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. part a what is the strength of the field at a distance of 1.2 mm

Answer 1

We can see the axon as a current-carrying wire. The magnetic field produced by a current-carrying wire is given by

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

`\mu_0 = 4 \pi \cdot 10^(-7) Tm/A` is the vacuum permeability
I is the current in the wire
r is the radial distance from the wire at which the field is calculated

The current in the axon is

`I=0.040 \mu A=0.040 \cdot 10^(-6) A`,
therefore the magnetic field strength at distance

`r=1.2 mm=1.2 \cdot 10^(-3)m`
from the axon is

`B= (\mu_0 I)/(2 \pi r)= ((4 \pi \cdot 10^(-7) Tm/A)(0.040 \cdot 10^(-6) A))/(2 \pi (1.2 \cdot 10^(-3) m))=6.67 \cdot 10^(-6) T = 6.67 \mu T`