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March 29, 2015

March 29, 2015

Posted by **M** on Monday, May 14, 2012 at 1:22pm.

help is greatly appreciated!

- Physics -
**Elena**, Monday, May 14, 2012 at 3:37pmε = 0.63V. I = 2.7 A.

ε = -dÔ/dt =-d(B•S•cosα)/dt.

α= 0 => cos α =1,

ε = - S •dB/dt.

If S is the area of the circle,

then S = π •r², => r=sqrt(S/π).

Perimeter (the length of wire) is

L = 2 •π•r = 2 •π• sqrt(S/π).

the side of the square is

L/4 =2 •π• sqrt(S/π)/4 = π• sqrt(S/π)/2

The area of the square is

S1 = {π• sqrt(S/π)/2}² = π²•S/ 4•π = =π•S/4.

ε1= - S1 •dB/dt= -(π•S/4)• dB/dt,

ε1/ε = [(π•S/4) • (dB/dt)] / {S •dB/dt}= =π/4.

ε1 = ε• π/4 = 0.63• π/4 = 0.495 V.

ε = I•R, => R = ε/I = 0.63/2.7 = 0.23 Ω,

ε1 = I1•R,

I1 = ε1/R = 0.495/0.23 = 2.12 A.

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