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Blood contains positive and negative ions and therefore is aconductor. A blood vessel, therefore, can be viewed as anelectrical wire. We can even picture the flowing blood as a seriesof parallel conducting slabs whose thickness is the diameterd of the vessel moving with speed{v}A) If the blood vessel is placed in a magnetic field B perpendicular to the vessel, as in the figure,show that the motional potential difference induced across it is{\cal{E}}\: =\:{vBd}.B) If you expect that the blood will be flowing at 14.8 {\rm cm}/{\rm s} for a vessel4.80 {\rm mm} in diameter, what strengthof magnetic field will you need to produce a potential differenceof 1.00 {\rm mV}?C)Show that the volume rate of flow (R) of the blood is equal to {R\:=\:}{{\pi {{\cal{E}}{d}}}\over {{{4}}{B}}}

Respuesta :

Answer:

Magnetic field required for the given induced EMF is 1.41 T

Explanation:

Potential difference across the blood vessel is given as

[tex]E = vBd[/tex]

here we know that the speed is given as

[tex]v = 14.8 cm/s[/tex]

[tex]d = 4.80 mm[/tex]

[tex]E = 1 mV[/tex]

now we have

[tex]1 \times 10^{-3} = (14.8 \times 10^{-2})B(4.80 \times 10^{-3})[/tex]

[tex]B = 1.41 T[/tex]

Now volume flow rate of the blood is given as

[tex]Q = Av[/tex]

[tex]Q = \frac{\pi d^2v}{4}[/tex]

from above equation we have

[tex]v = \frac{E}{Bd}[/tex]

Now we have

[tex]Q = \frac{\pi d^2\frac{E}{Bd}}{4}[/tex]

[tex]Q = \frac{\pi E d}{4B}[/tex]

"1.41 T" would be the required magnetic field.

According to the question,

Speed,

  • v = 14.8 mm

Diameter,

  • d = 4.80 mm

Potential difference,

  • E = 1 mV

Potential difference across the blood vessel will be:

→           [tex]E = vBd[/tex]

By substituting the values, we get

→ [tex]1\times 10^{-3} = (14.8\times 10^{-2})B (4.80\times 10^{-3})[/tex]

→            [tex]B = 1.41 \ T[/tex]

Now,

The volume flow rate will be:

→ [tex]Q = Av= \frac{\pi d^2 v}{4}[/tex]

from the above equation, we get

→ [tex]v = \frac{E}{Bd}[/tex]

then

→ [tex]Q = \frac{\pi d^2\frac{E}{Bd} }{4}[/tex]

      [tex]= \frac{\pi Ed}{4B}[/tex]

Thus the above approach is right.

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