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 = (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 = (pL)/(A)`

`R = (2*0.02)/(7.854*10^(-11))`

`R = 509.3*10^6\Omega`

`R = 509.3M\Omega`