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Magnetic effects of currents

  1. Oersted (1820) discovered that an electric current of magnitude $i$ flowing in a closed loop of area, ${\bf dS}$, where ${\bf dS}$ is normal to the plane of the loop, produces the same magnetic field as a magnetic dipole of moment
    \begin{displaymath}
{\bf m} = i{\bf dS}.
\end{displaymath} (3.19)

    This is shown is Figure 3.7

    Figure 3.7: A current loop with area ${\bf dS}$ produces the same magnetic field as a dipole.
    \includegraphics [scale=0.7]{fundfig35.eps}

  2. Outside regions where currents flow we have a potential magnetic field. Thus,

    \begin{displaymath}
\nabla \cdot {\bf B} = 0\qquad \hbox{and}\qquad \nabla \times {\bf B} = 0.
\end{displaymath}

    Hence, the magnetic field can be expressed in terms of a magnetic potential as ${\bf B} = - \nabla F$ and we have
    \begin{displaymath}
\nabla^{2}F = 0.
\end{displaymath} (3.20)

    In cylindrical polars $(R, \phi , z)$ (3.20) becomes

    \begin{displaymath}
{1\over R}{\partial \over \partial R}\left (R{\partial F \o...
...artial
\phi^{2}} + {\partial^{2}F\over \partial z^{2}} = 0.
\end{displaymath}

    A particularly simple solution is

    \begin{displaymath}
F = -{\mu_{0}\over 2\pi} I \phi, \qquad I = \hbox{ const},
\end{displaymath}

    for which

    \begin{displaymath}
B_{R}= -{\partial F \over \partial R} = 0, \hbox{ }B_{\phi}...
...I\over R}, \hbox{ }B_{z} = -{\partial F\over \partial z} = 0.
\end{displaymath}

    The field lines are circles on cylinders about the z-axis. The total current through a circle of radius $R$, that is an open surface, is

    \begin{displaymath}
\int {\bf J}\cdot {\bf dS} = \int {\nabla \times {\bf B}\ov...
...bf dS} = \oint_{C}^{}{1\over \mu_{0}}{\bf B}\cdot
{\bf dl},
\end{displaymath}

    on using Stokes' theorem. $C$ is a closed curve around the open surface $S$. Thus,

    \begin{displaymath}
\int {\bf J}\cdot {\bf dS} = {B_{\phi}\over \mu_{0}}\int dl...
...0}}2\pi R = {\mu_{0}\over 2\pi}{I\over
\mu_{0}R}2\pi R = I.
\end{displaymath}

    Thus, the total current is simply $I$ and it flows along the z-axis. This is a line current.

  3. Inside regions where currents flow. Here we must satisfy the equations
    $\displaystyle \nabla \cdot {\bf B}$ $\textstyle =$ $\displaystyle 0,$ (3.21)
    $\displaystyle {\nabla \times {\bf B}\over \mu_{0}}$ $\textstyle =$ $\displaystyle {\bf J},$ (3.22)

    where ${\bf J}$ is the current density. Thus, we can solve (3.21) by writing ${\bf B} = \nabla \times {\bf A}$, where ${\bf A}$ is the vector potential. If we have $\nabla
\cdot {\bf A} = 0$, (3.22) becomes

    \begin{displaymath}
\nabla \times \nabla \times {\bf A} = \mu_{0}{\bf J}.
\end{displaymath}

    The left hand side may be simplified by using a vector identity

    \begin{displaymath}
\nabla \times \nabla \times {\bf A} = \nabla (\nabla \cdot {\bf A}) - \nabla^{2}{\bf A} = -\nabla^{2}{\bf A}.
\end{displaymath}

    Therefore,
    \begin{displaymath}
\nabla^{2}{\bf A} = -\mu_{0}{\bf J}.
\end{displaymath} (3.23)

    This is a vector form of Poisson's equation. In the same way that the potential due to a volume charge $\rho _{c}$ satisfying $\nabla^{2}F =
-{\rho_{c}/ \epsilon}$ is

    \begin{displaymath}
F = {1\over 4\pi \epsilon}\int {\rho_{c}\over r} dV,
\end{displaymath}

    where

    \begin{displaymath}
r = \sqrt{(x - x_{1})^2 + (y - y_{1})^2 + z - z_{1})^2},
\end{displaymath}

    so the general solution of (3.23) is
    \begin{displaymath}
{\bf A} = {\mu_{0}\over 4\pi}\int {{\bf J}\over r}dV.
\end{displaymath} (3.24)

    For a wire carrying a current I in an element ${\bf ds}$, this reduces to

    \begin{displaymath}
{\bf A} = {\mu_{0}I\over 4\pi}\int {{\bf ds}\over r}.
\end{displaymath}

    Therefore, the magnetic field is given by

    \begin{displaymath}
{\bf B} = \nabla \times {\bf A} = {\mu_{0}I\over 4\pi}\int \nabla
\left ({1\over r}\right ) \times {\bf ds}.
\end{displaymath}

    Thus, we have
    \begin{displaymath}
{\bf B} = {\mu_{0}I\over 4\pi} \int {{\bf ds} \times {\bf r}\over
r^{3}}.
\end{displaymath} (3.25)

    (3.25) is called the Biot-Savart Law.

Example 3. .4For example, consider the magnetic field due to a current $I$ flowing in an infinite straight wire. Using the Biot-Savart law, (3.25), in cylindrical coordinates, we have

\begin{displaymath}
B_{\phi} = {\mu_{0}I\over 4\pi}\int {\sin \theta \over r^{2}} dz.
\end{displaymath} (3.26)

The situation is shown in Figure 3.8.

Figure 3.8: The general point $P$ in relation to a current element ${\bf
ds}$ of an infinite straight wire.
\includegraphics [scale=0.7]{fundfig36.eps}

The field at the general point $P$ is obtained by relating $z$, $\theta $ and $r$ with $R$ held fixed. From Figure 3.8 we have
\begin{displaymath}
\tan \theta = {R\over z}, \qquad \Rightarrow \qquad z = R {\cos
\theta \over \sin \theta}.
\end{displaymath} (3.27)


\begin{displaymath}
\sin \theta = {R\over r}, \qquad \Rightarrow \qquad r = {R\over
\sin \theta}.
\end{displaymath}

Finally, we have

\begin{displaymath}
{\bf ds} = dz {\bf e}_{z}, \qquad \Rightarrow \qquad {\bf ds} \times {\bf r} = r \sin \theta dz {\bf e}_{\phi}.
\end{displaymath}

Hence, using (3.27), we have

\begin{displaymath}
{dz\over d\theta} = R{\left(-\sin^{2}\theta -
\cos^{2}\th...
...uad
\Rightarrow \qquad dz = - {R\over \sin^{2}\theta}d\theta.
\end{displaymath}

Therefore, using the Biot-Savart law we have

\begin{displaymath}
B_{\phi} = {\mu_{0}I\over 4\pi}\int_{\pi}^{0}-{R\over
\si...
...
{\mu_{0}I\over 4\pi R}\int_{\pi}^{0} - \sin \theta d\theta.
\end{displaymath}

Integrating we obtain

\begin{displaymath}
B_{\phi} = {\mu_{0}I\over 4 \pi R}\left [\cos \theta \right
]_{\pi}^{0} = {\mu_{0}I\over 2\pi R},
\end{displaymath}

as obtained before.
next up previous
Next: Summary Up: Magnetic Fields Previous: Magnetic Fields
Prof. Alan Hood
2000-11-06