My problem arises in Griffiths's book chapter 6, when he discusses the auxiliary field $\mathbf{H}$. At first he claims that Ampere's law inside a magnetized material is as follows:

$$\nabla \times \left( \dfrac{1}{\mu_0} \mathbf{B} - \mathbf{M} \right) = \mathbf{J}_f$$

And that is of course understandable since the bound surface current is not inside the Amperian loop. Now, when we read on we reach an example (in that same section):

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And then his solution for both the field inside the material and outside of it:

Applying Eq. 6.20 to an Amperian loop of radius $s < R$, $$H(2 \pi s) = I_{f_{\text{enc}}} = I \dfrac{\pi s^2}{\pi R^2},$$ so, inside the wire, $$\mathbf{H} = \dfrac{I}{2 \pi R^2} s \hat{\phi} \ \ \ (S \le R). \tag{6.21}$$ Outside the wire $$\mathbf{H} = \dfrac{I}{2 \pi s} \hat{\phi} \ \ \ (S \ge R). \tag{6.22}$$ In the latter region (as always, in empty space) $\mathbf{M} = \mathbf{0}$, so $$\mathbf{B} = \mu_0 \mathbf{H} = \dfrac{\mu_0 I}{2 \pi s} \hat{\phi} \ \ \ (s \ge R),$$

Now, as can be seen, for the outside case, he uses the equation he earlier defined for the inside case only - which doesn't contain the term expressing the contribution of the bounded surface current to the magnetic field, I was under the impressiong that the definition

$$\mathbf{H} \equiv \dfrac{1}{\mu_0} \mathbf{B} - \mathbf{M}$$

is only valid for Ampere's law inside a material. For the outside case I would except to see the term

$$\mathbf{K}_b = \mathbf{M} \times \mathbf{\hat{n}}$$

as well but that is not the case.

Can someone please clarify this for me?

  • $\begingroup$ Please write out your equations using LaTeX rather than posting them as images. Figures are fine as images. $\endgroup$ Jan 25, 2021 at 18:07
  • $\begingroup$ No problem, Thanks. $\endgroup$ Jan 25, 2021 at 18:25
  • $\begingroup$ Please see my edit $\endgroup$ Jan 25, 2021 at 18:26

2 Answers 2


In addition to the bound surface current you've identified, there are also bound volume currents inside the wire. But we can show that the net effects of all these bound currents cancel out for points outside the wire.

Specifically, imagine any loop $L$ encircling the wire. The net amount of bound current passing through this loop is $$ I_b = \iint_S \vec{J}_b \cdot d \vec{a}, $$ where $S$ is a surface spanning our loop $L$. Since $\vec{J}_b = \vec{\nabla} \times \vec{M}$, we can apply Stokes's theorem to find that $$ I_b = \iint_S \vec{J}_b \cdot d \vec{a} = \iint_S \left( \vec{\nabla} \times \vec{M} \right) \cdot d \vec{a} = \oint_L \vec{M} \cdot d\vec{l}. $$ But $\vec{M} = 0$ everywhere on the loop (since it sits outside the wire), and so $I_b = 0$ as well.

This means that when we apply Ampere's Law, we have $$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I = \mu_0 (I_b + I_f) = \mu_0 I_f, $$ i.e., only the free current contributes to the integral. If we further assume that the problem has axial symmetry (which is reasonable in this problem, assuming that the copper rod is homogenous), then Griffiths's derivation of the magnetic field follows.

Note that the above method applies to bound surface currents as well. We just have to view a surface current $\vec{K}$ as a volume current "with a delta function in it", similarly to how we can treat point charges or surface charges as density functions over all of space by inserting delta-functions judiciously. For example, if we assume that the copper rod is a linear medium, then we have $\vec{M} = \chi_m \vec{H}$, and so $$ \vec{J}_b = \chi_m \vec{\nabla} \times \vec{H} = \frac{\chi_m I}{\pi R^2} \hat{z}, \qquad \vec{K}_b = \chi_m \vec{H} \times \hat{n} = - \frac{\chi_m I}{2 \pi R} \hat{z}. $$ These two terms can be combined, if we wish, into a single distributional bound current density: $$ \vec{J}_b = \left[ \frac{\chi_m I}{\pi R^2} - \frac{\chi_m I}{2 \pi R} \delta(s - R) \right] \hat{z}. $$ Defined in this way, we can simply talk about a bound current density which satisfies $\vec{J}_b = \vec{\nabla} \times \vec{M}$ in a distributional sense, without worrying about the split between volume and surface currents. Note also that neither term (bound volume current or bound surface current) has zero net current on its own; it's only their net bound current, combined, that vanishes.

  • $\begingroup$ Your last paragraph is exactly what i was looking for. I understand why the bound volume currents don't affect the field outside the cylinder, what i couldn't understand is why griffith ignores completely the bound surface currents... Would you mind elaborating on the surface currents zero contribution to the outside magnetic field? $\endgroup$ Jan 25, 2021 at 18:38

The relation

$$ \mathbf B = \mu_0(\mathbf H + \mathbf M) $$

is general, it is valid both inside body, outside the body (in vacuum), and also at the boundary of the two.

Outside the body magnetization vanishes, $\mathbf M = 0$ so there the relation simplifies into $$ \mathbf B =\mu_0 \mathbf H. $$


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