Two coils each of 100 turns are held such that one lies in vertical plane and the other in the horizontal plane with their centres coinciding. The radius of the vetical coil is 0.20m and that of the horizontal coil is 0.30m. How will you neutralize the magnetic field of the earth at their common centre? What is the current to be passed through each coil? Horizontal component of earth's magnetic field `=0.35xx10^-4T` and angle of dip =`30^@`.

Answer

A, B, D

Explanation

As the field due to a current-carrying coil is along its axis.the vertical coil will produce horizontal field not horizontal coil vertical, i.e.,
`mu_(0)/(4pi)(2piN_(V)l_(V))/(R_(V))=B_(H)` and `mu_(0)/(4pi)(2piN_(H)l_(H))/(R_(H))=B_(V)`
But as `tanphi=(B_(V))/(B_(H)),B_(V)=B_(H) tan phi=(B_(H))/sqrt3 [as phi=(phi//6)]`
and `1A/m=4pixx10^(-7)(Wb)/m^(2)`
so `10^(-7)(2pixx100xxI_(v))/0.2=4pixx10^(-7)xx27.8`
i.e.,`I_(v)=1112xx10^(-4) A`
and `10^(-7)(2pixx100xxI_(H))/0.3=4pixx10^(-7)xx27.8/sqrt3`
i.e.,`I_(H)=556sqrt6xx10^(-4) A`