SOLUTIONS FOR SUPERCONDUCTIVITY PHYSICS PROBLEMS

  1. V=IR, therefore R=V/I so 4.2 millivolts/300 miliamps = 0.014 {Omega}
  2. R x (area / length) = resistivity ({rho}) {rho} = (0.014 {Omega}) x (2.5mm x 3.4mm)/2.5cm = 0.000476 {Omega}-cm
  3. R = {rho} x (2.5cm)/(2.5mm x 3.4mm) = 5.3 x 10^-^7{Omega} for copper and 3.5 x 10^-^6{Omega} for stainless steel.

  4. If the contacts are excellent, the current is simply I=(1.5 V)/(10 {Omega})=0.15 ampere when the two are in series. There is no voltage drop across the superconductor until I exceeds the sample's critical current, of density J_c=I_c/(2.5mm x 3.4mm) >= .018A/mm^2.
  5. According to Ohm's Law, E = {rho}J. If {rho} (resistivity) is equal to zero inside the superconductor while the current is flowing, the electric field must be zero. Similarly, V=IR, the voltage is a result of a potential difference which infers the existence of an electric field.
  6. Magnetic moments are expected to be the saturation magnetizations (per unit mass) multiplied by sample masses. The forces must be equal to sample weights (masses multiplied by gravitational acceleration, g=9.8 m/s^2). Solving for the magnet's separation x from the reflector, twice the distance, r, from the image gives x=1/2[3M^2 x (10^-^3 kg)^2/g]^1^/^4 = 1/2[3 x (1.0 A-m^2/kg)^2 x (10^-^3 kg)^2/(9.8 m/s^2)]^1^/^4 = 12 mm for Nd_2Fe_1_6B. For Fe or Ni no levitation is observed because x = 1/2[3 x (5 mA-m^2/kg)^2 x (10^-^3 kg)^2/(9.8 m/s^2)]^1^/^4 = 0.8mm and 1/2[3 x (0.17A-m^2/kg)^2 x (10^-^3 kg)^2/(9.8 m/s^2)]^1^/^4 = 5 mm respectively, both less than 6 mm, their minimum sizes, the diameters 2[(3 x (1 g)/ x {pi} x (8 or 9 g/cm^3)]^1^/^3 of spheres.
  7. The current beginning at I_o when t=o decays to I_o/e = 0.368I_o after the characteristic time L/R=(4{pi} x 10^-^7 V-s/A-m) x (.03 m) x [ln(8 x .03m/2 x 10^-^4m) -7/4]/(0.01{Omega} )=20 microseconds or (4{pi} x 10^-^7 V - s/A - m) x (0.03 m) x [ln(8 x 0.03m/2 x 10^-^4m)-7/4]/(2.5 x 10^-^5{Omega})=8 milliseconds.
  8. The current I= {pi}b^2B{omega} [Rcos({omega} t)-i{omega} Lsin(t)]/(R^2+ {omega}^2L^2)= {pi}(0.03m)^2B x 2{pi} x 60s^-^1x [(10 m{Omega} or 2.5 {mu}{Omega})cos({omega} t)-i(2{pi} x 60s^-^1 x 2 x 10^-^7 Henry x sin({omega} t)]/[(10m{Omega} or 2.5{mu}{Omega} )^2 + (240{pi} x 10^-^7 {Omega})^2] = Bsin({omega} t)x 106 A/T or x 1.9 MA/T. One Tesla (T) equals 10,000 Gauss (G) of magnetic field. 
  9. 
 Voltage    T (K)   R (Ohm)      Voltage    T (K)    R (Ohm)

0.0010370   118.2   0.010368    0.0008440   93.5     0.008439
0.0010270   116.1   0.010266    0.0007830   93.2     0.007831
0.0010600   114.8   0.010597    0.0006390   93       0.006393
0.0010490   112.9   0.010493    0.0005050   92.6     0.005046
0.0010350   110.9   0.010351    0.0003790   92.3     0.003789
0.0010220   109.1   0.010216    0.0002430   92.1     0.002431
0.0010090   106.9   0.010095    0.0000930   91.7     0.000934
0.0010010   105     0.010011    0.0000100   91.4     0.000097
0.0009890   103.5   0.009894    0.0000030   91       0.000030
0.0009750   102.2   0.009755    0.0000002   90.8     0.000002
0.0009670   100     0.009673   -0.0000002   90.1    -0.000002
0.0009510    97.9   0.009511   -0.0000001   98.9    -0.000001
0.0009440    95.8   0.009439    0.0000003   89.5     0.000003
0.0009180    95     0.009176   -0.0000001   88.8    -0.000001
0.0009110    94.3   0.009112    0.0000001   88.5     0.000001
0.0008920    93.8   0.008916
  10. Graph
  11. The first derivative is represented by the slope of the graph in the normal phase. slope = dR/dT = {Delta}rise/ {Delta}run = (0.01065 - 0.009263)/(118.2 - 95) slope = 5.98 x 10^-^5{Omega}/K
  12. T_c = 92 K. This is experimental data that shows the curved low-temperature tail of a transition. Within the narrow temperature range of the tail, individual grains of YBa_2Cu_3O_7 superconductive. Resistance in not zero because the grain boundaries, surface area where the particles were sintered together, must be cooled further to become superconductive.
  13. See the Applications of Superconductors section. 

    

    

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    Date posted 04/01/96 (ktb)