Study Notes on the Biot-Savart Law and Magnetic Fields

The Biot-Savart Law describes the magnetic field dold{B} at a point $P$ generated by a length element dold{l} of a wire carrying a steady current $I$ .
The fundamental observations derived from the Biot-Savart experiment include: * The vector dold{B} is perpendicular to both the length element dold{l} (which points in the direction of the current flow) and the unit vector old{\text{\hat{r}}} directed from the element dold{l} toward point $P$ . * The magnitude of the magnetic field dold{B} is inversely proportional to the square of the distance between the source and the point of measurement: |dold{B}| \propto \frac{1}{r^2}. * The magnitude of dold{B} is directly proportional to the current $I$ and the magnitude of the length element dold{l}. * The magnitude of dold{B} is proportional to $\sin(\theta)$ , where $\theta$ is defined as the angle between the vector dold{l} and the unit vector old{\text{\hat{r}}}.

The observations above are summarized in the vector form of the Biot-Savart Law: dold{B} = \frac{\mu_0}{4\pi} \frac{I \, dold{l} \times \bold{\hat{r}}}{r^2}
In this expression, $\mu_0$ is the constant known as the Permeability of free space.
The value of the permeability of free space is defined as: $\mu_0 = 4\pi \times 10^{-7}\,\text{T} \cdot \text{m/A}$
The specific magnetic field defined in this equation represents the field created at a point solely by a small length segment $d\bold{l}$ of the conductor.

To find the total magnetic field $\bold{B}$ created at a point by a current $I$ in a conductor of finite size, one must integrate the contributions of all infinitesimal elements: $\bold{B} = \frac{\mu_0 I}{4\pi} \int \frac{d\bold{l} \times \bold{\hat{r}}}{r^2}$
The magnitude of this total magnetic field is calculated using the following integral: $B = \frac{\mu_0 I}{4\pi} \int \frac{\sin(\theta)\,dl}{r^2}$
While the law is typically discussed in the context of current-carrying wires, its validity extends to any current consisting of charges flowing through space.
A specific real-world example provided is the electron beam within a television picture tube.

Consider a section of a straight current-carrying conductor of length $2a$ . Using the Biot-Savart Law, the magnitude of the magnetic field can be expressed as: $B = \frac{\mu_0 I}{4\pi} \int_{-a}^{a} \frac{dl \sin(\phi)}{r^2}$
Based on trigonometry, the relationship $\sin(\pi - \phi) = \sin(\phi)$ holds.
Consequently, the expression becomes: $B = \frac{\mu_0 I}{4\pi} \int_{-a}^{a} \frac{dl \sin(\pi - \phi)}{r^2}$
Applying the Pythagorean theorem to the geometry of the system, where $x$ is the perpendicular distance to point $P$ and $y$ is the position along the wire: $r^2 = x^2 + y^2$
The trigonometric relationship for $\sin(\pi - \phi)$ is expressed as: $\sin(\pi - \phi) = \frac{x}{\sqrt{x^2 + y^2}} = (x^2 + y^2)^{-1/2}$

By substituting the geometric expressions into the magnetic field integral and letting $dl = dy$ : $B = \frac{\mu_0 I}{4\pi} \int_{-a}^{a} \frac{x \, dy}{(x^2 + y^2)(x^2 + y^2)^{1/2}}$ $B = \frac{\mu_0 I x}{4\pi} \int_{-a}^{a} \frac{dy}{(x^2 + y^2)^{3/2}}$
Utilizing special integrals, specifically: $\int \frac{dy}{(x^2 + y^2)^{3/2}} = \frac{1}{x^2} \frac{y}{(x^2 + y^2)^{1/2}}$
The evaluation of the integral from $-a$ to $a$ results in: $B = \frac{\mu_0 I x}{4\pi} \left[ \frac{y}{x^2(x^2 + y^2)^{1/2}} \right]_{-a}^{a}$ $B = \frac{\mu_0 I x}{4\pi} \left( \frac{2a}{x^2(x^2 + a^2)^{1/2}} \right)$
The simplified formula for the magnitude of the field of a finite conductor is: $B = \frac{\mu_0 I \, 2a}{4\pi x(x^2 + a^2)^{1/2}}$

When the half-length of the conductor $a$ is much larger than the distance $x$ (denoting an infinitely long wire where $a \rightarrow \infty$ ): $(x^2 + a^2)^{1/2} \cong a$
Under this condition, the magnitude of the field near the long, straight conductor is: $B = \frac{\mu_0 I}{2\pi x}$
Due to axial symmetry about the y-axis, the field at all points on a circle of radius $r$ around the conductor yields a magnitude: $B = \frac{\mu_0 I}{2\pi r}$
This equation defines the magnetic flux density $B$ near a long, straight current-carrying conductor.

Problem: A long straight conductor carries a $1.0\,\text{A}$ current. At what distance from the conductor axis does the resulting magnetic field have a magnitude of $B = 0.5 \times 10^{-4}\,\text{Tesla}$ ? (Note: This is approximately the magnitude of the Earth's magnetic field in Pittsburgh).
Solution Procedure: * Identify the formula for distance (radius): $r = \frac{\mu_0 I}{2\pi B}$ . * Input given values: $\mu_0 = 4\pi \times 10^{-7}\,\text{T} \cdot \text{m/A}$ , $I = 1.0\,\text{A}$ , and $B = 0.5 \times 10^{-4}\,\text{T}$ . * Calculation equation: $r = \frac{4\pi \times 10^{-7} \times 1.0}{2 \times 3.142 \times 0.5 \times 10^{-4}}$ * Final result: $r = 4 \times 10^{-3}\,\text{m} = 4\,\text{mm}$ .

A semi-infinite wire refers to a case where point $P$ is evaluated relative to only the upper or lower half of an infinite wire.
The magnetic field at point $P$ due to one half of an infinite wire is exactly half of the total field calculated for an infinite wire.
The resulting magnitude formula for a semi-infinite wire is: $B = \frac{\mu_0 I}{4\pi r}$