WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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The force on a charge moving in
the presence of a magnetic field
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is the charge in Coulombs times
its velocity in meters per second
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times the magnetic field strength
times sine of the angle
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between the velocity of the magnetic field.
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And we can solve for *B* by dividing
both sides by *qv* sine theta.
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And so the magnetic field strength is
the force divided by *qv* sine theta.
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So that's 1.7 times ten to the minus 16
newtons divided by the elementary charge
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which is the charge on a proton,
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because we are told that this
cosmic array consist of a proton.
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So that's 1.6 times ten to
the minus 19 Coulombs
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times five times ten to the seven
meters per second times sine 45
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which is the angle between the
velocity of the magnetic field.
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This gives a magnetic field strength of
3.01 times ten to the minus five Tesla.
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Now, the magnetic field
strength of the earth
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is about five times ten
to the minus five Tesla.
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And I say approximately equal because
this magnetic field strength really varies
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depending on where your
position is around the earth.
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And so we'll take the magnetic field we
determined in part a and this is part b.
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And we'll take the magnetic
field we determined in part a
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and divided by magnetic field of the
earth in order to compare them.
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And we have three times
ten to the minus five
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divided by five times ten to
the minus five which is 0.60.
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So this means that our magnetic field
that we calculated is 60 percent that of
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what is generally accepted to be
the field strength of the earth
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and since this number is approximately
can change, depending where you are.
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This means yes, our answer in part a is
consistent with the earth's magnetic field.