Animated Solution for Physics - Magnetic Effects of Current: Which one of the following options represents the magnetic field $\vec{B}$ at O due to the current flowing in the given wire segments lying on the xy plane?

When you first look at this circuit, it might seem like a tangled mess of wires. But in physics, complex systems are just a combination of simple parts. Let's break this wire down into six distinct segments: 1. A left vertical wire. 2. A top horizontal wire. 3. A semicircular arc. 4. A bottom horizontal wire. 5. A quarter-circular arc. 6. A right vertical wire.

Our goal is to find the net magnetic field $\vec{B}$ at the origin $O$. The principle of superposition tells us that we can calculate the magnetic field for each segment individually and then add them up as vectors.

Before we dive into heavy calculations, let's look for shortcuts. The Biot-Savart law tells us that the magnetic field $d\vec{B}$ produced by a small current element $d\vec{l}$ is proportional to the cross product $d\vec{l}\times \vec{r}$.

If a wire segment lies on a line that passes directly through the origin, the position vector $\vec{r}$ is parallel (or anti-parallel) to the current element $d\vec{l}$. The cross product of parallel vectors is zero!

Looking at our diagram, the top horizontal wire, the bottom horizontal wire, and the right vertical wire all point directly towards or away from the origin. Therefore, their contributions to the magnetic field at the origin are exactly zero:

This brilliant realization cuts our work in half! We only need to calculate the fields for the remaining three segments.

Let's analyze the left vertical wire. It is located at $x=-L$ and extends from $y=-L$ to $y=0$. The perpendicular distance from the origin is $d=L$.

To use the formula for a finite straight wire, we need the angles subtended by its ends at the origin. The top end is at $y=0$, which lies on the perpendicular, so ${\theta }_2=0^{\circ }$. The bottom end is at $(-L,-L)$, forming an isosceles right triangle with the origin. Thus, ${\theta }_1=45^{\circ }$.

Using the Biot-Savart formula for a finite wire:

Using the right-hand thumb rule, the current is flowing upwards, so the magnetic field at the origin points into the page, which is the $-\hat{k}$ direction.

Next, we have the semicircular arc. It has a radius of $r=L/2$ and subtends an angle of $\theta =\pi$ radians at the origin. The formula for the magnetic field at the center of a circular arc is:

Substituting our values: The current flows clockwise, so the right-hand rule again gives a direction of $-\hat{k}$.

Finally, we look at the quarter-circular arc. It has a radius of $r=L/4$ and subtends an angle of $\theta =\pi /2$ radians. Once again, the clockwise current means the field points in the $-\hat{k}$ direction.

Now, we simply add the non-zero contributions together. Since they all point in the same direction ($-\hat{k}$), we can just add their magnitudes:

Combining the terms for the arcs gives $\frac{{\mu }_{0}I}{L}$. Factoring out $\frac{{\mu }_{0}I}{L}$ from the entire expression, we get:

This perfectly matches option (C). By systematically breaking the problem into manageable pieces and leveraging symmetry, we turned a daunting diagram into a straightforward calculation!