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A short bar magnet of magnetic moment 5.25xx10^(-2)JT^(-1) is placed with its axis perpendicular to the earth's field 's field direction . At what distance from the centre of the magnetthe resultant field is inclined at 45^@ with earth's field on (a) its normal bisector and (b) its axis. Magnitude of te earth's field at the place is given to be 0.42 G, Ignore the length of the magnet in comparison to the distance involved. |
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Answer» SOLUTION :`m=5.25xx10^(-2)JT^(-1),theta=90^@` `B_E=0.42G=0.42xx10^(-4)T` Since RESULTANT at P makes `45^@, tan 45 = (B_("axial")/(B_E))` i.e., `l=(B_("axial")/B_E) "":.B_("axial")=B_E` But `B_("axial")=10^(-7)xx(2M)/d^3:. (10^(-7)xx2m)/d^3=B_E` or `d^3=(10^(-7)xx2m)/B_E=(10^(-7)xx2xx5.25xx10^(-2))/(0.42xx10^(-4))` `:.d=[(10^(-3)xx2xx5.25)/42]^(1/3)=10^(-1)xx((5.25)/21)^(1/3)=10^(-1)xx0.63m=6.3cm` Similarly `10^(-7)xxm/(d.3)=B_E" or "d.^3=(10^(-7)xxm)/B_E" or " d.=(6.3)/(2^(1/3))=5.0cm` 
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