Animated Solution for Physics - Magnetic Effects of Current: A pair of stationary and infinitely long bent wires are placed in the $xy$ plane as shown in figure. The wires carry current of $i=10\text{ A}$ each as shown. The segments $L$ and $M$ are along the $x$-axis. The segments $P$ and $Q$ are parallel to the $y$-axis such that $OS=OR=0.02\text{ m}$. Find the magnitude and direction of the magnetic induction at the origin $O$.

The study of electromagnetism is filled with elegant symmetries and surprising cancellations. When we first encounter a complex circuit with multiple bends and infinite segments, it can feel overwhelming. But the true beauty of physics lies in breaking down the complex into the simple.

In this problem, we are presented with two infinitely long bent wires carrying a steady current. Our mission? To find the net magnetic field at the origin. Let's embark on this journey step by step, using the powerful Biot-Savart law as our guiding light.

Imagine you are standing at the origin of the coordinate system. Around you, two wires are carrying a current of $10\text{ A}$. The first wire comes in from negative infinity along the $x$-axis, reaches the point $R$, and then takes a sharp $90^{\circ }$ turn to plunge down parallel to the negative $y$-axis towards negative infinity. The second wire mirrors this behavior: it comes from positive infinity along the $x$-axis, reaches the point $S$, and shoots straight up parallel to the positive $y$-axis.

To find the total magnetic field at the origin, we must invoke the principle of superposition. We will calculate the magnetic field contribution from each of the four straight segments—$L$, $M$, $P$, and $Q$—and then add them together as vectors.

A Beautiful Cancellation

Let's start with the horizontal segments, $L$ and $M$. These segments lie perfectly on the $x$-axis.

According to the Biot-Savart law, the magnetic field $d\mathbf{B}$ produced by a small current element $d\mathbf{l}$ is proportional to the cross product $d\mathbf{l}\times \mathbf{r}$, where $\mathbf{r}$ is the position vector from the element to the point of interest.

For any point lying exactly on the extended line of a straight wire, the angle between $d\mathbf{l}$ and $\mathbf{r}$ is either $0^{\circ }$ or $180^{\circ }$. Since the cross product of parallel or anti-parallel vectors is exactly zero, these segments produce absolutely no magnetic field at the origin!

This is a massive simplification. Half of our problem just vanished into thin air!

Semi-Infinite Power

Now, we turn our attention to the vertical segments, $P$ and $Q$. These are not just any wires; they are semi-infinite wires. They start at a specific point and extend to infinity in one direction.

Let's look at segment $P$. It starts at $R(-0.02,0)$ and goes down to $-\infty$. The perpendicular distance from the origin to this wire is $OR=0.02\text{ m}$. The magnetic field at a perpendicular distance $d$ from the end of a semi-infinite wire is given by:

Substituting our values, the magnitude of the field due to segment $P$ is:

But what about the direction? Here is where the Right-Hand Thumb Rule comes into play. Point your right thumb downwards, in the direction of the current in segment $P$. As you curl your fingers towards the origin, they point straight out of the screen. Therefore, ${\mathbf{B}}_1$ is directed perpendicularly outwards from the page.

Next, let's analyze segment $Q$. It starts at $S(0.02,0)$ and goes up to $+\infty$. The perpendicular distance is $OS=0.02\text{ m}$. The magnitude of its magnetic field is identical:

Applying the Right-Hand Thumb Rule again—thumb pointing upwards along the current—your fingers will once again curl outwards at the origin. Fascinating! Both fields are pointing in the exact same direction.

Since both ${\mathbf{B}}_1$ and ${\mathbf{B}}_2$ are pointing outwards, we can simply add their magnitudes to find the net magnetic field at the origin.

Because $OR=OS=0.02\text{ m}$, the two terms are perfectly equal. Adding them together gives:

Now, for the grand finale. Let's substitute the known values. We know the current $i=10\text{ A}$, and the permeability of free space ${\mu }_0=4\pi \times 10^{-7}\text{ T}\cdot \text{m/A}$.

Notice how elegantly the $\pi$ terms cancel out. The $4\pi$ divided by $2\pi$ leaves us with a clean factor of $2$.

And there we have it! The magnitude of the net magnetic induction at the origin is $10^{-4}\text{ T}$, and its direction is perpendicular to the paper, outwards.

This problem is a beautiful reminder of how symmetry and vector properties can turn a seemingly daunting configuration into a smooth, logical sequence of calculations. Keep practicing, keep visualizing, and the magic of physics will always reveal itself to you!