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A conductingcurrent-carryingplaneis placedin an externalunifrom magneticfield. Asa result, themagneticinduction becomes equal to B_(1) on oneof the plane and too B_(2) on the other. Find the magnetic force actingper unit area ofteh plane in hte cases illiustrated in FIg. Determinethe directionof the current in the plane in each case. |
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Answer» Solution :(a) The externalfield MUST be`(B_(1) + B_(2))/(2)` , whichwhensuperposedwith the internal FIELD `(B_(1) - B_(2))/(2)` (of opposite sign on the two sides of the plates) must giveactual field. Now `(B_(1) - B_(2))/(2) = (1)/(2) mu_(0) i` or, `i = (B_(1) - B_(2))/(mu_(0))` Thus, `F = (B_(1)^(2) -B_(2)^(2))/(2 mu_(0))`  (b) Here, the EXTERNAL field must be `(B_(1) - B_(2))/(2)` upward with an internal field, `(B_(1) + B_(2))/(2)`, upward on the left and downward on the RIGHT. Thus, `i = (B_(1) + B_(2))/(mu_(0))` and `F = (B_(1)^(2) - B_(2)^(2))/(2 mu_(0))`  (c) Our boundary condition following from Gauss's law is `B_(1) cos theta_(1)= B_(2) cos theta_(2)`. Also , `(B_(1) sin theta_(1) + B_(2) sin theta_(2)) = mu_(0) i` where `i` = current per unit length The externalfied parallelto the platemust be`(B_(1) sin theta_(1) - B_(2) theta_(2))/(2)` (The perpendicularcomponent`B_(1) cos theta_(1)`, does not matter SINCE the correspondingforce is tangetial) Thus, `F = (B_(1)^(2) sin^(2) theta_(1) - B_(2)^(2) sin^(2) theta_(2))/(2mu_(0))` per unit area `(B_(1)^(2) - B_(2)^(2))/(2 mu_(0))` per unit area. The direction of the current in the planeconductoris perpendicular to the paperand beyondthe drawing . 
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