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Current through ABC A'B'C' is I. What is the magnetic field at P? BP= PB' = r (Here C'B' PBC are collinear) |
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Answer» `B= (1)/(4pi) (2I)/(r )` `B_(AB) = (mu_(0))/(4pi) (I)/(r ) Ox` The magnetic field at point P due to current I in BC is `B_(BC)=0` (As the point P is along the BC). The magnetic field at point P due to current I in A.B. is `B_(A.B.)=(mu_(0))/(4pi) (I)/(r ) ox` The magnetic field at point P due to current I in B.C. is `B_(B.C.)= 0` (as the point P is along the BC)  `:.` The net magnetic FIELDAT P is `B= B_(AB) + B_(BC) + B_(A.B.) + B_(B.C.)` `=(mu_(0)I)/(4pi r) +0 + (mu_(0)I)/(4pi r) + 0= 2 ((mu_(0)I)/(4pi r))= (mu_(0))/(4pi) ((2I)/(r ))` |
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