Two parallel wires PQ, RS of resistance 10Omega and 20Omega are seperated by a distance of 10 cm and connected in parallel across a cell of emf200V and negligible intemal resistance. A wire AB of mass I g and length J cm is balanced exactly, between them. What must be the current in it.

Two parallel wires PQ, RS of resistance 10Omega and 20Omega are sepera

1.	Two parallel wires PQ, RS of resistance 10Omega and 20Omega are seperated by a distance of 10 cm and connected in parallel across a cell of emf200V and negligible intemal resistance. A wire AB of mass I g and length J cm is balanced exactly, between them. What must be the current in it.
Answer» Solution : AB experience a FORCE of attraction due to PQ and RS. ApplyV=iR `200=i_(1)xx10rArri_(1)=20A` `200=i_(2)xx20rArr=10A` AB will be in EQUILIBRIUM if `(mu_(0))/(2pia)i_(2)i_(3)I+mg=(mu_(0)i_(1)i_(3))/(2pia)` `(mu_(0))/(2pia)xx(i_(1)i_(3)-i_(2)i_(3))I=mg` `(4pixx10^(-7))/(2pixx5xx10^(-2))(20i_(3)-10i_(3))xx1/100=10^(-3)xx9.8` `(2xx10^(-5)xx10i_(3))/(5xx100)=9.8xx10^(-3)` `i_(3)=24500A`.

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Two parallel wires PQ, RS of resistance 10Omega and 20Omega are seperated by a distance of 10 cm and connected in parallel across a cell of emf200V and negligible intemal resistance. A wire AB of mass I g and length J cm is balanced exactly, between them. What must be the current in it.

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