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Tems11 [23]
3 years ago
14

For a magnetic field strength of 2T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical

nerve that has a diameter of 1.5mm.
Physics
1 answer:
nydimaria [60]3 years ago
7 0

Answer:

Incomplete question

This is the complete question

For a magnetic field strength of 2 T, estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of 1.5 mm. Assume that the entire nerve carries a current due to an applied voltage of 100 mV (that of a typical action potential). The resistivity of the nerve is 0.6ohms meter

Explanation:

Given the magnetic field

B=2T

Lenght of rod is 1mm

L=1/1000=0.001m

Diameter of rod=1.5mm

d=1.5/1000=0.0015m

Radius is given as

r=d/2=0.0015/2

r=0.00075m

Area of the circle is πr²

A=π×0.00075²

A=1.77×10^-6m²

Given that the voltage applied is 100mV

V=0.1V

Given that resistive is 0.6 Ωm

We can calculate the resistance of the cylinder by using

R= ρl/A

R=0.6×0.001/1.77×10^-6

R=339.4Ω

Then the current can be calculated, using ohms law

V=iR

i=V/R

i=0.1/339.4

i=2.95×10^-4 A

i=29.5 mA

The force in a magnetic field of a wire is given as

B=μoI/2πR

Where

μo is a constant and its value is

μo=4π×10^-7 Tm/A

Then,

B=4π×10^-7×2.95×10^-4/(2π×0.00075)

B=8.43×10^-8 T

Then, the force is given as

F=iLB

Since B=2T

F=iL(2B)

F=2.95×10^-4×2×8.34×10^-8

F=4.97×10^-11N

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Time required : 3 s

<h3>Further explanation </h3>

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3 years ago
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Answer:

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3 years ago
A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
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Explanation:

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2 years ago
In a charging process, 1 × 10^13 electrons are removed from one small metal sphere and placed on a second identical sphere. Init
liraira [26]

Answer:

r = 0.303m

= 30.3cm

Explanation:

Given that,

The number of electrons transferred from one sphere to the other,  

n  = 1 ×10 ¹³e le c t r o n s

The electrostatic potential energy between the spheres,  

U = − 0.061 J

The charge on an electron,  

q = − 1.6 × 10 ⁻¹⁹C

The coulomb constant,  

K = 8.98755 × 10 ⁹ N ⋅ m ² / C 2²

Due to the transfer of electrons, both spheres become equally and oppositely.

The charge gained by the sphere due to the excess of the electron is:  

q ₁ = n q

   = 1 ×10 ¹³ *  − 1.6 × 10 ⁻¹⁹

   = -1.6  × 10⁻⁶C

The charge left on the first sphere is =

q ₂ = -q₁ = 1.6  × 10⁻⁶C

The electric potential energy between two point charges is given by the following equation:

U = K q ₁q ₂/r

q ₁ and  q ₂ are the two charges.

r  is the distance between the charge and the point.

K  =  8.98755  ×  10 ⁹ N ⋅ m ² / C ²

we have:

-0.061 =  (8.98755  ×  10 ⁹ * (-1.6  × 10⁻⁶)²) / r

r = (18.41 ×  10 ⁻³) / 0.061

r = 0.303m

= 30.3cm

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3 years ago
Any tips on how to get a good grasp of physics and chemistry?
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Answer: Take notes and research into the subject more get extra help like tutoring if needed.

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