A short magnet oscillates in an oscillation magnetometer with a time period of 0.10s where the earth's horizontal magnetic field is `24 muT`. A downward current of `18A` is established in a vertical wire placed 20cm east of the magnet. Find the new time period.

Explanation

`9T_(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`
`therefore (T_(2))/(0.1)=sqrt((24xx10^(-6))/(6xx10^(-6))=2 rArrT_(2)=0.2s`