## 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

1. Michele_Laino

the magnetic flux change is: $\Large \Delta \Phi = S \times \Delta B = {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times 0.044 = ...Weber$

2. Michele_Laino

$\large \Delta \Phi = S \times \Delta B = {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times 0.044 = ...Weber$

3. anonymous

1.76E-5? choice D is the solution?

4. Michele_Laino

no, since we have to find the emf

5. Michele_Laino

more explanation:

6. anonymous

ohh ok! how do we do taht?

7. Michele_Laino

before 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|

8. Michele_Laino

so the requested emf is: $\Large E = \frac{{\Delta \Phi }}{{\Delta t}} = ...volts$ that is the Faraday-Neumann law

9. anonymous

oh ok! what do we plug in? :/

10. rvc

emf= flux/area

11. Michele_Laino

no, emf= flux/time

12. rvc

oh ye

13. Michele_Laino

14. anonymous

what are we plugging in? :/

15. rvc

yes faradays law applies here :)

16. rvc

i just messed the equation :)

17. Michele_Laino

next step, is: $\Large E = \frac{{\Delta \Phi }}{{\Delta t}} = \frac{{0.176 \times {{10}^{ - 4}}}}{{0.1}} = ...volts$

18. Michele_Laino

no worries!! :) @rvc

19. anonymous

1.76 E-4? so the solution is choice D?

20. Michele_Laino

that's right!

21. anonymous

yay! thanks!:)

22. rvc

you are the BEST @Michele_Laino

23. Michele_Laino

:) @rvc