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Magnetism question 2

  • Thread starter tayi
  • Start date
  • #1
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Homework Statement



What speed would a proton need to achieve
in order to circle Earth 1670 km above the
magnetic equator, where the Earth’s mag-
netic field is directed on a line between mag-
netic north and south and has an intensity of
4.1 × 10−8 T?
The mass of a proton is 1.673 × 10−27 kg.
Answer in units of m/s.

Homework Equations



f1=qvbsin(theta)
f2=mv^2/(d+R)
f3=Gm(proton)m(earth)/(d+R)^2 where d is 1670 km, and R is Earth distance


The Attempt at a Solution



I tried doing f1+f2=f3, b ut im not sure if this is right and i dont know how to solve for v
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
I think you'd want to equate f2 to f3 as the gravitational force of attraction will provide the centripetal force.
 

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