Let `B_P and B_Q` be the magnetic field produced by wire P and Q which are placed symmetrically in a rectangular loop ABCD. Current in wire P is directed I inward and in Q is 2I outward. If `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` Then find I.

Let `B_P and B_Q` be the magnetic field produced by wire P and Q which

1.	Let `B_P and B_Q` be the magnetic field produced by wire P and Q which are placed symmetrically in a rectangular loop ABCD. Current in wire P is directed I inward and in Q is 2I outward. If `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` Then find I.
Answer» `PtoAB, CD,DA,BC` `QtoAB, CD,DA,BC` Given `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` we know `int vecB_(n et). Dvecl=mu_0(2I-I)=mu_0I` Using superpostion taking wire P only `intvecB_P.dvecl=-mu_0I and int vecB_Q.dvecl=mu_0 2I` `int_A^B vecB_Pdvecl+int_B^C vecB_Pdvecl+int_A^D vecB_Pdvecl+int_D^A vecB_Pdvecl=-mu_0I` From symmetry it is clear `int_B^C vecB_Pdvecl=int_D^A vecB_Pdvecl` As `int_A^B vecB_Qdvecl=2mu_0I` Hence, `int_B^C vecB_Pdvecl=-mu_0I` Also, `int_D^A vecB_Q dvecl=-2mu_0I` Then, `int_A^D vecB_Q dvecl=4 mu_0I=int vecB_Qdvecl` `int_A^B vecB_P dvecl=-mu_0I implies int_D^C vecB_Q dvecl=2mu_0I` `int vecB_("net") dvecl=int vecB_Pdvecl+int vecB_Q dvecl=mu_0I` `=[-mu_0I-mu_0I-2mu_0I-2mu_0I]` `+[2mu_0I+2mu_0I+4mu_0I+4mu_0I]` `=-6mu_0I+12mu_0I=mu_0(6I)` Hence, `I=6A`.

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Let `B_P and B_Q` be the magnetic field produced by wire P and Q which are placed symmetrically in a rectangular loop ABCD. Current in wire P is directed I inward and in Q is 2I outward. If `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` Then find I.

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