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I'm having some trouble with this exercise:

An infinite sheet of parallel faces with thickness $d$ is uniformly magnetized. Its magnetization is 10 A / m and forms an angle of 60º with the normal to the sheet. Calculate B inside and outside the sheet.

Attempt:

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Firstly, I've calculated the magnetization vector $\vec{M}= \frac{M}{2}(0,\sqrt{3},1)$ in the figure's frame. Then I've obtained the current densities $\vec{J_m} = 0$ (as it is a constant magnetization) and $\vec{K_m}(z=0)=-\frac{\sqrt{M}}{2}M \vec{u_x}, \vec{K_m}(z=d)=+\frac{\sqrt{M}}{2}M \vec{u_x}$

Then I hesitate in calculating the magnetic field vector $\vec{B}$:

  • If I apply the Ampère's Integral Form Law, as the total magnetic current is $0$, then I'd get $\vec{B}=0$.
  • However, if I take the formula: $$\vec{B} = \frac{\mu_0}{4\pi}(\int\int_S \frac{\vec{K_m}(z=0) \times \vec{R}}{R^3}da + \int\int_S \frac{\vec{K_m}(z=d) \times \vec{R}}{R^3}da)$$

my integral becomes divergent as the sheet is infinite and I have to take the limits of the integral from $-\infty$ to $+\infty$. So, is $\vec{B}=0$? Why in the second case is it not possible to calculate the integral?

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  • $\begingroup$ Your magnetization vector need to be multiplied by 10 (amps/m). $\endgroup$
    – R.W. Bird
    May 18 at 13:32
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The integrals are (conditionally) convergent. To see this, Specifically, let the point we are evaluating the field be $(0,0,z)$, and let $(x', y', 0)$ be the coordinates of a current element on the sheet. The contribution to the field from this element will be $$ d \vec{B} = \frac{\mu_0}{4 \pi} \frac{(K \hat{x}) \times (z \hat{z} - x' \hat{x} - y' \hat{y})}{[z^2 + (x')^2 + (y')^2]^{3/2}} = -\frac{\mu_0 K}{4 \pi} \frac{z \hat{y} + y' \hat{z}}{[z^2 + (x')^2 + (y')^2]^{3/2}} $$ This must then be integrated over $x' \in (-\infty, \infty)$ and $y' \in (-\infty, \infty)$.

It is not hard to see that the $z$-component of $\vec{B}$, $$ B_z = -\frac{\mu_0 K}{4 \pi} \iint \frac{y'}{[z^2 + (x')^2 + (y')^2]^{3/2}} \, dx' \, dy' $$ will vanish when integrated over all $y'$, since it is an odd function of $y'$. And the $y$-component of the integral, $$ B_y = -\frac{\mu_0 K}{4 \pi} \iint \frac{z}{[z^2 + (x')^2 + (y')^2]^{3/2}} \, dx' \, dy' $$ falls off sufficiently fast that the integral converges to a finite value. (To actually perform this integral, it is easiest to switch to polar coordinates in the $x'y'$-plane.)

Strictly speaking, the integral for the $z$-component is conditionally convergent, and the value of such an integral depends on how we take the limit as the region of integration goes to infinity. The symmetry argument allows us to sidestep this by effectively choosing a "natural" way of taking the region of integration to infinity.

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  • $\begingroup$ Thanks, I understand now. So I have solved the integrals (which were not divergent at all) and I've got $B_y=\frac{\sqrt{3}}{2}\mu_0M_0$ inside the region. So then, if we consider a region out of the sheet, $\vec{B}$ will be $0$, right? $\endgroup$
    – conradDell
    May 19 at 9:38
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The magnetic dipole moment per unit volume associated with the y component of M should be the same as that which would be generated by two sheets of current (8.66 a/m) flowing in opposite directions on the surfaces of the magnetized sheet (in the +x direction on top, and the -x direction on the bottom). Use those currents with Ampere's law to find $B_y$, and the angle to get B.

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