Rate of Field Change for Induced Current of Loop

Click For Summary

Homework Help Overview

The problem involves a conducting loop of wire with a specified area and resistance, placed in a magnetic field. The objective is to determine the rate of change of the magnetic field required to induce a specific current in the loop.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between magnetic flux and the magnetic field, particularly questioning the angle between the area vector and the magnetic field. There are attempts to apply Faraday's Law and clarify the assumptions regarding the magnetic field's constancy.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the implications of the given information. Some guidance has been offered regarding the calculations, but there is no clear consensus on the correct approach or answer.

Contextual Notes

There is confusion regarding the nature of the magnetic field provided in the problem, as it is described as a constant field while the question asks for a rate of change. This has led to speculation about the intentions behind the problem's wording.

tevatron1
Messages
11
Reaction score
0

Homework Statement



A single conducting loop of wire has an area of 8.0×10−2 m^2 and a resistance of 110 Ω. Perpendicular to the plane of the loop is a magnetic field of strength 0.37 T.

At what rate (in T/s) must this field change if the induced current in the loop is to be 0.33 A?

Homework Equations



Trying to solve for ΔB/Δt

The Attempt at a Solution



We have a change in magnetic flux, but it seems to be due to ΔB rather than ΔA, so Δflux = ΔB*Acosθ (However since cos(90) = 0, I'm not sure if this will be correct).

Using Faraday's Law |ε| = N|(Δflux/Δt)|
Substituted Δflux = ΔB*A into Faraday's Law --> ε = N(ΔB*A/Δt)
Substituted the equation for the induced EMF into I = |ε|/R and solved for ΔB/Δt
ΔB/Δt = I*R/A (N=1 because the loop has 1 turn)

453.75 T/s = (.33A)*(110Ω)/(8.0*10^-2 m^2)
asks for 2 significant digits -> 450 T/s = Final Answer

I'm not sure where I'm going wrong...
 
Physics news on Phys.org
tevatron1 said:

Homework Statement






We have a change in magnetic flux, but it seems to be due to ΔB rather than ΔA, so Δflux = ΔB*Acosθ (However since cos(90) = 0, I'm not sure if this will be correct).
[/b]

θ would not be 90 as Area vector and B make 0 angle
 
kushan said:
θ would not be 90 as Area vector and B make 0 angle

Okay, then cos(0) = 1, which would prove that my assumption Δflux = ΔB*A*cos(0) = Δflux = ΔB*A

How does the rest of my thought process look?
 
as as i know you are doing correctly ,
where are you stuck ?
 
kushan said:
as as i know you are doing correctly ,
where are you stuck ?

Mastering Physics tells me this is the wrong answer, though it does not give any explanation as to why.
 
what is " mastering physics "
 
kushan said:
what is " mastering physics "

It's the online website for our physics homework. When you enter your answer, you are told whether or not your answer is correct. You are given 5 attempts, during which the points you earn diminish.
 
Even I am in confusion as they have given magnitude of magnetic field
(which means a constant field ) but asking find at what rate it would vary
 
kushan said:
Even I am in confusion as they have given magnitude of magnetic field
(which means a constant field ) but asking find at what rate it would vary

Yes, I noticed this as well. I purposely left that out of my equation as we are attempting to solve for ΔB/Δt. I assume it's the authors of the homework trying to trick us into thinking the value is necessary. Though I cannot see how it could be.
 
  • #10
I am sorry as I fail to help you further with this problem
 
  • #11
You know the resistance of the loop and you know the current so you should be able to find the induced emf
 
  • #12
"I am sorry as I fail to help you further with this problem "

Thanks, Kushan. I appreciate your help. I think there might be an issue with the website. It wouldn't be the first time.

Technician: You might notice that is exactly what I attempted to do... No worries though.
 
  • #13
As it turns out, the website wanted a different number of significant figures. Albeit, an incorrect number with the given information...
 
  • #14
Thank god
 

Similar threads

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