Derive expression for induced voltage and current

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SUMMARY

The discussion focuses on deriving expressions for the induced voltage and current in a circular conducting loop subjected to a sinusoidal voltage. The induced voltage in the inner loop is expressed as Vind = -∏a²μ₀V₀ωcos(ωt)/(2bR₁), where a is the radius of the inner loop, μ₀ is the permeability of free space, V₀ is the peak voltage, and R₁ is the resistance of the outer loop. The current through the inner loop is given by I₂ = -Vind/R₂. The participants clarify the role of π in the magnetic field equation and confirm its absence in the denominator.

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  • Basic proficiency in calculus, specifically differentiation
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hopkinmn
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Homework Statement



A circular conducting loop with radius a and resistance R2 is concentric with a circular conducting loop with radius b much greater than a and resistance R1. A Resistance dependent voltage is applied to the larger loop, having a slow sinusoidal variation in time given by V(t) = V0 sin ωt, where V0 and ω are constants with dimensions of voltage and inverse time, respectively. Assuming that the magnetic field throughout the inner loop is uniform (constant in space) and equal to the field at the center of the loop, derive expressions for the potential difference induced in the inner loop and the current i through that loop.

Homework Equations



Vind=(d/dt)AB
where A is the area: A=∏a^2
and B is magnetic field

I1=V0/R1
I2=-Vind/R2

The Attempt at a Solution



My understanding is that d/dt=ωcos(ωt)
and B=μ0*I1/(2*∏*b)=μ0*V0/(2*b*∏*R1)

The answer for Vind=-∏*a^2*μ0*V0*ω*cos(ωt)/(2*b*R1)
What I don't understand is what happened to ∏ in B=μ0*V0/(2*b*R1)?
 

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hopkinmn said:
My understanding is that d/dt=ωcos(ωt)
and
B=μ0*I1/(2*∏*b)=μ0*V0/(2*b*∏*R1)
The answer for Vind=-∏*a^2*μ0*V0*ω*cos(ωt)/(2*b*R1)
What I don't understand is what happened to ∏ in B=μ0*V0/(2*b*R1)?

Check the formula. There should be no pi in the denominator.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

ehild
 

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