We use the general form of Faraday's law fē· ²-d and consider a circular integration path of radius r₂. Due to symmetry, Faraday's la dB dt E(2₂) - (BA) (BA) = -A dt Solving for the magnitude of the electric field, we have Ad ²+1.40 2R²₂ at ( |E| = dt 2 #r₂² 2x²₂ -30 N/NT 0.0220 (t) (t).

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We use the general form of Faraday's law
þē·aší
and consider a circular integration path of radius r₂. Due to symmetry, Faraday's law becomes
dB
E(2₂) (BA) = -
= -A-
dt
dt
Solving for the magnitude of the electric field, we have
t² + 1.40
|EI =
=
A d
2πr 2
dt
B
2
AT 2
2x²2
= 1/2/(1
0.0220
a Ca
t).
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Transcribed Image Text:We use the general form of Faraday's law þē·aší and consider a circular integration path of radius r₂. Due to symmetry, Faraday's law becomes dB E(2₂) (BA) = - = -A- dt dt Solving for the magnitude of the electric field, we have t² + 1.40 |EI = = A d 2πr 2 dt B 2 AT 2 2x²2 = 1/2/(1 0.0220 a Ca t). Submit Skip (you cannot come back)
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