What is the magnitude of the loop current at t = 1.3s?

AI Thread Summary
The discussion revolves around calculating the magnitude of the loop current in a conducting loop subjected to a time-varying magnetic field. The magnetic field is defined by the equation Bz = at² - b, with specific values for a and b. The user initially calculates the EMF using the change in magnetic flux but arrives at an incorrect current value of 0.0789 A, while the correct answer is 0.16 A. A suggestion is made to use the derivative of the magnetic flux to accurately determine the EMF, emphasizing the importance of correctly accounting for the non-linear change in the magnetic field. The conversation highlights the challenges in applying Faraday's Law to non-linear magnetic fields.
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Homework Statement



A conducting loop with area 0.13m2 and resistance 6.0Ω lies in the x-y plane. A spatially uniform magnetic field points in the z direction. The field varies with time according to Bz=at2−b, where a = 2.8T/s2 and b = 8.0T .

(a) Find the (magnitude of the) loop current when t = 1.3s.

Homework Equations



V = I * R
EMF = Δ(BA)/Δt * n

The Attempt at a Solution



I calculated the EMF using the second formula listed, with n = 1 as there is only a single loop.

EMF = Δ(BA)/Δt
Binitial = 2.8(02) - 8 = -8
Bfinal = 2.8(1.32) - 8 = -3.268
=> ΔB = 4.732
=> Δ(BA) = 4.732 * 0.13 = 0.6152

=> EMF = 0.6152 / 1.3 = 0.4732 V

Then I used the first formula to find a value for I...

0.4372 = I * 6
0.0789 A = I

This is incorrect however, and the correct answer given is 0.16 - roughly twice my solution. Am I leaving out a multiplier somewhere? Any help is appreciated, this seemingly simple problem has frustrated me for too long.
 
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The magnetic field is not changing linearly, so a linear approximation of the change is not going to be too accurate.

Since you have the expression for B(t), I'd suggest using the derivative to find the change in flux since (Faraday's Law):
$$EMF = \frac{d \Phi}{dt}~~~~~~\text{and}~~~~~~\Phi(t) = Area\;B(t)$$
 
Got it, thanks gneill!
 
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