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Find V of a partical in B field :O!

  • #1

Homework Statement


A particle with a charge of 5.40nC is moving in a uniform magnetic field of B= -(1.20 )Tz^. The magnetic force on the particle is measured to be F=3.50×10−7N X^ + 7.60×10−7N Y^ .

Calculate the y component of the velocity of the particle.

Homework Equations


F= qv x B
F = qvBSin(theta)


The Attempt at a Solution


lost trying to use the magnitude to find also
i tried
F/qb=Vsin(theta) doesnt seem to yield results yet :/
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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Hi SoulofLoneWlf! :smile:
A particle with a charge of 5.40nC is moving in a uniform magnetic field of B= -(1.20 )Tz^. The magnetic force on the particle is measured to be F=3.50×10−7N X^ + 7.60×10−7N Y^ .

Calculate the y component of the velocity of the particle.
Hint: call the velocity (Vx,Vy,Vz) …

then what is the x-component of the force? :wink:
 
  • #3
Hi SoulofLoneWlf! :smile:


Hint: call the velocity (Vx,Vy,Vz) …

then what is the x-component of the force? :wink:
so would it be possible for me to say v = Vx + Vy + Vz
...
Fy=qVyBsin(thetha)?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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  • #5
erm :redface:

right idea :smile:, wrong result :cry:
hahaha felt that was wrong hmm let me try again i guess ><"
 
  • #6
erm :redface:

right idea :smile:, wrong result :cry:
F=x^(5.4Vy*1.2Z^T)-Y^(5.4Vx*1.2TZ^)
then place value for F as 3.50x10^-7 etc and equal x to x and y to y??
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
249
F=x^(5.4Vy*1.2Z^T)-Y^(5.4Vx*1.2TZ^)
then place value for F as 3.50x10^-7 etc and equal x to x and y to y??
Yup! :biggrin:
 
  • #8

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