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
   

   $${\bf m} = i{\bf dS}.$$ (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.

   

2. Outside regions where currents flow we have a potential magnetic field. Thus,
   
   $$\nabla \cdot {\bf B} = 0\qquad \hbox{and}\qquad \nabla \times {\bf B} = 0.$$
   
   
   Hence, the magnetic field can be expressed in terms of a magnetic potential as ${\bf B} = - \nabla F$ and we have
   

   $$\nabla^{2}F = 0.$$ (3.20)

   
   
   In cylindrical polars $(R, \phi , z)$ (3.20) becomes
   
   $${1\over R}{\partial \over \partial R}\left (R{\partial F \o... ...artial \phi^{2}} + {\partial^{2}F\over \partial z^{2}} = 0.$$
   
   
   A particularly simple solution is
   
   $$F = -{\mu_{0}\over 2\pi} I \phi, \qquad I = \hbox{ const},$$
   
   
   for which
   
   $$B_{R}= -{\partial F \over \partial R} = 0, \hbox{ }B_{\phi}... ...I\over R}, \hbox{ }B_{z} = -{\partial F\over \partial z} = 0.$$
   
   
   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
   
   $$\int {\bf J}\cdot {\bf dS} = \int {\nabla \times {\bf B}\ov... ...bf dS} = \oint_{C}^{}{1\over \mu_{0}}{\bf B}\cdot {\bf dl},$$
   
   
   on using Stokes' theorem. $C$ is a closed curve around the open surface $S$. Thus,
   
   $$\int {\bf J}\cdot {\bf dS} = {B_{\phi}\over \mu_{0}}\int dl... ...0}}2\pi R = {\mu_{0}\over 2\pi}{I\over \mu_{0}R}2\pi R = I.$$
   
   
   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
   

   $$\nabla \cdot {\bf B} = 0,$$ (3.21)

   $${\nabla \times {\bf B}\over \mu_{0}} = {\bf J},$$ (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
   
   $$\nabla \times \nabla \times {\bf A} = \mu_{0}{\bf J}.$$
   
   
   The left hand side may be simplified by using a vector identity
   
   $$\nabla \times \nabla \times {\bf A} = \nabla (\nabla \cdot {\bf A}) - \nabla^{2}{\bf A} = -\nabla^{2}{\bf A}.$$
   
   
   Therefore,
   

   $$\nabla^{2}{\bf A} = -\mu_{0}{\bf J}.$$ (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
   
   $$F = {1\over 4\pi \epsilon}\int {\rho_{c}\over r} dV,$$
   
   
   where
   
   $$r = \sqrt{(x - x_{1})^2 + (y - y_{1})^2 + z - z_{1})^2},$$
   
   
   so the general solution of (3.23) is
   

   $${\bf A} = {\mu_{0}\over 4\pi}\int {{\bf J}\over r}dV.$$ (3.24)

   
   
   For a wire carrying a current I in an element ${\bf ds}$, this reduces to
   
   $${\bf A} = {\mu_{0}I\over 4\pi}\int {{\bf ds}\over r}.$$
   
   
   Therefore, the magnetic field is given by
   
   $${\bf B} = \nabla \times {\bf A} = {\mu_{0}I\over 4\pi}\int \nabla \left ({1\over r}\right ) \times {\bf ds}.$$
   
   
   Thus, we have
   

   $${\bf B} = {\mu_{0}I\over 4\pi} \int {{\bf ds} \times {\bf r}\over r^{3}}.$$ (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

$$B_{\phi} = {\mu_{0}I\over 4\pi}\int {\sin \theta \over r^{2}} dz.$$ (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.

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

$$\tan \theta = {R\over z}, \qquad \Rightarrow \qquad z = R {\cos \theta \over \sin \theta}.$$ (3.27)

$$\sin \theta = {R\over r}, \qquad \Rightarrow \qquad r = {R\over \sin \theta}.$$

Finally, we have

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

Hence, using (3.27), we have

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

Therefore, using the Biot-Savart law we have

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

Integrating we obtain

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

as obtained before.