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www.GOALias.blogspot.comAnswers6.14 (a) ε = vBl = 0.12 × 0.50 × 0.15 = 9.0 mV;(b)(c)(d)(e)(f)(g)P positive end and Q negative end.Yes. When K is closed, the excess charge is maintained by thecontinuous flow of current.Magnetic force is cancelled by the electric force set-up due tothe excess charge of opposite signs at the ends of the rod.Retarding force = IBl= 9mV × 0.5 T × 0.15 m9mΩ= 75 × 10 –3 NPower expended by an external agent against the above retardingforce to keep the rod moving uniformly at 12 cm s –1= 75 × 10 –3 × 12 × 10 –2 = 9.0 × 10 –3 WWhen K is open, no power is expended.I 2 R = 1 × 1 × 9 × 10 –3 = 9.0 × 10 –3 WThe source of this power is the power provided by the externalagent as calculated above.Zero; motion of the rod does not cut across the field lines. [Note:length of PQ has been considered above to be equal to the spacingbetween the rails.]6.15µ NI 0B =l(Inside the solenoid away from the ends)µΦ =0NI AlTotal flux linkage = NΦ2µ 0N A= Il(Ignoring end variations in B)dε = ( )dtN Φε =avtotal change in fluxtotal timeεav4π× 10 × 25×10=−30.3 × 10= 6.5 V−7 −4× (500) 2 × 2.56.166.17Mµ aln 12π0= +axε = 1.7 × 10 –5 V2Bπaλ− kˆMR301