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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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Answer» Solution :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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