Magnetic forces acting on conducting fluids provide a convenient means of pumpin

Question

Magnetic forces acting on conducting fluids provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that can be caused by a mechanical pump. A horizontal tube with rectangular cross section (height h, width w) is placed at right angles to a uniform magnetic field with magnitude B so that a length l is in the field (the figure (Figure 1) ). The tube is filled with a conducting liquid, and an electric current of density J is maintained in the third mutually perpendicular direction.

Alternative Exercise 27.110 Magnetic forces actIng on conducting flukds provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that car be caused by a mechanical pump. A horizontal tube with rectangular cross section (height h, width w) is placed at right angles to a unitorm magnetic field with magnitude B so that a length ils in the fle d (the figure (Elgure 1) The tube is filled with a conducting liquids and an electric current of density J is maintained in the third mutually perpendicular direction. Figure Liquid metal or blood Part A Find the difference of pressure between a point in the liquid on a vertical plane through ab and a point in the liquid on another vertical plane through cd, under conditions In which the liquid Is prevented from fowing Express your answer in terms of the variables J, l, B, w, and h AP J B My Answers Give Up Submit Correct Part B What current density is needed to provide a pressure difference of 1.00 atm between these two points if B 2.20T and 35.0rrun? Submit My Answer a Give up incorrect; Try Again; 4 attempts remaining Provide Feedback Continue

Explanation / Answer

Part A

current will be I = J*(cros section area) = J* w*l

force on current carring conductor in magnetic fiel B will be F = I*h*B = J*w*l*h*B

pressure difference= force/area = J*w*l*h*B/w*h = J*l*B

Part B pressure difference = 1.00 atm = 101325 Pa = 101325 N/m2

From part A J= (pressure difference)/l*B = 101325 N/m2/(2.2T)*(35*10-3m) = 1316 *103 A/m2

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