Question

An average nerve axon is about 4x10-6m in radius, and the axoplasm that composes the interior of the axon has a resistivity of about 2.3 Ω ⋅m.

What is the resistance of just 2-cm length of this axon?

Provide your answer in mega-ohms (1 mega-ohm = 106 ohms or "millions of ohms").

(Note, this value is so large -- it corresponds to the resistance of tens of thousands of miles of the thinnest copper wire normally manufactured(!) -- that it explains why a nerve pulse traveling down an axon CANNOT simply be a current traveling along the axon. The voltage required to achieve a perceptible current in the axon would have to be gigantic! We will investigate how voltage pulses -- not current -- travel down axons in a future lab.)

Answer 1

For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship

$R = \frac{pL}{A}$

Where

p = Specific resistance of material

L = Length

A = Area

The area of nerve axon is given as

$A = \pi r^2$

$A = \pi (5*10^{-6})^2$

$A = 7.854*10^{-11}m^2$

The rest of values are given as

$p= 2 \Omega\cdot m$

$L = 2cm = 0.02m$

Therefore the resistance is

$R = \frac{pL}{A}$

$R = \frac{2*0.02}{7.854*10^{-11}}$

$R = 509.3*10^6\Omega$

$R = 509.3M\Omega$