Two Point Charges and Magnetic Field

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


Find the magnetic forces acting on each electron in the attached image
BChAPee.png



Homework Equations


B = [itex]\frac{μ0}{4pi}[/itex]*q*[itex]\frac{v x \hat{r}}{r^{2}}[/itex]
F[itex]_{m}[/itex] = qv x B

The Attempt at a Solution


B[itex]_{1}[/itex] = 10[itex]^{-7}[/itex] * 1.6*10[itex]^{-19}[/itex] * [itex]\frac{5000*10^{3}*sin(45)}{(10*10^{-10})(10*10^{-10})}[/itex]
B[itex]_{1}[/itex] = .057

Fm[itex]_{2}[/itex] = (1.6*10[itex]^{-19}[/itex])*1000*10^3*.057
Fm[itex]_{2}[/itex] = 9.12*10^{-15}

Now at this point I know something is wrong, but I'm not sure what. I thought the force on the 2nd electron would be the force created by the 1st electron, but that number is definitely screaming incorrect to me. I figure it I understand how to figure out the force on 1 I can figure out the force on the other. (Electron 1 is at the bottom, electron 2 at the top). Where is my mistake?
 

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Answers and Replies

  • #2
TSny
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In calculating B1 note that r should be squared in the denominator. kilo means 103. Are you sure you want to use 45o for the angle in the sine function? (What is the angle between ##\vec{v}## and ##\hat{r}##?)
 
  • #3
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Woops, don't try and do physics half asleep!:rolleyes: Those first two errors aside, I think I see what angle I am supposed to use. The velocity of B[itex]_{1}[/itex] is 45° from the x axis, but 75° from [itex]\hat{r}[/itex], since [itex]\hat{r}[/itex] is along r, correct?
 
  • #4
TSny
Homework Helper
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Looks good.
 

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