Using an electromagnetic flow meter, see Fig, a heart surgeon monitors the flow

Question

Using an electromagnetic flow meter, see Fig, a heart surgeon monitors the flow rate of blood through an artery. The blood contains positive and negative ions or charged particles. Electrodes A and B make contact with the outer surface of the blood vessel, which has interior diameter 3 mm. For a magnetic field magnitude of 0.04 T, a potential difference of 160 mu Volt appears between the electrodes. Calculate the speed of the blood? 1mu = 1 times 10^-6m a. 0.76 m/s b. 4.25 m/s c. 1.33 m/s d. 2.06 m/s

Explanation / Answer

Diameter of blood vessels, d=3mm = 3 * 10-3m

Magnetic field, B=0.04 T

Potential difference between electrodes, V=160µ Volt =160*10-6 Volt

Let speed of blood , v=?

As         

                qvB=qE               where E= electric field between electrodes=V/d.

so           qvB=qV/d

                  v=V/Bd=160*10-6 / (0.04*3*10-3 ) =1.33 m/s.

                  v=1.33 m/s.

Hence speed of blood is 1.33m/s.

So correct option is (C)

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