Derive expression for induced voltage and current

AI Thread Summary
A circular conducting loop with radius a and resistance R2 is influenced by a larger loop with radius b and resistance R1, where a sinusoidal voltage V(t) = V0 sin ωt is applied. The induced voltage in the inner loop is derived using the formula Vind = -∏a^2μ0V0ωcos(ωt)/(2bR1). There is confusion regarding the presence of π in the magnetic field equation B = μ0V0/(2bR1), which should not include π in the denominator. The correct understanding is critical for accurately calculating induced voltage and current through the inner loop. Clarification on the formulas is essential for resolving the misunderstanding.
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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