# NCEES Practice Exam problem 516



## SteveR

I'm stumped on this one, and I don't have any good Electromagnetic books. Any tips? I would post the question, but it would make no sense without the diagram.

I don't understand with the path for leg B is "d", and leg C is "3d". To me, it seems like B should be "3d", and C should be "5d".


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## Flyer_PE

SteveR said:


> To me, it seems like B should be "3d", and C should be "5d".



All of the flux generated by the coil has to pass through the first and last segments of length "d". It splits to go through legs B and C. The problem is to determine how much goes through each leg. The math works the same as for current division. Leg C has three times as much "resistance" as leg B. It will then carry 1/4 of the total flux.

Does this help?

Jim


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## grownupsara

I've read through your explanation a few times, but still can't quite get it, which is driving me crazy since this doesn't seem like that complicated of a problem. I still can't manage to get Leg C having 3x the resistance of B, or 3x the path distance. For the B path, I'm counting 1 horiz. d + 1 vert. d + 1 horiz. d to get back to the coil, which is 3d. For the C path, I'm counting 2 horiz. d + 1 vert. d + 2 horiz. d = 5d. That gives me a ratio of 5:3, which is nowhere close to 3:1. Thanks for your help!



IFR_Pilot said:


> All of the flux generated by the coil has to pass through the first and last segments of length "d". It splits to go through legs B and C. The problem is to determine how much goes through each leg. The math works the same as for current division. Leg C has three times as much "resistance" as leg B. It will then carry 1/4 of the total flux.
> Does this help?
> 
> Jim


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## Flyer_PE

grownupsara said:


> I've read through your explanation a few times, but still can't quite get it, which is driving me crazy since this doesn't seem like that complicated of a problem. I still can't manage to get Leg C having 3x the resistance of B, or 3x the path distance. For the B path, I'm counting 1 horiz. d + 1 vert. d + 1 horiz. d to get back to the coil, which is 3d. For the C path, I'm counting 2 horiz. d + 1 vert. d + 2 horiz. d = 5d. That gives me a ratio of 5:3, which is nowhere close to 3:1. Thanks for your help!



I think the only thing you are missing is that ALL of the flux travels through the first and last segments. Try re-drawing the problem as an electrical circuit with a 3mA DC source where the coil is and 1-ohm resistors where each of the legs.

Let me know if this helps.

Jim

Edit: Changed source to a 3mA current source.


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## grownupsara

Hey, that finally made it click! I scanned in the work I did using your advice, with each d segment = 1ohm resistance. Thanks!


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## Ilan

Thanks guys,

I finally got this one!

arty-smiley-048:


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## Nik

Could you please explain it electrically ? I saw your posts of replacing each leg with 1 Ohm and using 3mA as current source.

Still cant figure it out :-/

Nik



IFR_Pilot said:


> SteveR said:
> 
> 
> 
> To me, it seems like B should be "3d", and C should be "5d".
> 
> 
> 
> 
> All of the flux generated by the coil has to pass through the first and last segments of length "d". It splits to go through legs B and C. The problem is to determine how much goes through each leg. The math works the same as for current division. Leg C has three times as much "resistance" as leg B. It will then carry 1/4 of the total flux.
> 
> Does this help?
> 
> Jim
Click to expand...


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## Flyer_PE

How's this:


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## BamaBino

IFR_Pilot said:


> I think the only thing you are missing is that ALL of the flux travels through the first and last segments.


Doesn't 3/4 of the flux travel thru the middle segment?


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## DK PE

BamaBino said:


> IFR_Pilot said:
> 
> 
> 
> I think the only thing you are missing is that ALL of the flux travels through the first and last segments.
> 
> 
> 
> Doesn't 3/4 of the flux travel thru the middle segment?
Click to expand...

Using current division, the current/flex divides as inverse of path resistance, therefore, 1/4 of the 3mA = 0.75mA through outside segment which leaves 3/4 * 3mA = 9/4 = 2.25mA through middle segment. They'd better add to 3mA.


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