Magnetic Field

Biot-Savart Law

It is found that current gives magnetic field $$\textbf{B}(\textbf{r}) = \frac{\mu_0}{4\pi}\int\frac{\textbf{I}\times\hat{d}}{d^2}dl$$ where $\textbf{d}$ is the displacement from the source of current to the point of observation. This is known as the Biot-Savart law. Magnetic field by a line of current is $$\textbf{B}(\textbf{r}) = \frac{\mu_0}{4\pi}\int \frac{Id\textbf{l}\times\hat{\textbf{d}}}{d^2}$$ Magnetic field by a sheet of current is $$\textbf{B}(\textbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\textbf{K}(\textbf{r}')\times\hat{\textbf{d}}}{d^2}da$$ Magnetic field by a volume of current is $$\textbf{B}(\textbf{r}) = \frac{\mu_0}{4\pi}\int \frac{\textbf{J}(\textbf{r}')\times\hat{\textbf{d}}}{d^2}d\tau$$

Example (Griffiths Third Edition Ex 5.5)

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

Solution:

$d\vec{l}'\times\hat{d}$ points out of the page with magnitude $$dl'\sin\alpha,$$ which is equivalent to $$dl'\cos\theta.$$ As $l'=s\tan\theta$, $$dl'=\frac{s}{\cos^2\theta}d\theta.$$ As $s=d\cos\theta$, $$\frac{1}{d^2}=\frac{\cos^2\theta}{s^2}$$ So, $$\begin{align} \vec{B}(\vec{r}) &= \frac{\mu_0}{4\pi}\int \frac{Id\vec{l}\times\hat{\vec{d}}}{d^2}\\ B &= \frac{\mu_0}{4\pi}\int^{\theta_2}_{\theta_1}\left(\frac{\cos^2\theta}{s^2}\right)\left(\frac{s}{\cos^2\theta}\right)\cos\theta d\theta\\ &= \frac{\mu_0}{4\pi s}\int^{\theta_2}_{\theta_1}\cos\theta d\theta\\ &= \frac{\mu_0}{4\pi s}(\sin \theta_2 - \sin \theta_1), \end{align}$$ pointing out of the page.

Checkpoint (Griffiths Third Edition Q5.10)

Find the magnetic field at point $P$ on the axis of a tightly wound solenoid (helical coil) consisting of $n$ turns per unit length wrapped around a cylindrical tube of radius $a$ and carrying current $I$. Express your anser in terms of $\theta_1$ and $\theta_2$.

For a ring of width $dz$, the current is $I\to nIdz$. The x,y components are cancelled, and z component is given by $$\begin{align} dB &= \frac{\mu_0}{4\pi}nIdz\int \frac{ad\phi}{d^2}\cos\theta \\ &= \frac{\mu_0nIdz}{4\pi}\left(\frac{\cos\theta}{d^2}\right)2\pi a\\ B&= \frac{\mu_0nI}{2}\int\frac{a^2}{(a^2+z^2)^{3/2}}dz \end{align}$$ Since $z=a\cot\theta$, we have $$dz=-\frac{a}{\sin^2\theta}d\theta$$ and $$\frac{1}{(a^2+z^2)^{3/2}}=\frac{\sin^3\theta}{a^3}$$ So, $$\begin{align} B &= \frac{\mu_0nI}{2}\int\frac{a^2\sin^3\theta}{a^3\sin^2\theta}(-a d\theta)\\ &=-\frac{\mu_0nI}{2}\int\sin\theta d\theta\\ &=\frac{\mu_0nI}{2}\left[\cos\theta\right]^{\theta_2}_{\theta_1}\\ &=\frac{\mu_0nI}{2}(\cos\theta_2-\cos\theta_1) \end{align}$$

Remarks: for an infinite solenoid, $\theta_2=0$ and $\theta_1=\pi$, $$B=\frac{\mu_0nI}{2}(1-(-1))=\mu_0nI$$

Curl of Magnetic Field

Curl of a magnetic field is given by applying curl on Biot-Savart law, assuming current is steady, i.e. $\nabla\cdot\textbf{J} = \frac{\partial\rho}{\partial t} = 0$ $$\bbox[5px,border:2px solid #666] { \nabla \times \textbf{B} = \mu_0\textbf{J} }$$ prove omitted as I failed to find to satisfying rigorous proof. By Stoke's Theorem, $$\oint \textbf{B} \cdot d\textbf{l} = \mu_0I_{enc}$$ This is known as the Ampere's law.

Example (Griffiths Third Edition Ex 5 .7)

Find the magnetic field a distance $s$ from a long straight wire, carrying a steady current $I$

Solution: By symmetry, the magnitude of $\vec{B}$ is constant around an amperian loop of radius $s$, centred on the wire. $$\begin{align} \oint\vec{B}\cdot d\vec{l} &= B\oint dl\\ &=B(2\pi s)\\ &= \mu_0 I_{\text{enc}}\\ &=\mu_0I\\ B&=\frac{\mu_0I}{2\pi s} \end{align}$$

Checkpoint (Griffiths Third Edition Q5.13(b))

A steady current $I$ flows down a long cylindrical wire of radius $a$. Find the magnetic field, both inside and outside the wire, if the current is distributed in such a way that $J$ is proportional to $s$, the distance from the axis.

Let $J = ks$. $$\begin{align} I &= \int^a_0 Jda\\ &= \int^a_0ks(2\pi s) ds\\ &= \frac{2\pi ka^3}{3}\\ k&=\frac{3I}{2\pi a} \end{align}$$ For $s < a$, $$\begin{align} I_{\text{enc}} &= \int^s_0 Jda\\ &= \int^s_0ks(2\pi s)ds\\ &= \frac{2}{3}\pi ks^3\\ &= I\frac{s^3}{a^3} \end{align}$$ For $s > a$, $$I_{\text{enc}} = I$$ So, $$\vec{B} = \begin{cases} \frac{\mu_0 I s^2}{2\pi a^3}\hat{\phi}, \text{ for }s < a \\ \frac{\mu_0 I}{2\pi s}\hat{\phi}, \text{ for }s > a \end{cases}$$

Divergence of Magnetic Field

By Biot-Savart law, for a general current density $$\nabla_{\textbf{r}} \cdot \textbf{B}(\textbf{r}) = \frac{\mu_0}{4\pi}\int \nabla_{\textbf{r}} \cdot (\textbf{J}(\textbf{r}') \times \frac{\hat{d}}{d^2})d\tau$$ The integral part $$\nabla_{\textbf{r}} \cdot (\textbf{J} \times \frac{\hat{d}}{d^2}) = \frac{\hat{d}}{d^2}\cdot(\nabla\times\textbf{J})-\textbf{J}\cdot(\nabla\times\frac{\hat{d}}{d^2})$$ $\nabla\times\textbf{J} = 0$ as $\textbf{J}(\textbf{r}')$ depends on $\textbf{r}'$ only, not $\textbf{r}$ while $\nabla\times\frac{\hat{d}}{d^2}$ is always 0, so we have $$\bbox[5px,border:2px solid #666] { \nabla \cdot \textbf{B} = 0 }$$