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 diameter d 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 ${\cal{E}}\: =\:$ .

B) If you expect that the blood will be flowing at 14.8 cm/s for a vessel4.80 mm in diameter, what strengthof magnetic field will you need to produce a potential differenceof 1.00 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

a) Thickness of the vessel \(=\mathrm{d}\) speed \(=v\) the blood which contains ions is like a conductor, when placed in a magnetic field a force on the ions is \(F_{s}=q v B\)

This force causes the charges to move, creating an excess charge at the upper and lower end, in turn creates an electric field. \(F_{\bar{z}}=q E\)

The charges are in equilibrium \(q E=q v \vec{B}\)

\(E=v B\)

The potential difference between the upper and lower end is \(V=E \cdot d\)

From the above equation substituting the value of \(\mathrm{E}\) and writing \(\varepsilon\) for \(\mathrm{v}\) \(\varepsilon=v B d\)

b) \(\quad\) Potential difference \(\mathrm{V}=1 \mathrm{mV}\) \(\begin{aligned} &=1 \times 10^{-1} \mathrm{v} \\ \mathrm{v} &=14.8 \mathrm{~cm} / \mathrm{s} \\ &=0.148 \mathrm{~m} / \mathrm{s} \end{aligned}\)

\(\begin{aligned} \text { Speed } & \mathrm{v}=14.8 \mathrm{~cm} \mathrm{~s} \\ &=0.148 \mathrm{~m} / \mathrm{s} \\ \mathrm{d} &=4.8 \mathrm{~mm} \end{aligned}\)

Thickness \(=4.8 \times 10^{-3} \mathrm{~m}\)

The motional emf \(V=v B d\)

\(B=\frac{V}{v d}\)

\(=\frac{1 \times 10^{-3} \mathrm{v}}{\left(4.8 \times 10^{-3} \mathrm{~m}\right)(0.148 \mathrm{~m} / \mathrm{s})}\)

\(=1.407 \mathrm{~T}\)

c) From part a \(v=\frac{\varepsilon}{B d}\)

The volume rate of flow is \(R=\) Area \(\times\) speed \(R=A v\)

\(=A\left(\frac{\varepsilon}{B d}\right)\)

\(=\left(\frac{\pi d^{2}}{4}\right)\left(\frac{\varepsilon}{B d}\right)\)

\(\mathrm{R}=\frac{\pi d \varepsilon}{4 B}\)