at a complex wire loop carrying a steady current $I$. To find the total magnetic field at the origin $O$, we must break this loop down into simpler, manageable pieces.

The loop $ABCD$ consists of four distinct segments: two straight radial lines ($BC$ and $DA$) and two circular arcs ($AB$ and $CD$).

By the principle of superposition, the net magnetic field at the origin is simply the vector sum of the magnetic fields produced by each of these four segments individually.

Let's first tackle the straight segments, $BC$ and $DA$. Notice their geometry carefully. If you extend these line segments, they pass exactly through the origin $O$.

According to the Biot-Savart Law, the magnetic field $d\vec{B}$ is proportional to the cross product $d\vec{l}\times \vec{r}$. For any point lying on the axis of a current-carrying wire, the angle between the current element $d\vec{l}$ and the position vector $\vec{r}$ is either $0^{\circ }$ or $180^{\circ }$.

Since the cross product vanishes in both cases, these straight segments produce absolutely zero magnetic field at the origin.

Now, let's focus on the circular arcs. The magnetic field at the center of a circular arc of radius $r$ subtending an angle $\theta$ is given by the standard formula:

Crucial Step: The angle $\theta$ must strictly be in radians! Here, the angle is $30^{\circ }$, which converts to $\frac{\pi }{6}$ radians.

For the inner arc $AB$ (radius $a$): Using the Right-Hand Grip Rule, since the current flows counter-clockwise from $A$ to $B$, this magnetic field points outwards ($\odot$) from the plane of the screen.

For the outer arc $CD$ (radius $b$): Here, the current flows clockwise from $C$ to $D$. Applying the Right-Hand Grip Rule again, this magnetic field points inwards ($\otimes$) into the plane of the screen.

We now have two magnetic field vectors at the origin pointing in exactly opposite directions. To find the net field, we must subtract the smaller magnitude from the larger one.

Since the inner arc $AB$ is closer to the origin ($a<b$), it produces a stronger magnetic field. Therefore, the net magnetic field will point outwards.

Substituting our calculated values:

Let's factor out the common terms to simplify the expression:

Taking the common denominator, we arrive at our final, elegant result:

This perfectly matches option (b).