# ass24 - Solutions A.6.I DENTIFY Apply Eq(29.4 I R = E S ET...

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Unformatted text preview: Solutions, Assignment 24 29.6.I DENTIFY : Apply Eq.(29.4). / . I R = E S ET U P : / / . B d dt AdB dt = E XECUTE : (a) ( 29 5 4 4 ( ) (0.012 T/s) (3.00 10 T/s ) B Nd d d NA B NA t t dt dt dt- = = = + E ( 29 4 4 3 4 3 3 (0.012 T/s) (1.2 10 T/s ) 0.0302 V (3.02 10 V/s ) . NA t t-- = + = + E (b) At 5.00 s, t = 4 3 3 0.0302 V (3.02 10 V/s )(5.00 s) 0.0680 V.- = + = E 4 0.0680 V 1.13 10 A. 600 I R- = = = E E VALUATE : The rate of change of the flux is increasing in time, so the induced current is not constant but rather increases in time. 29.7. I DENTIFY : Calculate the flux through the loop and apply Faradays law. S ET U P : To find the total flux integrate B d over the width of the loop. The magnetic field of a long straight wire, at distance r from the wire, is 2 I B r = . The direction of B r is given by the right-hand rule. E XECUTE : (a) When 2 i B r = , into the page. (b) . 2 B i d BdA Ldr r = = (c) ln( / ). 2 2 b b B B a a iL dr iL d b a r = = = (d) ln( ) ....
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ass24 - Solutions A.6.I DENTIFY Apply Eq(29.4 I R = E S ET...

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