224 Faraday’s LawP31.29(a)The force on the side of the coil entering the field(consisting of Nwires) isFN ILBN IwB==afaf.The induced emf in the coil isε=NddtNdBwxdtNBwvBΦ.so the current isIRNBwvRcounterclockwise.The force on the leading side of the coil is then:FNNBwvRwBNBwvR=FHGIKJ=22 2to the left .(b)Once the coil is entirely inside the field,ΦBNBAconstant ,so=0,I=0 ,andF=0.FIG. P31.29(c)As the coil starts to leave the field, the flux decreasesat the rate Bwv, so the magnitude of thecurrent is the same as in part (a), but now the current is clockwise. Thus, the force exerted onthe trailing side of the coil is:FR=to the left again .P31.30Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. Theloop will create a field out of the page by carrying a counterclockwise current. Therefore, current
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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 .