Bo=4π∗10−7∗800∗51.2=4.18∗10−3Tr=D2=32=1.5mm=1.5∗10−3mq=2(4.18∗10−3(50)∗(π)(1.5∗10−3)2cos02000=1.47∗∗10−9Ck=1.47∗10−985=1.73∗10−11C/cmB_o=\frac{4\pi *10^{-7}*800*5}{1.2}=4.18*10^{-3}T\\ r= \frac{D}{2}=\frac{3}{2}=1.5 mm= 1.5*10^{-3}m\\ q=\frac{2(4.18*10^{-3}(50)*(\pi)(1.5*10^{-3})^2 cos 0}{2000}=1.47**10^{-9}C\\ k= \frac{1.47*10^{-9}}{85}=1.73*10^{-11} C/cmBo=1.24π∗10−7∗800∗5=4.18∗10−3Tr=2D=23=1.5mm=1.5∗10−3mq=20002(4.18∗10−3(50)∗(π)(1.5∗10−3)2cos0=1.47∗∗10−9Ck=851.47∗10−9=1.73∗10−11C/cm
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