ch30-p026 - 26. (a) We assume the flux is entirely due to...

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the possibility of an overall minus sign since we are asked to find the absolute value of the flux. /2 00 || () l n . 22 / 2 rb B ii a adr rr b µµ + + ⎛⎞ Φ= = ⎜⎟ ππ ⎝⎠ When 1.5 = , we have 8 (4 T m A)(4.7A)(0.022m) | | ln(2.0) 1.4 10 Wb. 2 B π ×⋅ = × −7 π1 0 (b) Implementing Faraday’s law involves taking a derivative of the flux in part (a), and recognizing that / dr dt v = . The magnitude of the induced emf divided by the loop resistance then gives the induced current: loop 3 42 5 ln 2/ 2 2 [ ( / 2 ) ] (4 T m A)(4.7A)(0.022m)(0.0080m)(3.2 10 m/s)
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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