The horizontal component of the earth's magnetic field at any place is

Explanation

Here r=0.2m,`theta=60^(@)`
`B_(1)[due to bar ]=mu_(0)/4pi M/r^(3) sqrt(3cos^(2)theta +1)`
`=4pixx10^(-7)/4pi xx 2/(0.2)^(3) sqrt(3cos^(2)60^(@)+1)=3.3xx10^(-5)T`
`tan Beta=tan theta/2=tan 601^(@)/2=sqrt(3)/2 rArr Beta = 40^(@)` At `P_(1)` direction of `B_(B)` as shown, whereas `B_(H)` (due to earth acts along North). So net field B is resultant of `B_(I) and B_(H)`.
`B=sqrt(B_(I)^(2)+ B_(H)^(2)+2B_(I)B_(H)cos 41^(@))`=`10^(-5)sqrt((3.3)^(2)+(3.0)^(2)+2xx3.3xx3.0xx0.756)=5.9xx10^(-5)T`