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Instructions 50 cm 10 cm 8.0 A R = 1.0 x 10-3

Instructions 50 cm 10 cm 8.0 A R = 1.0 x 10-3

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Instructions 50 cm 10 cm 8.0 A R = 1.0 × 10-3 N A small circular loop of wire of radius 5 cm and resistance 1E-3 Q is centered inside a large circular loop of wire of radius 50 cm. The larger loop, which initially carries a current of 8 A, is cut and its current is reduced to zero over a time interval of 1E-6 s. Our goal will be to find the average current in the smaller loop during this time interval. (The magnetic field of the larger loop is approximately constant over the smaller loop.)
Transcribed Image Text:Instructions 50 cm 10 cm 8.0 A R = 1.0 × 10-3 N A small circular loop of wire of radius 5 cm and resistance 1E-3 Q is centered inside a large circular loop of wire of radius 50 cm. The larger loop, which initially carries a current of 8 A, is cut and its current is reduced to zero over a time interval of 1E-6 s. Our goal will be to find the average current in the smaller loop during this time interval. (The magnetic field of the larger loop is approximately constant over the smaller loop.)
Magnitude of the cross sectional area A of the small loop: O 7.9E-3 m2 O 3.1E-2 m² O 79 m² O 3.1E2 m2
Transcribed Image Text:Magnitude of the cross sectional area A of the small loop: O 7.9E-3 m2 O 3.1E-2 m² O 79 m² O 3.1E2 m2
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