How Fast Must the Magnetic Field Change to Induce a 3.4 A Current in a Coil?

Click For Summary

Homework Help Overview

The problem involves a coil with a specific number of turns and dimensions, where a magnetic field is applied perpendicularly. The goal is to determine the rate at which the magnetic field must change to induce a certain current in the coil, using the properties of copper and relevant equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various equations related to magnetic fields, resistance, and induced electromotive force (EMF). There are attempts to calculate resistance using the resistivity of copper, and questions arise regarding the correct use of coil dimensions and wire properties.

Discussion Status

Participants are actively engaging with the problem, exploring different equations and their applicability. Some have provided calculations and are seeking verification or correction of their methods. There is a recognition of potential errors in previous calculations, and multiple interpretations of the problem setup are being considered.

Contextual Notes

There is a mention of homework constraints, including limited attempts on an online platform, which adds pressure to find the correct solution. Participants are also clarifying the definitions of variables and their relevance to the problem.

BuBbLeS01
Messages
602
Reaction score
0

Homework Statement


A 102-turn, 3.8-cm-diameter coil is made of 0.59-mm diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B increase to induce a 3.4 A current in the coil? The resistivity of copper is 1.7·10-8 Ω·m.


Homework Equations





The Attempt at a Solution


A = (0.00059/2)^2 * pi = 2.734 x 10^-7
I = 3.4 A
I don't what to do?
 
Physics news on Phys.org
Is there an answer to compare to at the back of the book?
 
no because its an even number :(
 
B=µoNI/2r, B is magnetic force, µo is magnetic constant, N is number of turns, I is current, r is radius

I think that's it but that R is unused in the problem, and it concerns me that I missed something.
 
is r the radius of the coil or th wire??
 
well i did the radius of the coil and its wrong but thanks for trying
 
How do you know its wrong?
 
its my homework problem and our homework is on lon-capa...so we have like 4 tries for each problem and tells you when you enter the answer.
 
oh my gosh this problem is driving me nuts I can't figure it out! I tried using all sorts of formulas like...
R = pL/A to get the resistance but I don't have L
So then I tried...
B = U*I*(N/L) to get L but I don't have B!
 
  • #10
Yes you do have L because you know the diameter of the coil and how many turns there are.
 
  • #11
Isn't the L for B=UIN/L the length of the solenoid not the length of the wire?
 
  • #12
oh yea...ugh I hate this problem!
 
  • #13
silvashadow said:
Isn't the L for B=UIN/L the length of the solenoid not the length of the wire?

I was referring to the L in the resistivity equation that Bubbles was asking about.
 
  • #14
hage567 said:
I was referring to the L in the resistivity equation that Bubbles was asking about.

Oh I see. What's the step after finding R?
 
  • #15
How do I find L with having the diameter and number of turns??
 
  • #16
I was going to find E using I = E/R then use that in the equation...
E = A * B/t to get the rate
 
  • #17
BuBbLeS01 said:
How do I find L with having the diameter and number of turns??

Well, circumference=pi*diameter, lol 102 turns = 102 circumferences
 
  • #18
oh lol...ok so my length is 9.8696 the when I use R = pL/A is it the area of the wire or coil??
 
  • #19
BuBbLeS01 said:
I was going to find E using I = E/R then use that in the equation...
E = A * B/t to get the rate

This is what I would try.
 
  • #20
BuBbLeS01 said:
oh lol...ok so my length is 9.8696 the when I use R = pL/A is it the area of the wire or coil??

The wire, since that is what you are finding the resistance of.
 
  • #21
BuBbLeS01 said:
oh lol...ok so my length is 9.8696 the when I use R = pL/A is it the area of the wire or coil??

Area of wire, cross section area.
 
  • #22
silvashadow said:
Isn't the L for B=UIN/L the length of the solenoid not the length of the wire?

The question did not actually state that this was a solenoid. It said it was a loop of 102 turns with a magnetic field perpendicular to it (so going through it). The magnetic field is not being produced by the coil. So I don't think the B = U*I*(N/L) equation is what you are looking for here, IMO.
 
  • #23
and the length would be the length of the wire not the coil right
 
  • #24
Yes, that's right.
 
  • #25
hage567 said:
The question did not actually state that this was a solenoid. It said it was a loop of 102 turns with a magnetic field perpendicular to it (so going through it). The magnetic field is not being produced by the coil. So I don't think the B = U*I*(N/L) equation is what you are looking for here, IMO.

Oh I see. Thanks for setting me straight. I'm still learning and your opinion is very helpful to me.
 
  • #26
then in the equation E = A * B/t do I use the area of the coil?
 
  • #27
oh noooo I got it wrong! My final answer was 53.6469 T/s
 
  • #28
any other ideas?
 
  • #29
BuBbLeS01 said:
oh noooo I got it wrong! My final answer was 53.6469 T/s

I get a different answer than this. Try checking your math over carefully and see what you get.
 
  • #30
hage567 said:
I get a different answer than this. Try checking your math over carefully and see what you get.
I did it twice but I am still getting the same answer let me show you what I did maybe I missed something...
I have...
I = 3.4 A
p = 1.7 e^-8
Dcoil = 0.0308 m
Dwire = 0.00059 m
Acoil = 7.4506 e^-4
Awire = 2.73397e6-7

C = pi*Dwire...

So L = pi*Dwire*102 = 0.18906m

R = p*Lwire/Awire = 0.0117559 Ohms

I = E/R...E = IR = 0.03997 V

E = Acoil * (B/t)...B/t = E/Acoil = 53.6469 T/s
 

Similar threads

Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
11K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
5K