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Let BP and BQ be the magnetic field prod...

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.

Text Solution

Verified by Experts

`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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