Question

Typical blood velocities in the coronary arteries range from 10 to 30 cm/s. An electromagnetic flowmeter applies a magnetic field of 0.25 T to a coronary artery with a blood velocity of 15 cm/s. This field exerts a force on ions in the blood, which will separate. The ions will separate until they make an electric field that exactly balances the magnetic force. This electric field produces a voltage that can be measured.

a. What force is felt by a singly ionized (+) sodium ion?
b. Charges in the blood will separate until they produce an electric field that cancels this magnetic force. What will be the resulting electric field?
c. What voltage will this electric field produce across an artery with a diameter of 3.0mm?

Answer 1

The magnetic field B = 0.25T

the velocity of the blood v = 15cm/s

(a) The force is felt by singly ionized ion

              F = qvB sinθ

here the velocity and magnetic field are perpendicular to each other

so      F = qvB

            = (1.6*10^-19C) ( 0.15m/s) (0.25T)

            = 0.06 *10^-19 N or 6*10^-21 N

(b) When the electric force due to electric field and magnetic force due to magnetic field are equal

               then   qE = qv B

                        

                          E = vB = (0.15m/s) (0.25T)

                              = 3.75*10^-2 N/C

(c) When the artery diameter is d = 3.0mm

       then the voltage   V = Ed

                                     = (3.75*10^-2) (3*10^-3m)

                                     = 11.25*10^-5 V

                                     = 112.5 μV