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