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

`i_(1) = 0.1110 A, i_(2) = 0.096 A`

Explanation

`B_(v) = B_(H) tan 30^(@) = 3.49 xx 10^(-5) xx (3)/sqrt3`
`B_(v) = 2.05 xx 10^(-5) T`
`B_(v)` should be cancel out by field due to current in coil placed in horizontal plane
`B_(v) = (mu_(0))/(4pi) (2pi NI_(1))/ (R )`
`2.05b xx 10^(-5) = 10^(-7) xx (2xx 3.14 xx 100 xx I_(1))/(0.3)`
`I_(1) = (2.05)/(31.4) = 0.097 Amp`
`B_(H)` should be cancle out by field due to current in vartical plane
`B_(H) (mu_(0))/(4pi) (2piNI_(2))/(r)`
`3.49 xx 10^(-7) = 10^(-7) xx (2 xx 3.14 xx 100 xx I_(2))/(0.2)`
`I-(20 =0.1110 Amp`
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