**NEED PHYSICS HELP: Lenz's Law***

[KennyH]

Hey guys, studying for an exam today while watching the Twin 125's at Daytona. Anyway got a question for you physics gurus out there. Here it is:

A bar magnet is falling through a loop of wire with constant velocity. The north pole enters first. The induced current will be greatest when the magnet is located so that:

a) the loop is near either the north or south pole
b) the loop is near the north pole only
c) the loop is near the middle of the magent
d) with the acceleration, the induced current is zero.


I believe the answer is A becuase the induced current on the loop should be the same in magnitude with either the north or south pole. It should not matter. However, I am bad about second guessing myself. Please LMK what you think and why. FAST

 

[Evadman]

<< a) the loop is near either the north or south pole
b) the loop is near the north pole only
c) the loop is near the middle of the magent
d) with the acceleration, the induced current is zero. >>



c) the loop is near the middle of the magent.

With a constant velocity you will end up with a induced current that looks roughly like this:

....--""""""""""""""""""""""""""""""""""--....

Cinda hard drawing in txt The wire will break the greatest amount of field lines towards the center of the magnet because of the way the field lines spread from the N and S poles.

 

[yomega]

I agree with Evadman, when the magnet is half way through the loop then you have the magnetic field lines as close to perpendicular (sp?) to the loop as possible.

 

[Sohcan]

It's been a while since Electrodynamics, but if I recall Lenz's Law correctly ("nature abhors a change in flux"), the answer is A. The induced EMF is proportional to the change of magnetic flux wrt to time, not magnetic field strength. As the magnet passes through the loop, the greatest change in flux is at the poles...while the middle of the magnetic is passing through the loop, there is no change in field strength, and thus no induced EMF. Magnetic field lines go from north to south IIRC, so as the north pole enters, the loop induces a current such that it creates a magnetic flux opposing the one created by the bar magnet. Thus, from the perspective of the magnet, the induced field points towards it, created by a counter-clockwise current. When the pole passes through, the current drops to zero, and as the south pole passes through, a clock-wise current is induced.

 

[PowerEngineer]

Sohcan (and you) are right. Current is induced by the rate of change of magnetic field strength, which should occur when either pole passes through the loop.

 

[Evadman]

Assuming you guys are correct (Sohcan and PowerEngineer) Then I humbly ask forgiveness It has been too long since Physics.

<edit>
Awww crap. You guys are right. Linky

 

[highwire]

I'd say it was C.
The question was at what point does the max CURRENT occurs. Since the loop is a super-conductor ( nobody mentioned resistance ), the current will be the integral of the emf and will be max at the mid point. No?

 

[Sohcan]

<< Since the loop is a super-conductor ( nobody mentioned resistance ) >>

(Rubs eyes) Where does the problem say that the loop is superconducting?:Q Faraday's Law says -d(flux) / dt = (line integral)(E*dl), which yields the same conclusion as Lenz's Law, but in a more quantitative sense (Lenz's Law only tells you the direction of the induced current): EMF is induced by a time rate of change in the flux. There is in fact an example identical to this problem in my Electrodynamics textbook, with a nice figure plotting magnetic flux and EMF vs. time, with the EMF at a maximum where there is a max change in flux (at the poles), and zero where there is no change in flux (in the middle).

 

[KennyH]

Yep, found a great example from Eastern Illinois University in .pdf form last night. The induced current is indeed MAX at both poles due to a changing flux. No flux in the middle so the current drops to zero. So I believe I am right with the answer A.

 

[Agent004]

Sohcan is correct .

 

[Mday]

i agree with sohcan.