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the flux of the E=3i+24j+6k through a portion of 5m^2 of the plane the normal to which is 2i+4j+6k

the flux of the E=3i+24j+6k through a portion of 5m^2 of the plane the normal to which is 2i+4j+6k

Grade:12th pass

1 Answers

Arun
25750 Points
4 years ago
 
Dear student 
E = 3i + 24j + 6k 
So, area vector magnitude = 5 
and it is normal to 2i+ 4j +6k
area vector = 5k 
Flux = E.dA 
Flux = (3i+24j+6k ). 5k 
=> 30k 
 
Regards
Arun

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