714190122(410 T m/A)(2.66 10A)(0.0200 m)1.18 10T22 (0.0300 m)dirBRµπππ−−−×⋅×===×. (b) Outside we have (by Eq. 32-17) 02/2dB=where r2 = 0.0500 cm. Here we obtain 41902(410 T m/A)(2.66 10A)1.06 10T22 (0.0500 m)diBr−×=×9. (a) Inside we have (by Eq. 32-16) 2dBR=, where 10.0200 m,r=0.0300 m,R=
This is the end of the preview. Sign up
access the rest of the document.
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.