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Measuring Blood Flow. Blood contains positive and negative ions and thus is a conductor. A blood vessel, therefore, can be viewed as an electrical wire. We can even picture the flowing blood as a series of parallel conducting slabs whose thickness is the diameter d of the vessel moving with speed v. (See Fig. E29.34.) (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 E = vBd. (b) If you expect that the blood will be flowing at 15 cm/s for a vessel 5.0 mm in diameter, what strength of magnetic field will you need to produce a potential difference of 1.0 mV? (c) Show that the volume rate of flow (R) of the blood is equal to R = πEd/4B. (Note: Although the method developed here is useful in measuring the rate of blood flow in a vessel, it is limited to use in surgery because measurement of the potential E must be made directly across the vessel.)

Step by step solution

01

Proving ε=Bvd

The blood slab has a flowing blood which has a width of d and is perpendicular to the field with a speed of v, which is equivalent to a rod of length d, and is moving in a magnetic field B, hence the induced emf is:$\mathrm{Î\mu }=Bvd$

02

Calculating the magnetic field

The blood is flowing at a speed of v = 15cm/s , vessel with diameter d = 5.0mm, and the potential difference of $\mathrm{Î\mu }=1.mV$. Therefore, the magnetic field is given by:

$$\begin{gathered} \mathrm{B}=\frac{\mathrm{Î\mu }}{\mathrm{vd}} \\ =\frac{1.0×{10}^{-3}\mathrm{V}}{\left(0.15\mathrm{m}/\mathrm{s}\right)\left(5.0×{10}^{-3}\mathrm{m}\right)} \\ =1.3\mathrm{T} \end{gathered}$$

03

Rate of flow of blood

The cross-sectional area of blood vessel is:

$\mathrm{A}=\frac{{\mathrm{Ï€»å}}^2}{4}$

Now, the volume of blood flowing past the cross-section of the vessel equals to the distance that is travelled in time t is x (let), therefore:

$v=\frac{\mathrm{Ï€}{d}^{2}x}{4}$

Since the blood is moving with a velocity of , then therefore the equation becomes:

$\mathrm{V}=\frac{\left({\mathrm{Ï€»å}}^2\mathrm{vt}\right)}{4}$

Therefore, the volume rate is:

$$\begin{gathered} R=\frac{V}{t} \\ =\frac{\mathrm{Ï€}{d}^{2}v}{4} \\ =\frac{\mathrm{Ï€}{d}^{2}v}{4}\frac{\mathrm{Î\mu }}{Bd} \\ =\frac{\mathrm{Ï€}\mathrm{Î\mu }d}{4B} \\ =\frac{\mathrm{Ï€}\mathrm{Î\mu }d}{4B} \end{gathered}$$

Hence, proved that blood flow is$\frac{\mathrm{Ï€}\mathrm{Î\mu }d}{4B}$

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