Because of a wartime shortage of copper, the electromagnets at Oak Ridge Nationa
ID: 1899177 • Letter: B
Question
Because of a wartime shortage of copper, the electromagnets at Oak Ridge National Lab used to separate U-235 isotopes were wound with pure silver (atomic volume 10.3 cc/mole). If 100 meters of silver wire 2 mm in diameter required 8.1 watts when carrying 4 amps of current, calculate for silver the (a) resistivity, (b) mobility, (c) collision time, (d) mean free path, (e) Hall coefficient, and (f) the Hall field in a transverse magnetic field of 9 tesla. Assume silver has one free electron per atom, and a temperature of 300K. (Note: Electrical power equals the product of the current and the voltage or the product of the current squared and the resistance.)Explanation / Answer
P=I^2*R
8.1=4^2*R
R=0.506 ohm
a) R = l/A
>>>> 0.506 = *100/(*0.001^2)
=1.59*10^-8 m
b)1/ = n*q*
Now free electron density of silver = 5.86*10^28 per m^3
= 1/ ( *n*q)
= 1/(1.59*10^-8 * 5.86*10^28 * 1.6*10^-19) = 149.078 cm^2/Vs
c) =(v*m)/ qE
v---drift velocity = E
m-----mass of electron
=( E*m)/qE = m/q
=149.078*9.1*10^-31 / (1.6*10^-19)
=8.478*10^-10 sec
d)Now time between collision is related as=d/vf
vf-----fermi velocity for silver is 1.39*10^6 m/sec
d------distance between collision = *vf = 8.478*10^-10*1.39*10^6 =1.178*10^-3 m
e)Hall coefficient
= 1/(q*n) = 1/(5.86*10^28 * 1.6*10^-19)
= 1.067* 10^-10 m^3/C
f)Hall field = IB/nql
=JB*Hall coefficient (as I/l = J)
=(2/100) *9*1.067* 10^-10
=1.92*10^-11 v