Question

pls answer question 3-4

40. Heart-lung machines and artificial kidney machines BIO employ electromagnetic blood pumps. The blood is con- QIC fined to an electrically insulating tube, cylindrical in prac- S tice but represented here for simplicity as a rectangle of interior width w and height h. Figure P28.40 shows a rect- angular section of blood within the tube. Two electrodes fit into the top and the bottom of the tube. The potential difference between them establishes an electric current through the blood, with current density Jover the section of length L shown in Figure P28.40. A perpendicular magnetic field exists in the same region. (a) Explain why this arrange- ment produces on the liquid a force that is directed along the length of the pipe. (b) Show that the section of liquid in the magnetic field experiences a pressure increase JLB. (c) After the blood leaves the pump, is it charged? (d) Is it carrying current? (e) Is it magnetized? (The same electro- magnetic pump can be used for any fluid that conducts elec- tricity, such as liquid sodium in a nuclear reactor.) Figure P28.40 41. Review

Answer 1

Question (1), Part (a)

When a current carrying conductor is placed in a magnetic field so that direction of current is perpendicular to the direction of magnetic field, the conductor experience an electromagnetic force . Blood contains small amount of ionic salts like Na and Ca compounds. Presence of salts in the blood gives electrical conductivity .

Hence when current is passed through the blood in a direction perpendicular to the direction of magnetic field, part of blood that conducts current is subjected to J x B force , where J is current density i.e, current per unit area and B is magnetic field induction. Direction of this J x B force is perpendicular to the direction of current and also perpendicular to direction of magnetic field as shown in figure given below .

Z JxB B -B -y -- y X -Z

Hence blood is directed to flow along the channel direction due to the force J x B .

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Question (1), Part (b)

The force ( J x B ) mentioned in part-(a) is the force per unit volume .

Total force acting on blood = ( J $\times$ B ) $\times$ A $\times$ L

where A is the area of cross section and L is elecrode length

Pressure = force /area = J $\times$ B $\times$ L

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Question (1), Part (c), Part(d) , Part(e)

After blood leaving the pump , it is not charged , not carrying current and not magnetized

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Second question

e . . R

Radius R of circular path of charged particle in magnetic field is obtained from the following relation

q $\times$ v $\times$ B = m $\times$ ( v² / R ) .........................(1)

where q is the magnitude of charge on charge particle, v is speed of charged particle, B is magnetic field induction and m is mass . Left hand side of above equation is magnetic force acting on charged particle and right hand side is centripetal force required for circular motion.

Hence we get from above equation , R = ( m v ) / ( q B )   ................... (2)

As seen from figure , ( h / R ) = sin$\theta$ or sin$\theta$ = ( h q B ) / ( m v ) ...............(3)

Let us calculate the RHS of above eqn.(3) using the given values

sin$\theta$ = ( 0.11 $\times$ 10^-6$\times$ 0.4 ) / ( 2 $\times$ 10^-13$\times$ 2 $\times$ 10⁵ ) = 1.1

If we use the given exact values for h, q, B, m and v , we get sin$\theta$ > 1 .

Hence saying the experiementally measured deflection matching with the prediction is wrong .

Even if small variation in the given values sin$\theta$ may approach 1 , i.e. $\theta$ $\approx$ 90^o , hence charge particle may not come out of magnetic field region in the forward direction of charged particle.