To construct a non-mechanical water meter, a 0.500-T magnetic field is placed across the supply water pipe to a home and the Hall voltage is recorded. Required:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?

a.  4 m/s b. 0.2 V

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

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