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3 years ago

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Two electrons 1 and 2 move along antiparallel paths separated by a distance of 10 nm, traveling at speeds 4.5 x 10^7 m/s and 8.0

x 10^6 m/s.
Part A) What is the magnitude of the magnetic force exerted by electron 2 on electron 1? Express your answer with the appropriate units.

1 answer:

fomenos3 years ago

5 0

Answer:

The magnetic force on electron 1 by electron 2= $9.22\times 10^{-15} N$

Explanation:

We are given that

Distance between two electrons=10 nm

Speed of electron 1= $4.5\times 10^7$ m/s

Speed of electron 2= $8.0\times 10^6$ m/s

We have to find the magnitude of the magnetic force exerted by electron 2 on electron 1.

Magnetic force on electron 1 by electron 2

$F=\frac{\mu_0e^2v_1v_2}{4\pi r^2}$

$\frac{\mu_0}{4\pi}=10^{-7}$

$e=1.6\times 10^{-19} C$

$v_1=4.5\times 10^7 m/s$

$v_2=8.0\times 10^6 m/s$

$r=10nm=10\times 10^{-9}m$ $(1nm=10^{-9} m$ )

Substitute the values in the given formula

The magnetic force on electron 1 by electron 2= $\frac{10^{-7}\times (1.6\times 10^{-19})^2\times 4.5\times 10^7\times 8\times 10^6}{(10\times 10^{-9})^2}$

The magnetic force on electron 1 by electron 2= $9.22\times 10^{-15} N$

Hence, the magnetic force on electron 1 by electron 2= $9.22\times 10^{-15} N$

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a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

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$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

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$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

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c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

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$cos \theta = \frac{r}{R_{E}}$

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$r= R_{E}cos \theta$

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$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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