A short magnet oscillates in a vibration magnetometer with a time period of 0.10 s where the horizontal component of earth's magnetic field is `24 mu T` . An upward current of 18 A is established in the vertical wire placed 20 cm east of the magnet . Find the new time period.

Explanation

`T_(2)/T_(1)=sqrt(B_(1)/B_(2))" Where "B_(1)=B_(H)=24xx10^(-6)T`
and `B_(2)=B_(H)~B=B_(H)~(mu_(0)i)/(2pir)`
`=24xx10^(-6)~(4pixx10^(-7)xx18)/(2pixx0.2)=6xx10^(-6)T`
`thereforeT_(2)/0.1=sqrt((24xx10^(-6))/(6xx10^(-6)))=2rArrT_(2)=0.2s`