Solving the Speed of Blood Flow in a Magnetic Field

[Steve F]

1. The question...

"Blood is a conducting fluid. When flowing through an artery of diameter 10mm which is subject to a constant magnetic field of strength 60mT, an e.m.f. of 0.3mV appears across the width of the artery. Calculate the speed of the blood"

3. The solution...

The solution given is
E = BA/t (I get that bit)
and
E = BLv (don't get this bit)
where v is speed of the blood and L is??

It continues...
v = E/BL = 0.0003/(0.060 x 0.01) = 0.5m/s

So it appears L is apparently 0.01m (10mm) which is the same as the diameter of the artery.

Any help greatly appreciated.

 

[kuruman]

Better start with $|E|=\frac{d\Phi}{dt}=B\frac{dA}{dt}$ for Faraday's Law in a steady field. Consider a rectangle of blood of height $d$ (diameter of artery) and length $dx$ ...

 

[Steve F]

Better start with $|E|=\frac{d\Phi}{dt}=B\frac{dA}{dt}$ for Faraday's Law in a steady field. Consider a rectangle of blood of height $d$ (diameter of artery) and length $dx$ ...

Thanks for the reply, but i still don't follow how he gets E = BLv

 

[kuruman]

A rectangle of height $L$ (I renamed the variable) and length $dx$ has area $dA=Ldx$. What is $\frac{dA}{dt}~?$ Note: $L$ is the diameter of the artery.

 

[Steve F]

A rectangle of height $L$ (I renamed the variable) and length $dx$ has area $dA=Ldx$. What is $\frac{dA}{dt}~?$ Note: $L$ is the diameter of the artery.

But A isn't dependent on t, so dA/dt doesn't make sense. Sorry if I'm being stupid!

 

[kuruman]

Look at the picture. It shows a piece of the artery and an element of blood of length Δx flowing with speed v left to right. Imagine the magnetic field in a direction perpendicular to the screen. The flux through that element of blood is BΔA. If the element takes time Δt to cross the dotted line, the rate of change of flux with respect to time is BΔA/Δt. Look at the picture again. What is ΔA in terms of L and Δx? What is BΔA/Δt?

 

[Steve F]

Ah I see it now. I was taking A to be the area of cross-section of the artery...but of course the field can't be parallel to the blood flow.

Thanks for your help