Animated Solution for Physics - Magnetic Effects of Current: A long straight wire along the z-axis carries a current $I$ in the negative z-direction. The magnetic vector field $\mathbf{B}$ at a point having coordinate $(x,y)$ on the $z=0$ plane is

Magnetic Field of a Long Straight Wire

Imagine you are standing on the $xy$-plane, looking straight down into the abyss of the $z$-axis. A long, infinitely straight wire is piercing right through the origin, carrying a steady current $I$ away from you, deep into the negative $z$-direction. Our mission? To find the exact mathematical expression for the magnetic field vector $\mathbf{B}$ at any arbitrary point $P(x,y)$ on the plane you are standing on.

This is a classic problem that tests not just your knowledge of Ampere's Law, but your ability to translate physical intuition into rigorous vector mathematics. Let's break it down step-by-step.

First, let's establish the magnitude of the magnetic field. According to Ampere's Law, the magnetic field produced by a long straight wire at a perpendicular distance $r$ is given by:

If our point $P$ has coordinates $(x,y)$, the distance $r$ from the origin is simply the magnitude of the position vector $\mathbf{r}=x\hat{i}+y\hat{j}$. Using the Pythagorean theorem, we know:

Now comes the crucial part: the direction. Physics isn't just about magnitudes; it's about spatial reality. Point your right thumb in the direction of the current—which is into the screen ($-z$ direction). Notice how your fingers curl? They curl in a clockwise direction.

This tells us that the magnetic field lines are concentric circles centered at the origin, flowing clockwise. Therefore, at point $P$, the magnetic field vector $\mathbf{B}$ must be perfectly tangent to this circle. In vector terms, $\mathbf{B}$ is perpendicular to the position vector $\mathbf{r}$.

Let the position vector $\mathbf{r}$ make an angle $\theta$ with the positive $x$-axis. From basic trigonometry:

Since $\mathbf{B}$ is perpendicular to $\mathbf{r}$ and points clockwise, it will point downwards and to the right (if $P$ is in the first quadrant). By analyzing the geometry, the angle $\mathbf{B}$ makes with the negative $y$-axis is exactly $\theta$. Resolving $\mathbf{B}$ into its Cartesian components gives us:

Notice the negative sign! The $y$-component points downwards, which is a common trap where many students make a silly mistake.

Now, we simply substitute our expressions for $B$, $\sin \theta$, and $\cos \theta$ into the vector equation:

Factoring out the common terms, we get:

Finally, substituting $r^2=x^2+y^2$, we arrive at our elegant final answer:

The Ninja Technique (Cross Product)

If you want to bypass the trigonometry entirely and solve this like a true physicist, use the cross product! The direction of the magnetic field is given by the cross product of the current direction vector and the position vector: ${\hat{u}}_B={\hat{u}}_I\times {\hat{u}}_r$.

Here, the current is along $-\hat{k}$. So:

Since $\hat{k}\times \hat{i}=\hat{j}$ and $\hat{k}\times \hat{j}=-\hat{i}$, we get:

Multiply this unit vector by the magnitude $B=\frac{{\mu }_{0}I}{2\pi r}$, and you instantly get the same result. Beautiful, isn't it?