Deep Dive: Applications of the Biot-Savart Law

The Biot-Savart law is a universal tool for calculating magnetic fields from steady currents. Below are the detailed derivations for the two most fundamental geometries in magnetostatics.

Example 1: The Infinite Straight Wire

Problem: Find the magnetic field at a distance $s$ from a long straight wire carrying a steady current $I$.

1. Geometric Setup

Let the wire lie along the $z$-axis, carrying current $I$ in the $+\hat{z}$ direction. We calculate the field at a point $P$ located distance $s$ away on the $x$-axis.

Source element: $d\vec{l}' = dz \hat{z}$
Separation vector: $\vecrsep = s \hat{x} - z \hat{z}$
Distance: $\rsep = \sqrt{s^2 + z^2}$

2. The Cross Product

The Biot-Savart law requires $d\vec{l}' \times \hatrsep$. The magnitude of this cross product is given by $|d\vec{l}'| \sin(\alpha)$, where $\alpha$ is the angle between the current and the position vector.

From geometry, if we define $\theta$ as the angle from the perpendicular distance $s$, then $\alpha = 90^\circ + \theta$, and $\sin(90^\circ + \theta) = \cos \theta$. The direction is circumferential ($\hat{\phi}$).

dB = \frac{\mu_0 I}{4\pi} \frac{dz \cos \theta}{\rsep^2}

3. Trigonometric Substitution

We express variables $z$ and $r$ in terms of the angle $\theta$:

$$ z = s \tan \theta \implies dz = s \sec^2 \theta \, d\theta $$ $$ \rsep = \frac{s}{\cos \theta} $$

Substituting these into the integral:

B = \frac{\mu_0 I}{4\pi} \int \frac{(s \sec^2 \theta \, d\theta) \cos \theta}{(s / \cos \theta)^2}

Simplifying the terms:

B = \frac{\mu_0 I}{4\pi s} \int \cos \theta \, d\theta

4. Integration Limits

For an infinite wire ($z \to \pm \infty$), the angle $\theta$ spans from $-\pi/2$ to $+\pi/2$.

B = \frac{\mu_0 I}{4\pi s} \left[ \sin \theta \right]_{-\pi/2}^{\pi/2} $$ $$ B = \frac{\mu_0 I}{4\pi s} (1 - (-1)) = \frac{\mu_0 I}{2\pi s}

Final Result

\vec{B} = \frac{\mu_0 I}{2\pi s} \hat{\phi}

The field decreases inversely with distance $s$ and curls around the wire (Right-Hand Rule).

Example 2: Circular Loop on Axis

Problem: Find the magnetic field a distance $z$ above the center of a circular loop of radius $R$, which carries a steady current $I$.

1. Geometric Setup

The loop lies in the $xy$-plane. We calculate the field at point $P(0,0,z)$. The distance from any point on the rim to $P$ is constant: $\rsep = \sqrt{R^2 + z^2}$.

The current element $d\vec{l}'$ is always perpendicular to the vector $\vecrsep$, so $|d\vec{l}' \times \hatrsep| = dl$.

2. Symmetry Argument

The magnetic field vector $d\vec{B}$ produced by a segment $d\vec{l}'$ has a vertical component ($dB_z$) and a horizontal component ($dB_\perp$).

For every element on one side of the loop, there is a diametrically opposite element whose horizontal field component exactly cancels the first. Therefore, only the vertical components survive integration.

B = \int dB_z = \int dB \cos \theta

Where $\theta$ is the angle between $\vecrsep$ and the plane of the loop.

3. Integration

From geometry, $\cos \theta = \frac{R}{\rsep} = \frac{R}{\sqrt{R^2 + z^2}}$.

Substituting into the Biot-Savart expression:

$$ dB_z = \frac{\mu_0 I}{4\pi} \frac{dl}{\rsep^2} \left( \frac{R}{\rsep} \right) = \frac{\mu_0 I R}{4\pi \rsep^3} dl $$

Since $I, R,$ and $\rsep$ are constants, we simply integrate $dl$ around the circumference ($2\pi R$):

B_z = \frac{\mu_0 I R}{4\pi (R^2 + z^2)^{3/2}} \int_0^{2\pi R} dl $$ $$ B_z = \frac{\mu_0 I R}{4\pi (R^2 + z^2)^{3/2}} (2\pi R)

Final Result

\vec{B} = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \hat{z}

At the center ($z=0$), this simplifies to $B = \frac{\mu_0 I}{2R}$.