Question

1. This problem investigates the use of the Hall effect to study blood flow. When a current / travels along a conducting plate of width w, which lies perpendicular to a magnetic field B, a potential difference V develops across the plate as shown in the figure below. + + + + + + a. Show the direction of the electric field on the diagram. (Either copy the diagram from this document, or redraw it in your solution.) b. Draw two free body diagrams: one for the forces on a positive charge and one for the forces on a negative charge. c. Show that, in equilibrium, the potential difference across the plate is given by V = vBw. where v is the velocity of the charges moving in the conductor. This potential difference is known as the Hall Voltage or Hall emf and this phenomenon is known as the Hall effect. d. Suppose we wish to use the Hall effect to measure the flow of blood in an artery. If we apply a magnetic field transverse to the direction of blood flow, it will cause the positively and negatively charged ions in the blood to separate and result in a voltage across the artery. Assume that the artery has a square cross sectional area, where a =4.0 mm is the length of each side. Assume that the velocity is the same throughout the cross section and that the applied magnetic field has a magnitude 0.50 mt. If a voltage difference of 1.0 uV is produced, what is the magnitude of the velocity of the blood in the artery? e. What is the volume flowrate of blood in the artery, expressed in units of mL/s?

please answer d & e with full explanations, thanks!

Answer 1

d) we need to use hall effect equation which is given as

V = BvL

where V is emf, B is field, v is velocity, L is length

so,

v = V / BL

v = 1e-6 / 0.5e-3 * 4e-3

v = 0.5 m/s

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e) To find volume flow rate, we need the equation

Q = vA

where A is area of cross section, as we have a square cross section , area = L²

Q = vL²

Q = 0.5 * (4e-3)²

Q = 8e-6 m³ / s

we need this in mL/s

so,

1 m³ / s = 100000 mL/s

so,

Q = 8 mL/s