Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Part A What force is felt by a singly ionized (positive) sodium ion? Submit My A

ID: 1325632 • Letter: P

Question

Part A

What force is felt by a singly ionized (positive) sodium ion?

SubmitMy AnswersGive Up

Part 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?

SubmitMy AnswersGive Up

Part C

What voltage will this electric field produce across an artery with a diameter of 3.6mm ?

SubmitMy AnswersGive Up

F_B = N Typical blood velocities in the coronary arteries range from 10.0 to 30.0 cm/s . An electromagnetic flowmeter applies a magnetic field of 0.234T to a coronary artery with a blood velocity of 10.1cm/s . As we see in the figure, this field will exert 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.(Figure 1) Part A What force is felt by a singly ionized (positive) sodium ion? SubmitMy AnswersGive Up Part 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? SubmitMy AnswersGive Up Part C What voltage will this electric field produce across an artery with a diameter of 3.6mm ? SubmitMy AnswersGive Up

Explanation / Answer

An electro magnetic flowmeter applies a magnetic field of B = 0.234 T

Artery blood velocity of v = 10.1 cm /s

                                           = 0.101 m/s

a ) Force is felt by singly ionized sodium

        F = q v B

            = 1.6*10^-19 C * 0.101 m/s * 0.234 T

            = 3.78*10^-21 N

b ) resulting electric field is E_H = v * B

                                                       = 0.101 m/s * 0.234 T

                                                       = 0.0236 N / C

  c ) Hall voltage is V_H = E_H * d

                                        = 0.0236 N / C * 3.6*10^-3 m

                                        = 0.8496*10^-4 V