Metal bar on conducting rails

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



This is a Problem 7.7 fom Griffiths Introduction to Electrodynamics (3ed)

A metal bar of mass m slides frictionlessly on two parallel conducting rails a
distance l apart. A resistor R is connected across the rails and a uniform magnetic field B, pointing into page, fills the entire region.

If the bar moves to the right at speed v, what is the current in the resistor ?


Homework Equations



[tex]\Phi[/tex]=BACos[tex]\phi[/tex]
[tex]E[/tex]=[tex]\frac{d\Phi}{dt}[/tex]


The Attempt at a Solution



my reasonig is:

magnetic flux is:
[tex]\Phi[/tex]=BACos[tex]\phi[/tex]=BA=B(A0+A1).

A0 is initial surface, and A1 is surface which bar makes moving to the right with spead v.
so:

A1=x*l=v*t*l , but v is also function of t, so: A1=v(t)*t*l

I know that equation for A1 is wrong, becouse when I try to get electromotive froce
I get this:

E=d[tex]\Phi[/tex]/dt=B(0+v'(t)*t+v(t))

in solution manual it's:

E=Bl*dx/dt=Blv

Can someone explain why my reasoning is wrong, it seams logical to me.
 

Answers and Replies

  • #2
Doc Al
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A1=x*l=v*t*l , but v is also function of t, so: A1=v(t)*t*l
Setting x = v*t assumes that v is constant.

In any case, what matters is the rate at which flux changes, which depends on the speed at the moment in question:
d(A1) = l*v*dt
d(A1)/dt = l*v, even if v is changing.
 
Last edited:
  • #3
64
1
Ok, than you.
That will help me to answer other question from that problem.
 

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