The Biot-Savart law gives the magnetic field $d\vec B$ produced by a small wire segement $d \vec l$ carrying current $I$.

• $d\vec l$ is the line segment pointing in the direction of the current.
• $r$ is the separation between the segment and the observer.
• $\hat r$ is the unit vector (i.e. $|\hat r| = 1$) pointing from the segment to the observer.

For a long wire (does not have to be straight), the total magnetic field is found by integrating over the whole wire: $$ \vec B = \int \frac{\mu_0}{4\pi} \frac{I d\vec l \times \hat r}{r^2} $$

Another form of the Biot-Savart law

Long straight wire

We now apply the Biot-Savart law to an infinitely long straight wire to find the magnetic field at distance $x$. We are going to simplify our work by looking at only the magnitude: $$ \begin{eqnarray} B = |\vec B| &=& |\int d\vec B| \end{eqnarray} $$

In general, $|\int d\vec B| \neq \int |d\vec B|$, just like $|\vec u + \vec v| \neq |\vec u| + |\vec v|$ (try $\vec u = 2\hat i$ and $\vec v = -2\hat i$). There is one exception, which is when $\vec u$ and $\vec v$ are parallel. In our Biot-Savart integral, you can check that $d\vec B \propto d\vec l \times \hat r$ all point in the same direction (into the screen) for every segment of the wire. Because of this, we can safely assert: $$ \begin{eqnarray} B &=& \int |d\vec B| \\ &=& \int \frac{\mu_0}{4\pi} \frac{I |d\vec l \times \hat r|}{r^2} \end{eqnarray} $$

Looking inside the integrand: $$ \begin{eqnarray} |d\vec l \times \hat r| &=& |d\vec l| |\hat r| \sin \phi \\ &=& \sin \phi dy & \text{ because $|\hat r|=1$} \\ &=& \sin (180^\circ - \phi) dy & \text{ trig identity}\\ &=& \sin \beta dy & \text{ because $\beta=180^\circ - \phi$}\\ &=& \cos (90^\circ - \beta) dy & \text{ trig identity}\\ &=& \cos \alpha dy & \text{ because $\alpha = 90^\circ - \beta$} \end{eqnarray} $$

We can also write $r$ and $dy$ in terms of $\alpha$: $$ \begin{eqnarray} \cos \alpha &=& \frac{x}{r} \\ \Rightarrow r &=& \frac{x}{\cos \alpha} \\ &=& x \sec \alpha \end{eqnarray} $$ $$ \begin{eqnarray} y &=& x \tan \alpha \\ \Rightarrow dy &=& x \frac{d \tan \alpha}{d\alpha} d\alpha \\ &=& x \sec^2 \alpha d\alpha \\ \Rightarrow |d\vec l \times \hat r| &=& \cos \alpha dy \\ &=& \cos \alpha (x \sec^2 \alpha d\alpha) \\ &=& x \sec \alpha d\alpha \end{eqnarray} $$

Putting everything into the integral: $$ \begin{eqnarray} B &=& \int \frac{\mu_0}{4\pi} \frac{I |d\vec l \times \hat r|}{r^2} \\ &=& \frac{\mu_0}{4\pi} \int \frac{I (x \sec \alpha d\alpha)}{(x \sec \alpha)^2} \\ &=& \frac{\mu_0 I}{4\pi x} \int_{-\pi/2}^{+\pi/2} \cos\alpha d\alpha \\ &=& \frac{\mu_0 I}{4\pi x} (\sin \alpha)\bigg|_{-\pi/2}^{+\pi/2} \\ &=& \frac{\mu_0 I}{4\pi x} (1-(-1)) \\ &=& \frac{\mu_0 I}{2\pi x} \end{eqnarray} $$

Ampere's law is used to relate the magnetic field around a loop to the current enclosed by the loop:

• Integral taken over a one-dimensional loop.
• $I_{enclosed}$ is the total amount of current inside the loop.
• The integration can computed in either directions (clockwise or counterclockwise).
• Similar to the idea of Gauss' law $\oint \vec E \cdot d\vec A = \frac{q_{enclosed}}{\epsilon_0}$ (although Gauss' law intergrates over a two-dimensional surface).
• The sign of the current is determined by the right hand grip rule:
  • Wrap your four fingers in the direction of the path integration.
  • Your thumb determins the positive direction for $I$.
  • Currents flowing opposite to your thumb are considered negative.

Straight wire

Applying Ampere's law to a single current, in the counterclockwise loop shown in the figure, we have: $$ \begin{eqnarray} \oint \vec B \cdot d\vec s &=& \mu_0 I_{enclosed} = \mu_0 I & \text{ because $I_{enclosed} = +I$ by right hand rule} \\ \Rightarrow \mu_0 I &=& \oint B \cos 0^\circ ds & \text{ because $\vec B \parallel d\vec s$} \\ &=& \oint B ds \\ &=& B \oint ds & \text{ because $B=constant$ by symmetry}\\ &=& B (2\pi r) \\ \Rightarrow B &=& \frac{\mu_0 I}{2\pi r} \end{eqnarray} $$ This is the magnetic field of a straight long wire that we used before.

Toroid

Suppose a toroid has a total of $N$ turns. The total current enclosed by the Amperian loop as shown in the figure is $I_{enclosed} = NI$. $$ \begin{eqnarray} \oint \vec B \cdot d\vec s &=& \mu_0 I_{enclosed} = \mu_0 NI \\ \Rightarrow \mu_0 N I &=& \oint B \cos 0^\circ ds & \text{ because $\vec B \parallel d\vec s$} \\ &=& \oint B ds \\ &=& B \oint ds & \text{ because $B=constant$ by symmetry}\\ &=& B (2\pi r) \\ \Rightarrow B &=& \frac{\mu_0 N I}{2\pi r} \end{eqnarray} $$

Solenoid

Suppose a solenoid has $n$ turns per unit length (i.e. turns density is $n$). Choose a rectangular Amperian loop as shown in the figure, and push the top edge to infinity where $B\approx 0$. This means the top edge does not contribute to $\oint \vec B \cdot d\vec s$. The right and left edge also do not contribute because $\vec B \perp d\vec s$ so $ \vec B \cdot d\vec s = 0$. Only the bottom edge is left in the integral. $$ \begin{eqnarray} \oint \vec B \cdot d\vec s &=& \mu_0 I_{enclosed} = \mu_0 I n w & \text{ because number of wires inside loop is $nw$} \\ \Rightarrow \mu_0 I n w &=& \int_{bottom} \vec B \cdot d\vec s + \int_{right} \vec B \cdot d\vec s + \int_{top} \vec B \cdot d\vec s + \int_{left} \vec B \cdot d\vec s \\ &=& \int_{bottom} \vec B \cdot d\vec s + 0 + 0 + 0 \\ &=& \int_{bottom} B \cos 0^\circ ds & \text{ because $\vec B \parallel d\vec s$} \\ &=& \int_{bottom} B ds \\ &=& B \int_{bottom} ds & \text{ because $B=constant$ by symmetry}\\ &=& B w \\ \Rightarrow B &=& \mu_0 n I \end{eqnarray} $$ This is the same equation that was given before for solenoid.