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anonymous
 one year ago
Each side of a square loop of wire measures 2.0 cm. A magnetic field of 0.044 T perpendicular to the loop changes to zero in 0.10 s. What average emf is induced in the coil during this change?
A. 1.8 V
B. 0.088 V
C. 0.88 V
D. 0.00018 V
anonymous
 one year ago
Each side of a square loop of wire measures 2.0 cm. A magnetic field of 0.044 T perpendicular to the loop changes to zero in 0.10 s. What average emf is induced in the coil during this change? A. 1.8 V B. 0.088 V C. 0.88 V D. 0.00018 V

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the magnetic flux change is: \[\Large \Delta \Phi = S \times \Delta B = {\left( {2 \times {{10}^{  2}}} \right)^2} \times 0.044 = ...Weber\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\large \Delta \Phi = S \times \Delta B = {\left( {2 \times {{10}^{  2}}} \right)^2} \times 0.044 = ...Weber\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01.76E5? choice D is the solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, since we have to find the emf

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2more explanation:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok! how do we do taht?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2before magnetic flux chnaging, the flux of the magnetic field through the square loop is: area*magnetic field=0.02*0.02*0.044 after the magnetic field changing the new magnetic flux through the square loop is: area*magnetic field=0.02*0.02*0=0 dw:1434436897217:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the requested emf is: \[\Large E = \frac{{\Delta \Phi }}{{\Delta t}} = ...volts\] that is the FaradayNeumann law

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok! what do we plug in? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, emf= flux/time

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please apply the FaradayNeumann law @rvc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are we plugging in? :/

rvc
 one year ago
Best ResponseYou've already chosen the best response.1yes faradays law applies here :)

rvc
 one year ago
Best ResponseYou've already chosen the best response.1i just messed the equation :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next step, is: \[\Large E = \frac{{\Delta \Phi }}{{\Delta t}} = \frac{{0.176 \times {{10}^{  4}}}}{{0.1}} = ...volts\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no worries!! :) @rvc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01.76 E4? so the solution is choice D?

rvc
 one year ago
Best ResponseYou've already chosen the best response.1you are the BEST @Michele_Laino
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