Consider two thin sheets at z = 0 and z = a, respectively, each with uniform sur

Question

Consider two thin sheets at z = 0 and z = a, respectively, each with uniform surface current density: K_0 cap x at z = 0, and -K_0 cap x at z = a. What is vector B between the two sheets, and also outside (above and below) both sheets. Does this remind you of a familiar electrostatics problem at all? How? Griffiths derives a formula for the vector B field from a solenoid (Example 5.9). Rewrite his answer (which is in terms of I) so it is expressed in terms of K (see his Figs. 5.34 and 5.35 for help with this). Can you comment on the similarity to the situation in part 3a?

Explanation / Answer

Here we will be using amperes law to calculate B, just like we used Gauss law to calculate E.

By amperes law,

integral Bdr = uo i enclosed

Lets take a amperes rectangle in yz plane symmetrical to both side of origin.

B*2y = 2uo* integral Jy dz

B*2y = 2uo yJo integral zdz

B = uoJo z^2/2....for |z| <h i.e. inside slab

B = uoJoh^2/2 for |z| > h i.e. above amd below the slab

Total B = uo Jo (h^2+z^2) /4

b) For solenoid B= uo N i/L, can be found by amperes law.

    since figure is not given, we can not say anything further.

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