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Induced emf in various rectangular loops

  • #1
62
1

Homework Statement


[/B]
Three conducting loops, all with the same resistance ##R## move towards a uniform and constant magnetic field, all with the same velocity ##v##. Their relative sizes can be identified by the grid. As the loops move into the magnetic field an induced current begins to flow.

a) Will the induced current be the same [in the loops] or not? If they differ- in which loop is the current the highest. Motivate your answer. [see picture below]

b) Will the current flow counterclockwise or clockwise? Motivate your answer.

loops.jpg


Homework Equations


Faraday's law: ##ε=(-dΦ_B/dt)##
Magnetix flux: ##Φ_B=∫B⋅dA####=BAcosθ##
Lenz's law: (the minus sign in Faraday's law)
Ohm's law: ##V=RI##

The Attempt at a Solution


So I take it that it's the moment as the loops just enter the magnetic field. Time or orientation is not specified. Since the B-field and the dA is parallel ##(cos(0)=1)## we have that the magnetic flux ##Φ_B## is ##B*dA##.

If we take the height (in the y-direction) to be ##l## and the width (in the x-direction) to be ##x## then we can write the area ##A## as ##x*l## and since ##l## is constant the area depends on how far the loop has entered the magnetic field. We take the amount in the x-direction that has entered the field to be ##dx##.

Now we have ##Φ_B=B*dA=B*dx*l##. And when we take the derivative with respect to time to calculate the Emf we get: ##ε=(-dΦ_B/dt)=B*l*(dx/dt)##

We note that dx/dt is just velocity and we can rewrite ##ε=(-dΦ_B/dt)=Blv##

So my answers are:

This means that the length of the wire doesn't matter, only the height ##l##. This would mean that the induced current is greater in ##C## and lower but equally so in ##A## and ##B##.

Due to Lenz's law the current flows counterclockwise in the loops as they're drawn to create a magnetic field that opposes the B-field in the drawing.

------------------------

Now here's where I get worried if I missed out on anything important.

When the loops have moved JUST over one whole grid unit into the magnetic field there's no change in magnetix flux through two of the loops so there's no induced emf in them. Loop ##B## still has one grid unit to move before this happens so at this moment the current ##I## in ##A## and ##C## is 0. But not in ##B##.

Since the height in ##C## is ##2l## (compared to the other two loops) i take it that when all three loops have moved one grid unit into the field (just at the moment where there's still change in magnetic flux in all three loops) the current in ##C## is twice as large as in both ##A## and ##B##.

Did I do anything crazy here? Did I miss out on any details or cases which makes this false or did I get it right?


Thanks for reading!
 

Answers and Replies

  • #2
BiGyElLoWhAt
Gold Member
1,560
113
Looks good. C, however, is twice as large from the moment they enter to the moment just before C is fully inside the field. Assuming v is constant, the emf is constant of the domain of your emf function. (Your function is only valid from just after the loops enter the field to just before they are fully in, and the same with exiting [ - sign])
 
  • #3
149
11
You shoul use Faraday's law: [tex]ε=-\frac{d∅}{dt}[/tex] => In the same Δt, the loop sweeps an area which is the biggest, this loop has the highest ε. So the current the highest.
 
  • #4
BiGyElLoWhAt
Gold Member
1,560
113
You shoul use Faraday's law: [tex]ε=-\frac{d∅}{dt}[/tex]
Isn't that what he did?
 
  • #5
62
1
Looks good. C, however, is twice as large from the moment they enter to the moment just before C is fully inside the field. Assuming v is constant, the emf is constant of the domain of your emf function. (Your function is only valid from just after the loops enter the field to just before they are fully in, and the same with exiting [ - sign])
Cool, thanks! Yeah so the emf in ##C## is twice that of ##A## and ##B## during the whole period that all the loops still are subject to an induced emf, right? =)
 
  • #6
BiGyElLoWhAt
Gold Member
1,560
113
Si, senor(/ita).
 

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