Question

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

Answer 1

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

Answer 2

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

Answer 3

1.76E-5? choice D is the solution?

Answer 4

no, since we have to find the emf

Answer 5

more explanation:

Answer 6

ohh ok! how do we do taht?

Answer 7

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

Answer 8

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

Answer 9

oh ok! what do we plug in? :/

Answer 10

emf= flux/area

Answer 11

no, emf= flux/time

Answer 12

oh ye

Answer 13

please apply the Faraday-Neumann law @rvc

Answer 14

what are we plugging in? :/

Answer 15

yes faradays law applies here :)

Answer 16

i just messed the equation :)

Answer 17

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

Answer 18

no worries!! :) @rvc

Answer 19

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

Answer 20

that's right!

Answer 21

yay! thanks!:)

Answer 22

you are the BEST @Michele_Laino

Answer 23

:) @rvc