Chapter 32261P32.78P=IV∆IV==××=×∆100 10500 1093..W200 10 VA3From Ampere’s law,BrI20πµbg=enclosedor BIr=µπ02enclosed.(a)At ra00200. m,IenclosedA3.FIG. P32.73andB=×⋅×−4105 00 102002000050050073TmAAmT mTejej.....(b)At rbm,IIenclosedA×3.andB=×−4105 00 100500200AmmTej...(c)UudVrdrIdrrIbaabab==FHGIKJzzzaf2002022244AAAlnU=××FHGIKJ=−4105 00 101 000 104229 10229236Amcm2.00 cmJ MJejej.ln.(d)The magnetic field created by the inner conductor exerts a force of repulsion on the currentin the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a smallrectangular section of the outer cylinder of length Aand width w.It carries a current of
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .