Answer:
a. 4 m/s b. 0.2 V
Step-by-step explanation:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
The hall voltage V = vBd where v = flow-rate, B = magnetic field strength = 0.500 T and d = diameter of pipe = 3.00 cm = 0.03 m
Since V = vBd
v = V/Bd given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have
v = 0.060 V/(0.500 T × 0.03 m)
v = 0.060 V/(0.015 Tm)
v = 4 m/s
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?
Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m
Substituting the variables into the equation, we have
V = vBd
V = 4 m/s × 0.500 T × 0.10 m
V = 0.2 V