Question

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.

Answer 1

The magnitude of the maximum force on the 1-mm-long segment of the cylindrical nerve is 0.01884 N.

Step-by-step explanation:

To calculate the magnitude of the maximum force on the cylindrical nerve segment, we need to use the formula for the magnetic force on a current-carrying wire:

F = I * L * B * sinθ

where

F is the force

I is the current

L is the length of the wire segment

B is the magnetic field strength

θ is the angle between the current direction and the magnetic field.

In this case, the length of the wire segment is 1 mm, which is 0.001 m, and the diameter of the nerve is 1.5 mm, which means the radius is 0.75 mm or 0.00075 m.

We can calculate the length of the wire segment as follows:

L = 2 * π * r = 2 * 3.1416 * 0.00075 m = 0.00471 m

Given that the magnetic field strength is 2 T and the angle between the current and the field is 90°, the formula becomes:

F = 0.00471 * 2 * 2 * sin(90°) = 0.01884 N

Therefore, the magnitude of the maximum force on the 1-mm-long segment of the cylindrical nerve is 0.01884 N.

Answer 2

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

Step-by-step 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