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PolarNik [594]
3 years ago
13

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu

rrent of 0.040 μa. part a what is the strength of the field at a distance of 1.2 mm
Physics
1 answer:
kondor19780726 [428]3 years ago
6 0
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
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