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Question

https://s1.lite.msu.edu/enc/53/9ba24c2d549e3d687d061b863f16cd2a4f352f9ed41cc003720b774bf8a1674a0f3533ea197dbd4e1bcbb5686d1f10232b9a623062410188f19f29481939ab1af2df86094ce5c824.gif

Two long, parallel conductors are carrying currents in the same direction, as seen in the figure below.

Conductor A carries a current of IA = 180 A and is held firmly in position; conductor B carries current IB and is allowed to slide freely up and down (parallel to A) between a set of non-conducting guides. If the linear mass density of conductor B is 0.104 g/cm, what value of current IB will result in equilibrium when the distance between the two conductors is 2.50 cm?

Explanation / Answer

field on B due to A is 2*10^-5 *(180/2.5) = 144*10^-5
force on B per unit length = 144*10^-5 *(I in b)
force due to gravity is 1.04*10^-2 * 10= 0.104
equating both I in B = 72.22 amperes

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