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

Answer:

$6.66\cdot 10^{-12}T$

Explanation:

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

$B=\frac{\mu_0 I}{2\pi r}$

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

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