Answer:
The peak current carried by the axon is 5.85 x 10⁻⁸ A
Explanation:
Given;
distance of the field from the axon, r = 1.3 mm
peak magnetic field strength, B = 9 x 10⁻¹² T
To determine the peak current carried by the axon, apply the following equation;
[tex]B = \frac{\mu I}{2\pi r}[/tex]
where;
B is the peak magnetic field
r is the distance of the magnetic field from axon
μ is permeability of free space = 4π x 10⁻⁷
I is the peak current
Re-arrange the equation and solve for "I"
[tex]B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A[/tex]
Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A
