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Particle in a mag field with linear drag

  • Thread starter Liquidxlax
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  • #1
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Homework Statement



A particle of mass m and electric charge q>0 is subject to a uniform, constant mag field of [tex]\vec{B} = B \hat{k}[/tex]. At t=0 its velocity vector is [tex]\hat{v}[/tex]naught which lies in the x-y plane. The particle is also subject to a linear drag force fd=-b[tex]\vec{v}[/tex]

a)Draw a free body diagram showing all the forces action on the particle. Use this to find expressions for F=ma in the x and y directions respectively, assuming that the force of gravity is negligible.

b) Use the equations from part a) and the complex variable technique discussed in class (see also taylor, section 2.5) to solve for the complex velocity n(t)= vx+ivy as a function of time, in terms of the initial condition [tex]\vec{v}[/tex]naught

c) use your solution to part b) to find an expression for the particle's speed as a function of time.

hate that my prof links the questions....


Homework Equations



[tex]\omega = \frac{qB}{m}[/tex]

ay = -[tex]\omega v_{x}[/tex]

ax = [tex]\omega v_{y}[/tex]

-bv = [tex]\beta Dv[/tex]


The Attempt at a Solution



Lorentz force law where E=0 qvxb = qB{i j k}{vx vy 0}{0 0 k}

=qB(vyi - vxj)


F = ma = qB(vyi - vxj) - b(vyi - vxj)

after some shuffling around dividing by mass subbing in ay and ax

[1-(b/m[tex]\omega[/tex])]*[ ay + ax ] = a

multiply by mass

F= ma = [m - (b/[tex]\omega[/tex])]* [ ay + ax ]

I'm not sure if i was supposed to make this a function of t or if that is with part b
 

Answers and Replies

  • #2
322
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anyone have an idea?
 
  • #3
322
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ehild your a smart guy. wanna help a guy out?
 
  • #4
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I figured it out, needed to clear my head and it was easy
 

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