Inductance of a loop in a nonuniform magnetic field

  • Thread starter sfdevil
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  • #1
sfdevil
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Homework Statement


A square loop made of wire with negligible resistance is placed on a horizontal frictionless table. The mass of the loop is m and the length of each side is b. a nonuniform vertical magnetic field B=B0(1+kx) exists in the region, where B0 and k are constants. The loo is given a quick push with initial velocity v along x-axis. The loop stops after a time interval T. Find the inductance of the loop.



Homework Equations




emf(ind) = -L*dI/dT
U=1/2*L*I^2
emf= -delta flux/delta t

The Attempt at a Solution



well, I am sort of in a loss for this one.
I tried to get the induced emf by finding dflux/dt:

flux
=integral ( B0*(1+kx) * b dX )
= b*B0*(b*2*k*x+b^2*k+2*b)/2

change of flux in regards with time = dphi/dx * dx/dt
= b^2*k*B0*v (because velocity= dx/dt)

so emf is b^2*k*B0

now I am not sure what to do, since the resistance is negligble and i can't find the current from it...

also the whole time interval thing, where does it come into play (kinematics ?)
and should I use conservation of energy here ?
1/2 * m * v^2 = 1/2 * L * I^2 ? for some reason I don't think its the right way

just thoroughly confused with this one. please help !
 

Answers and Replies

  • #2
sfdevil
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anyone ? any hints/suggestions ?
 
  • #3
Nick89
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I don't think you got the right equation for emf, you are trying to use the emf generated by a solenoid.

Also, you don't need to find any current (you can't, since there is no resistance), you only need to find the emf.
 
  • #4
sfdevil
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I don't think you got the right equation for emf, you are trying to use the emf generated by a solenoid.

Also, you don't need to find any current (you can't, since there is no resistance), you only need to find the emf.

can you possibly expand on your answers ?

what do you mean I don't have the right expression for the emf ? and what does a solenoid have to do with this problem ?

the textbook I use states that the self-induced emf in any closed loop of current is -L*dI/dT...

I saw a similar problem done in a portable ta guide the way I described... only there resistance wasnt negligible and the field was Bz=C/x and not B=B0*(1+kx)...
 

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