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A short magnet of magnetic moment 2.0 Weber-metre is lying in a horizontal plane with its north pole pointing 60^(@) east of north. Find the net magnetic field at a point north of the magnet 0.2 m away from it. Horizontal component of earth's magnetic field = 0.3 xx 10^(-4) T. |
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Answer» Solution :Herer=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` 
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