Question

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak current of 0.040 μa. what is the strength of the field at a distance of 1.2 mm

Answer 1

`6.66\cdot 10^(-12)T`

Step-by-step explanation:

The magnetic field produced by a current-carrying wire is given by

`B=(\mu_0 I)/(2\pi r)`

where

`\mu_0` is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

`I=0.040 \mu A=4\cdot 10^(-8)A`

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

`B=((4\pi \cdot 10^(-7) H/m)(4\cdot 10^(-8)A))/(2\pi (0.0012 m))=6.66\cdot 10^(-12)T`