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Next: Magnetostatics in Magnetic Media Up: Magnetostatic Fields Previous: Localized Current Distribution

Exercises

  1. Consider two thin current loops. Let loops $ 1$ and $ 2$ carry the currents $ I_1$ and $ I_2$ , respectively. The magnetic force exerted on loop 2 by loop 1 is [see Equation (616)]

    $\displaystyle {\bf F}_{21} = \frac{\mu_0\,I_1\,I_2}{4\pi}\oint_1\oint_2\frac{d{\bf r}_2\times (d{\bf r}_1\times {\bf r}_{12})}{\vert{\bf r}_{12}\vert^{\,3}},$

    where $ {\bf r}_{12} = {\bf r}_2-{\bf r}_1$ . Here, $ {\bf r}_1$ and $ {\bf r}_2$ are the position vectors of elements of loops $ 1$ and $ 2$ , respectively. Demonstrate that the previous expression can also be written

    $\displaystyle {\bf F}_{21} =- \frac{\mu_0\,I_1\,I_2}{4\pi}\oint_1\oint_2\frac{(d{\bf r}_1\cdot d{\bf r}_2)\,{\bf r}_{12}}{\vert{\bf r}_{12}\vert^{\,3}}.
$

    Hence, deduce that

    $\displaystyle {\bf F}_{12} = -{\bf F}_{21},
$

    in accordance with Newton's third law of motion.

  2. Consider the two current loops discussed in the previous question. The magnetic field generated at a general position vector $ {\bf r}$ by the current flowing around loop $ 1$ is [see Equation (614)]

    $\displaystyle {\bf B}({\bf r}) = \frac{\mu_0\,I_1}{4\pi}\oint_1 \frac{d{\bf r}_1\times ({\bf r}-{\bf r}_1)}{\vert{\bf r}-{\bf r}_1\vert^{\,3}}.$

    Demonstrate that

    $\displaystyle {\bf B} = \nabla\times {\bf A},$

    where

    $\displaystyle {\bf A}({\bf r})= \frac{\mu_0\,I_1}{4\pi}\oint_1\frac{d{\bf r}_1}{\vert{\bf r}-{\bf r}_1\vert}.
$

    Show that the magnetic flux passing through loop $ 2$ , as a consequence of the current flowing around loop $ 1$ , is

    $\displaystyle {\mit\Phi}_{21} = \frac{\mu_0\,I_1}{4\pi}\oint_1\oint_2 \frac{d{\bf r}_2\cdot d{\bf r}_1}{\vert{\bf r}_1-{\bf r}_2\vert}.
$

    Hence, deduce that the mutual inductance of the two current loops takes the form

    $\displaystyle M = \frac{\mu_0}{4\pi}\oint_1\oint_2 \frac{d{\bf r}_1\cdot d{\bf r}_2}{\vert{\bf r}_2-{\bf r}_1\vert}.
$

  3. The vector potential of a magnetic dipole of moment $ {\bf m}$ is given by

    $\displaystyle {\bf A}({\bf r}) = \frac{\mu_0}{4\pi}\,\frac{{\bf m}\times {\bf r}}{r^{\,3}}.$

    Show that the corresponding magnetic field is

    $\displaystyle {\bf B}({\bf r}) = \frac{\mu_0}{4\pi}\left[\frac{3\,({\bf r}\cdot{\bf m})\,{\bf r}- r^2\,{\bf m}}{r^{\,5}}\right].
$

  4. Demonstrate that the torque acting on a magnetic dipole of moment $ {\bf m}$ placed in a uniform external magnetic field $ {\bf B}$ is

       $ \tau$ $\displaystyle = {\bf m}\times {\bf B}.
$

    Hence, deduce that the potential energy of the magnetic dipole is

    $\displaystyle W = - {\bf m}\cdot{\bf B}.
$

  5. Consider two magnetic dipoles, $ {\bf m}_1$ and $ {\bf m}_2$ . Suppose that $ {\bf m}_1$ is fixed, whereas $ {\bf m}_2$ can rotate freely in any direction. Demonstrate that the equilibrium configuration of the second dipole is such that

    $\displaystyle \tan\theta_1 = - 2\,\tan\theta_2,
$

    where $ \theta_1$ and $ \theta_2$ are the angles subtended by $ {\bf m}_1$ and $ {\bf m}_2$ , respectively, with the radius vector joining them.


next up previous
Next: Magnetostatics in Magnetic Media Up: Magnetostatic Fields Previous: Localized Current Distribution
Richard Fitzpatrick 2014-06-27