Consider an electrophysiology experiment in which a 0.200 ~mm0.200 mm diameter round silver wire is used as a stimulation electr
ode. In this experiment, a current of 4.69~ mA4.69 mA is applied through the wire. The density of free electrons in silver is 5.86 \times 10^{28}~ \frac{\text{electrons}}{m^3}5.86×10 28 m 3 electrons . Calculate the magnitude of the drift velocity of the electrons in the silver wire under these conditions. Enter your answer in units of \frac{\text{meters}}{\text{second}} second meters using calculator scientific notation (e.g., 1.23e-41.23e−4 for 0.0001230.000123).
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Answer:
(a) The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.
Explanation:
(a) The amount of heat dissipated (), measured in joules, by the cylindrical resistor is the power multiplied by operation time (
), measured in hours. That is:
(1)
If we know that and
, then the amount of heat dissipated by the resistor is:
The resistor disspates 103680 joules during a 24-hour period.
(b) The heat flux (), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (
), measured in square meters:
(2)
(3)
Where:
- Diameter, measured in meters.
- Length, measured in meters.
If we know that ,
and
, the heat flux of the resistor is:
The heat flux of the resistor is approximately 4340.589 watts per square meter.
(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (), no unit, is the ratio of the top and bottom surfaces to total surface:
(3)
If we know that and
, then the fraction is:
The fraction of heat dissipated from the top and bottom surfaces is 0.045.