Question

D5.3. Find the magnitude of the current density in a sample of silver for
which σ = 6.17 × 107 S/m and μe = 0.0056m2/V · s if (a) the drift velocity
is 1.5μm/s ; (b) the electric field intensity is 1 mV/m; (c) the sample is a cube
2.5 mm on a side having a voltage of 0.4 mV between opposite faces; (d) the
sample is a cube 2.5 mm on a side carrying a total current of 0.5 A.

Answer 1

(a) let E = Electric field
\[E=\frac{v}{μe }\] where v is drift velocity
\[E=\frac{1.5\times10^{-6}}{0.0056 } \]
\[E=0.268mV.m\]
let J = curent density
\[J=\sigma E\]
\[J=(6.17\times10^{7})\times0.268\times10^{-3}\]
\[J=16535.6 A/m^{2}\]

b) \[J=\sigma E\]
\[J=(6.17\times10^{7})\times (1\times10^{-3})\]
\[J=61700A/m^{2}\]

c) \[E=\frac{V}{L}\]
\[E=\frac{0.4\times10^{-3}}{0.0025}\]
\[E=0.16V.m\]
\[J=(6.17\times10^{7})\times0.16\]
\[J=9872000A/m^{2}\]

d)\[J=\frac{I}{A}\]
\[A=2.5\times2.5\]\[A=6.25mm^{2}\]\[A=6.25\times10^{-6}m^2\]
\[J=\frac{0.5}{6.25\times10^{-6}}\] \[J=80000A/m^{2}\]

Answer 2

the first equation is
\[E=\frac{v}{\mu _{e} }\]where v is drift velocity

Answer 3

Interesting.

Answer 4

the equation is just E= v/μe i dont know why it repeated it self three times

Answer 5

(>. >)σ - - - - - - - - - - - ( x-x )

Answer 6

i'm horny