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

Here `B_(H) = 24 xx 10^(-6)T`
`T_(1) = 0.1 s`
Since ` B = B_(H) - B_(wire)`
`= 24 xx 10^(-4) - (mu_(0)i)/(2 pi r)`
`= 24 xx 10^(-6) - (2 xx 10^(-7) xx 18)/(0.2)`
`= (24 - 10) xx 10^(-4)`
`= 14 xx 10^(-6)`
`T = 2 pi sqrt((1)/(MB_(H)))`
`T_(1))/(T_(2)) = sqrt((B)/(M_(2)))`
`rArr (0.1)/(T_(2)) = sqrt((14 xx 10^(-8))/(24 xx 10^(-8)))`
`= rArr((0.1)/(T_(2)))^(2) = (14)/(24)`
`rArr T_(2)^(2) = (0.01 xx 14)/(24) `
`rArr T_(2) = 0.076`