정보저장재료 Report Solution

1.1

sol) [eqn: F=pH= {p _{1} p _{2}} over {d ^{2}}]

[eqn: H= {p} over {d ^{2}}] , [eqn: M= {m} over {V} = {pl} over {V} = {p} over {A}]

thus, the unit of H, M is [eqn: erg/Oe BULLET cm ^{3}]

1.2

a.

sol) [eqn: M`=` {m} over {V} = {7.45 TIMES 10 ^{3}} over {LEFT ( {1.25} over {2} RIGHT ) ^{2} TIMES 6 TIMES pi } =1011.8`emu/cm ^{3}]

b.

sol) [eqn: m= {nAi} over {10}]

thus, [eqn: i= {10m} over {nA} = {10 TIMES 7.45 TIMES 10 ^{3}} over {200 TIMES LEFT ( {1.25} over {2} RIGHT ) ^{2} TIMES pi } =303.54`(A)]

c.

sol) [eqn: m= {nAi} over {10}]

thus, [eqn: m= {200 TIMES LEFT ( {1.25} over {2} RIGHT ) ^{2} TIMES pi TIMES 1.5} over {10} =36.81`(erg/Oe)]

1.3

sol)  [eqn: 1.2T`=`1.2 TIMES 10 ^{4} Oe]

[eqn: chi (cgs)= {chi (SI)} over {4 pi } = {2.0 TIMES 10 ^{-6}} over {4 pi } =1.59 TIMES 10 ^{-7}]

[eqn: chi (cgs)= {M} over {H}]

→ thus, [eqn: M= chi H=1.59 TIMES 10 ^{-7} TIMES 1.2 TIMES 10 ^{4} =1.91 TIMES 10 ^{-3} Oe]

[eqn: M= {m} over {V}]

→ thus, [eqn: m=MV=1.91 TIMES 10 ^{-3} TIMES pi LEFT ( {1.25} over {2} RIGHT ) ^{2} TIMES 6=1.4 TIMES 10 ^{-2} emu]

1.4

a.

sol) [eqn: H`=` {4 pi n`i} over {10L} = {4 pi TIMES 250 TIMES 1.5} over {10 TIMES ( {5.5} over {2} ) TIMES 2 pi } =27.27Oe]

b.

sol) [eqn: PHI _{observed} =BA]

thus,

[eqn: B= {PHI _{observed}} over {A} = {8.25 TIMES 10 ^{-3} ``webber} over {1.2``cm ^{2}} = {8.25 TIMES 10 ^{-3} TIMES 10 ^{4} `Oe TIMES (10 ^{2} `cm) ^{2}} over {1.2``cm ^{2}}]

[eqn: B=H+4 pi M]

[eqn: M= {B-H} over {4 pi } = {6.875 TIMES 10 ^{5} -27.27`Oe} over {4 pi } =5.471 TIMES 10 ^{4} `emu/cm ^{3}]

c.

sol) [eqn: mu = {B} over {H} = {6.875 TIMES 10 ^{5}} over {27.27} =2.52 TIMES 10 ^{4}]