Rate of Field Change for Induced Current of Loopby tevatron1 Tags: emf, induced current, loop, magnetic flux 

#1
Feb2112, 11:14 AM

P: 11

1. The problem statement, all variables and given/known data
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? 2. Relevant equations Trying to solve for ΔB/Δt 3. 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... 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 



#2
Feb2112, 11:21 AM

P: 259





#3
Feb2112, 11:26 AM

P: 11

How does the rest of my thought process look? 



#4
Feb2112, 11:31 AM

P: 259

Rate of Field Change for Induced Current of Loop
as as i know you are doing correctly ,
where are you stuck ? 



#5
Feb2112, 11:32 AM

P: 11





#6
Feb2112, 11:33 AM

P: 259

what is " mastering physics "




#7
Feb2112, 11:35 AM

P: 11





#8
Feb2112, 11:37 AM

P: 259

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 



#9
Feb2112, 11:39 AM

P: 11





#10
Feb2112, 11:44 AM

P: 259

I am sorry as I fail to help you further with this problem




#11
Feb2112, 12:04 PM

P: 1,506

You know the resistance of the loop and you know the current so you should be able to find the induced emf




#12
Feb2112, 12:12 PM

P: 11

"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
Feb2112, 12:26 PM

P: 11

As it turns out, the website wanted a different number of significant figures. Albeit, an incorrect number with the given information...




#14
Feb2212, 12:26 AM

P: 259

Thank god



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