Question

An electromagnetic flowmeter is to be used to measure blood speed. A magnetic field of \(0.115 \mathrm{T}\) is applied across an artery of inner diameter \(3.80 \mathrm{mm}\) The Hall voltage is measured to be \(88.0 \mu \mathrm{V} .\) What is the average speed of the blood flowing in the artery?

Short answer

Answer: The average speed of the blood flowing in the artery is approximately 2.092 m/s.

Step-by-step solution

Identify given values

In this problem, we are given the following values: - The magnetic field strength \((B) = 0.115\:T\) - The inner diameter of the artery \((d) = 3.80\:mm\) - The Hall voltage \((V) = 88.0\:\mu V\)

Convert the units

Since the units in the problem are not consistent, we need to convert them before we can make use of the formula. Let's convert \(d\) from mm to meters (m) and \(V\) from \(\mu V\) to volts (V). - \(d = 3.80\:mm \cdot \frac{1\:m}{1000\:mm} = 3.80 \times 10^{-3}\:m\) - \(V = 88.0\:\mu V \cdot \frac{1\:V}{10^6\:\mu V} = 88.0 \times 10^{-6}\:V\)

Calculate the average speed

Now we can use the formula to calculate the average speed of the blood \((v)\) in the artery. $$ v = \frac{V}{Bd} $$ Plug in the values for \(V\), \(B\), and \(d\): $$ v = \frac{88.0 \times 10^{-6}\:V}{0.115\:T \times 3.80 \times 10^{-3}\:m} $$

Solve for the average speed

Now let's solve for the average speed \((v)\): $$ v = \frac{88.0 \times 10^{-6}\:V}{0.115\:T \times 3.80 \times 10^{-3}\:m} = 2.092\:m/s $$ Hence, the average speed of the blood flowing in the artery is approximately \(2.092\:m/s\).