Question

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. A typical axon carries a peak current of 0.040 ?A.

What is the strength of the field at a distance of 1.6 mm ?

Answer 1

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

Step-by-step explanation:

We can think the axons as current-carrying wires

The strength of the magnetic field produced by a current-carrying wire is

`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 = 0.04\cdot 10^(-6) A`

r = 1.6 mm = 0.0016 m

So the strength of the magnetic field is

`B=((4\pi \cdot 10^(-7)H/m)(0.04\cdot 10^(-6) A))/(2\pi (0.0016 m))=5.0\cdot 10^(-12)T`