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 $$dl'\sin\alpha = dl'\cos\theta$$ 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(2pi 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 $$ \bbox[5px,border:2px solid #666] { \nabla \cdot \textbf{B} = 0 } $$