Consider an interface between free space and an iron metal with $\mu = 1000\mu _{o}$ . In free space, $\pmb{B}_{1}$ makes an angle of $\theta _{1} = 1.0^{\circ }$ with the normal to the interface.

Find (a) direction of $\pmb{B}_{2}$ in the iron metal, and (b) ratio $\pmb{B}_{2}/ \pmb{B}_{1}$ .

Question

Consider an interface between free space and an iron metal with $\mu = 1000\mu _{o}$ . In free space, $\pmb{B}_{1}$ makes an angle of $\theta _{1} = 1.0^{\circ }$ with the normal to the interface.

Find (a) direction of $\pmb{B}_{2}$ in the iron metal, and (b) ratio $\pmb{B}_{2}/ \pmb{B}_{1}$ .

Find (a) direction of [latex]\pmb{B}_{2}[/latex] in the iron metal, and (b) ratio [latex]\pmb{B}_{2}/ \pmb{B}_{1}[/latex].

Accepted Answer

(a) From the boundary conditions for H and B, we get

[latex] \frac{1}{\mu } B_{2t} = \frac{1}{\mu_{o} } B_{1}\sin (1^{\circ }) (tangential component of \pmb{H}) (5-97a) \ B_{2n} = B_{1}\cos (1^{\circ }) (normal component of \pmb{B}) (5-97b)[/latex]

In the iron metal, we have
[latex] an heta _{2} = \frac{B_{2t}}{B_{2n} } = \frac{(\mu /\mu_{o} ) B_{1}\sin (1^{\circ })} {B_{1}\cos (1^{\circ })} = 17.46 \ Thus, heta _{2} =86.7^{\circ } [/latex]

(b) Combining Eqs. (5-97a) and (5-97b), we obtain

[latex]B_{2} =\sqrt{ (B_{2t})^{2} + (B_{2n} )^{2}} = B_{1}\sqrt {(1000\mu_{o} /\mu_{o} )^{2} \sin^{2} (1^{\circ }) + \cos^{2} (1^{\circ })} \ Thus, B_{2}/B_{1} =17.5 [/latex]

Question 5.13: Consider an interface between free space and an iron metal w......

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