91Ó°ÊÓ

The magnetic field probe to determine the current associated with a nerve impulse traveling along an axon. If the peak field strength$1.0mm$ from an axon is $8.0pT$, what is the peak current carried by the axon.

The axon behaves as current-carrying wire that produces a magnetic field and the Biot-savart law helps us to analysis the magnitude and direction of the magnetic field .At any point where the magnetic field due to the segment $∆\stackrel{→}{s}$of current-carrying wire is given by equation$\left(29.7\right)$

$B=\frac{{\mathrm{μ}}_{\mathrm{ο}}}{2\mathrm{π}}\frac{I}{r}$

Our target is find the current $I$, so we solve equation $\left(1\right)$for$I$to be in the form

$I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}}$

Where $r$is the distance between the axon and the point where the magnetic field is $8pT$and in $m$it is given by

$r=\left(1mm\right)\left(\frac{1m}{1000mm}\right)=1×{10}^{-3}m$

Now, we plug the values for $B,r$and ${\mathrm{μ}}_{\mathrm{ο}}$into equation $\left(2\right)$to get the current

role="math" localid="1649224759138" $$\begin{gathered} I=\frac{2\mathrm{π}Br}{{\mathrm{μ}}_{\mathrm{ο}}} \\ =\frac{2\mathrm{π}\left(8×{10}^{-12}T\right)\left(1×{10}^{-3}m\right)}{4\mathrm{π}×{10}^{-7}T·m/A} \\ =0.04×{10}^{-6}A \\ =0.04\mathrm{μ}A \end{gathered}$$