We use the general form of Faraday's lawþē·ašíand consider a circular integration path of radius r₂. Due to symmetry, Faraday's law becomesdBE(2₂) (BA) = -= -A-dtdtSolving for the magnitude of the electric field, we havet² + 1.40|EI ==A d2πr 2dtB2AT 22x²2= 1/2/(10.0220a Cat).Submit Skip (you cannot come back)

The following equation is to be simplified:

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).

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).

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter16: Electromagnetic Waves

Section: Chapter Questions

Problem 54P: An incandescent light bulb emits only 2.6 W of its power as visible light. What is the rms electric...

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