Exercise 4.2.3: Refraction of B at an interface

The normal component of $B$ is continuous: $B_{2z}=B_{1z}=5\mathrm{mT}$. With $J_{nS}=0$ the tangential component of $H$ is continuous, $B_{2\tau }/{\mu }_2=B_{1\tau }/{\mu }_1$:

($|B_2|=10.3\mathrm{mT}$. For comparison $H_1=B_1/2{\mu }_0=796a_x-1194a_y+1989a_{z}A/m$: the tangential part is indeed the same, the normal part jumps by ${\mu }_1/{\mu }_2=2/5$.)

$\tan \theta =|B_{\tau }|/B_n$:

in agreement with Eq. (4.2.13).

$M=B-{\mu }_{0}H=\left(1-1/{\mu }_r\right)B$:

The line of $B$ bends away from the normal on entering the medium of higher $\mu$, from $35.8^{\circ }$ to $61.0^{\circ }$. For iron against air (${\mu }_r\sim 10^3$) the bending is so strong that flux leaves the iron almost normally whatever its direction inside, the fact that makes a magnetic circuit behave like a wire circuit. Note that the tangential part of $M$ jumps at the interface, from $(1,-1.5)$ to $(4,-6)$ mT in its $x$ and $y$ components. This jump is the bound surface current $J_{ms}=(M_1-M_2)\times a_n/{\mu }_0=3581a_x+2387a_y$ A/m that the interface carries, although the free surface current is zero. (The jump of $M_n$, from $2.5$ to $4$ mT, adds nothing to it, because $a_n\times a_n=0$.)