The time period of the magnetic in an oscillation pmagnetometer in the earth magnetic field is 2s. A short bar magnet is placed to the north of the magnetometer, at a separation 10cm from the oscillation magnet, with its north pole pointing towards north. The time period beceomes half. Calculate the magnetic moment of this short magnet.
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Time period T=`=2pi sqrt((I)/(MB)) Rightarrow T alpha (1)/(sqrtH)B alpha (1)/(T^(2))` Let , is magnet moment due to short magnet and B' is a magnetic field due to short magnet, along south to north Given , `T_(2)=2s, B_(h)=12muT` `" " T_(2)=1s, B_(h)=B+B'=12+B'` ` " " (B+B')/(B)=(T_(1)^(2))/(T_(2)^(2))` ` " " (12+B')/(12)=((2)/(1))=4` `" "B'=36muT` `" "B'=(mu_(0))/(4pi)(2M)/(r^(3))` `Rightarrow " " 36xx10^(-6)=10^(-7)xx(2M)/((0.10)^(3))` `" " M=0.18Am^(2)`
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