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.

Explanation

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)`