Question

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}}}

Answer 1

Answer:

Magnetic field required for the given induced EMF is 1.41 T

Explanation:

Potential difference across the blood vessel is given as

$E = vBd$

here we know that the speed is given as

$v = 14.8 cm/s$

$d = 4.80 mm$

$E = 1 mV$

now we have

$1 \times 10^{-3} = (14.8 \times 10^{-2})B(4.80 \times 10^{-3})$

$B = 1.41 T$

Now volume flow rate of the blood is given as

$Q = Av$

$Q = \frac{\pi d^2v}{4}$

from above equation we have

$v = \frac{E}{Bd}$

Now we have

$Q = \frac{\pi d^2\frac{E}{Bd}}{4}$

$Q = \frac{\pi E d}{4B}$