Calculating Current in a Wire Loop Inside and Outside a Solenoid

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The discussion centers on calculating the current in wire loops positioned inside and outside a solenoid. For a 5.0-cm-diameter loop inside the solenoid, the calculated current is approximately 0.38 mA, while for a 15-cm-diameter loop outside, the current is about 1.5 mA, assuming the loop encircles the solenoid. Participants express confusion over the interpretation of the problem, particularly regarding whether the larger loop completely encircles the solenoid, which affects the induced current. Clarifications on the use of the solenoid's total turns and the significance of magnetic flux are emphasized. The conversation concludes with a consensus on the calculated current values, pending confirmation of the assumptions made.
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Homework Statement


A 2400-turn solenoid is 2.4 m long and 10 cm in diameter. The solenoid current is increasing at 1.4 kA/s .

Find the current in a 5.0-cm-diameter wire loop with resistance 9.0 Ω lying inside the solenoid and perpendicular to the solenoid axis.

Repeat for a similarly oriented 15-cm-diameter loop with the same resistance, lying entirely outside the solenoid.

Homework Equations


M = µo (N/l) N' A'
I = emf/R

The Attempt at a Solution


The first part I was able to figure out it's the second part that I am struggling with. I tried doing the same equation I used.

Part A: .38mA

Here is what I plugged in:
(4pi(10-7)(2400pi)(.15)2)/ (4(2.4))*1400= 31.089*10-3

I = emf/R
31.089*10-3/9= 3.45 mA which is wrong
 
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How much of the area of the larger ring has magnetic flux?
 
TSny said:
How much of the area of the larger ring has magnetic flux?
oh! 15cm -5cm. So I should be donig it with 10cm diameter. or .05radiius!

(4pi(10-7)(2400pi)(.10)2)/ (4(2.4))*1400= 13.81mV

13.81/9= 1.5mA
 
Noreturn said:
oh! 15cm -5cm. So I should be donig it with 10cm diameter. or .05radiius!
I believe so.
 
TSny said:
I believe so.
I updated that post I just made with the math. I only have one more chance to submit since I used all my chances. I'll wait until someone can confirm that is right.
 
Looks correct to me. The only thing that's not clear to me in the problem is the statement that the larger ring is "lying completely outside the solenoid". That could mean that it doesn't encircle the solenoid. But I think you are probably correct in assuming that it does.
 
So that was incorrect, any other ideas on where we went wrong? It doesn't show me the right answer or how they got it it. Someone asked a similar question in class and said that we shouldn't be multiplying by 2400 on part B since it's only asking for 1 coil, but it doesn't really sound like that so that does not make sense either.

Any other help is appreciated to help understand it.

If we just do one coil and do the math the same we get .001mA which in two sig figs is 0 which doesn't make sense.
 
Sorry that the answer was incorrect. I still don't see any mistake. You definitely want to use the 2400 turns in getting the number of turns per unit length, n, of the solenoid: n = (2400 turns)/(2.4 m) = 1000 turns per meter.

If the problem meant to take the larger ring as completely outside and not encircling the solenoid, then there would be negligible magnetic flux through the larger ring. So, the answer would be zero induced emf and zero induced current.

Did anyone claim that they got the correct answer?

Note: 0.001 has one significant figure; namely, the 1. None of the zeros written here are counted as significant. More examples:

0.000034 has two significant figures (the 3 and the 4)

.0000340 has three significant figures (the 3, the 4, and the last 0). It can be helpful to use scientific notation:

.001 = 1 x 10-3

.000034 = 3.4 x 10-5

.0000340 = 3.40 x 10-5
 
I have not heard of anyone else getting it right so far.

So just to confirm with our calculations we were getting 1.5mA?
 
  • #10
Noreturn said:
So just to confirm with our calculations we were getting 1.5mA?
Yes. I get 1.54 mA.
That is, 1.5 mA to 2 significant figures. This assumes that the solenoid passes through the loop.
 
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