A coil of wire with 205 circular turns of radius 3.90cm is in a from magnetic fi

Question

A coil of wire with 205 circular turns of radius 3.90cm is in a from magnetic field along the axis of the coil. The coil has R = 44.1ohm. At rate, in teslas per second, must the magnetic field be changing to induce current of 0.152A in the coil? A typical arteriole has a diameter of 0.080mm and carries blood at the rate of 9.6 times 10^-6 cm^3/s. What is the speed of the blood in an arteriole? Suppose an arteriole branches into 8800 capillaries, each with a diameter of 6.0 times 10^-6m. What is the blood speed in the capillaries? (The low speed in capillaries is beneficial: it promotes the diffusion of materials to and from the blood.)

Explanation / Answer

3)

a)

well if you really need the speed in units of cm/s, you can imagine that a tiny blood volume croos the section of the arteriole in a certain time. So, as we have the flow of blood, we have to divide this by the area of the section of the arteriol. Let's calculate the Area:

A = *(d/2)^2 Being d= diameter

Let's change units of length so we work only with cm

1mm = 0.1cm
0.080 mm=0.0080 cm

A = *(0.0080cm/2)^2 = 5.024 x 10^-5 cm^2

Then the speed you are trying to get is:

v = (9.6 x 10^-6)(cm3/s) / (5.024 x 10^-5 cm^2)

v = 0.1911 cm/s

b)

Here we do the same calculation:

Let's get the area of the section of the capillaries:

A = *(d/2)^2

d = 6 x 10^-6 m. Let's change to cm
1m = 100cm
6 x 10^-6 m = 6 x 10^-4 cm.

A = *((6 x 10^-4) / 2)^2

A = 0.2826 x 10^-6 cm^2

Now we have to remember that the flow is divide in equal parts, so the volume by seconds is:

q = (9.6 x 10^-6) / 8800 = 0.001090 x 10^-6 cm^3/s, so the speed in this case is:

v = (0.001090 x 10^-6)(cm^3/s) / 0.2826 x 10^-6 cm^2

v = 0.003860 cm/s

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