Metal bar on rails, induced emf

  • #1
A metal bar with length L, mass m, and resistance R is placed on frictionless metal rails that
are inclined at an angle α above the horizontal. The top end of the rails are connected with a
conducting wire. The resistance of the rails and wire are negligible. The rails are embedded in a
uniform magnetic field B perpendicular to the plane in which the rails sit. The bar is released from
rest and slides down the rails. Determine the magnitude of the induced emf on the loop after a generic time τ , shorter than the time required to reach the terminal velocity.




I know that [itex]|\epsilon|=\frac{d\phi}{dt}[/itex], where in this case [itex]d\phi=BdA=BL\frac{g}{2}sin(\alpha)\tau^{2}[/itex] and [itex]dt=\tau[/itex], so [itex]|\epsilon|=BL\frac{g}{2}sin(\alpha)\tau[/itex]. But it doesn't seem right to me because at some point, the bar will reach terminal velocity, which means the bar must be decelerating and I haven't taken that into account because I don't know how to. Any help will be appreciated.
 

Answers and Replies

  • #2
tiny-tim
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hi subzero0137! :smile:
… the bar must be decelerating and I haven't taken that into account …

the acceleration will be less than gsinα

because the constant gravitational component of force is opposed by the velocity-dependent electromagnetic force :wink:
 
  • #3
hi subzero0137! :smile:


the acceleration will be less than gsinα

because the constant gravitational component of force is opposed by the velocity-dependent electromagnetic force :wink:

I see. But how would I account for that? Would the acceleration simply be gsin(α)-F(v)? I don't know what F(v) is...
 
  • #4
tiny-tim
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I don't know what F(v) is...

you have mx'' equals a function of x' …

write the equation out, and solve it! :smile:
 
  • #5
you have mx'' equals a function of x' …

write the equation out, and solve it! :smile:

Sorry, I'm confused now. Is mx'' the overall force? If yes, shouldn't it be mx''=mgsin(α)-F(x')?
 
  • #6
tiny-tim
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…shouldn't it be mx''=mgsin(α)-F(x')?

your RHS is a function of x' :smile:
 
  • #7
your RHS is a function of x' :smile:

Ohh, of course :tongue:

But when you say "solve it", do you mean solve the differential equation? Because my class hasn't solved diff equations of that form yet.
 
  • #8
tiny-tim
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it's something like dv/dt = Av + B, or dv/(Av + B) = dt …

can you solve that? :smile:
 
  • #9
it's something like dv/dt = Av + B, or dv/(Av + B) = dt …

can you solve that? :smile:

It'll be (1/A)ln(Av+B)=t+C, but I don't know how to use that result in this question :confused:
 
  • #10
tiny-tim
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Determine the magnitude of the induced emf on the loop after a generic time τ , shorter than the time required to reach the terminal velocity.

It'll be (1/A)ln(Av+B)=t+C, but I don't know how to use that result in this question :confused:

(or Av+B = AoeAt)

that gives you v as a function of t, from which you can get the induced emf :smile:
 
  • #11
(or Av+B = AoeAt)

that gives you v as a function of t, from which you can get the induced emf :smile:

But what are the constants A, Ao and B, and how does v as a function of t give me F(v)? Sorry if I'm being difficult, but I genuinely don't understand. Wouldn't F(v)=BIL, where I=BLv/R?
 
  • #12
tiny-tim
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write out the equation properly, and solve it (using v(0) = 0), and all that should become clear :smile:

(and now i'm off to bed :zzz:)​
 
  • #13
write out the equation properly, and solve it (using v(0) = 0), and all that should become clear :smile:

(and now i'm off to bed :zzz:)​

Which equation, F=(v(BL)^2)/R or Av+B=Ao*e^AT? Before you go, can you please tell me what F(v) should be? I can't do other parts of the question if I don't get past this point....
 

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