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Unformatted text preview: (c) Consider Eq. 3228 with E = 1 2 E m . In order to satisfy this equation, we require sin( d t ) = 1 / 2. Now we note that the problem states that E is increasing in magnitude , which (since it is already negative) means that it is becoming more negative. Thus, dierentiating Eq. 3228 with respect to time (and demanding the result be negative) we must also require cos( d t ) < 0. These conditions imply that t must equal (2 n 5 / 6) [ n = integer] . Consequently, Eq. 3329 yields (for all values of n ) i = I sin 2 n 5 6 + 2 = ( 3 . 91 10 3 A ) 3 2 ! = 3 . 38 10 2 A ....
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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.
 Fall '08
 SPRUNGER
 Physics, Charge, Current

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