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3 years ago

What hall voltage is produced by a 0.200-t field applied across a 2.60-cm-diameter aorta when blood velocity is 60.0 cm/s?

1 answer:

amm18123 years ago

8 0

The hall voltage will be calculated using the formula:

E = Blv

where:

>Hall voltage: E = ?
>Magnetic field:
B = 0.200 Tesla or Wb/m^2
>Width of conductor or Diameter of Aorta:
l = 2.60 cm, converting to meter = .0260 m
>Velocity of charge flowing:
v = 60 cm/s, converting to meter = 0.6 m/s

Substituting the given :

E = (0.200 Wb/m^2) * (0.260 m) * (0.6 m/s)
E = (0.200 Wb/m^2) * (0.156 m^2/s)
E = 0.0312 Wb/s

Since 1 volt = 1 Wb/s then,

E = 0.0312 V or 31.2 mV

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(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

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So, the inlet velocity is 1.4035 m/s.

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$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

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