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Wewaii [24]
2 years ago
15

A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav

eling along an axon.
If the peak field strength 1.3mm from an axon is 9.0pT , what is the peak current carried by the axon?
Physics
1 answer:
fenix001 [56]2 years ago
7 0

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;

B = \frac{\mu I}{2\pi r}

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"

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

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

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Answer:

Explanation:

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Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

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a = - 1.67 m/s²

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According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
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Answer:

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Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

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The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

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