p36_066 - ~ R = (10 6 ◦ ) + (5 6 45 ◦ ) + (5 6 − 45...

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66. With phasor techniques, this amounts to a vector addition problem ~ R = ~ A + ~ B + ~ C where (in magnitude- angle notation) ~ A =(10 6 0 ), ~ B =(5 6 45 ), and ~ C =(5 6 45 ), where the magnitudes are understood to be in µ V/m. We obtain the resultant (especially efficient on a vector capable calculator in polar mode):
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Unformatted text preview: ~ R = (10 6 ◦ ) + (5 6 45 ◦ ) + (5 6 − 45 ◦ ) = (17 . 1 6 ◦ ) which leads to E R = (17 . 1 µ V / m) sin ( ωt ) where ω = 2 . × 10 14 rad / s....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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