Figure shows a short magnet executing small oscillations in an oscillation magnetometer in earth's magnetic field having horizontal component `24 muT`. The time period of oscillation is `0.10 s`. An upward electric current of `18 A` is established in the vertical wire placed 20cm east of the magnet by closing the switch `S`. Find the new time

Explanation

The magnetic field at the site of the short magnet due to the vertical current is
`B = (mu_0 i)/(2pid) = (2 xx 10^(-7) T m A^(-1)) (18 A)/(0.20 m)`
As the wire is east of the magnet, this magnetic field will be from north to south according to the right-hand thumb rule. The earth's magnetic field has horizontal component `24 muT` from south to north. Thus, the resultant field will be `6.0 muT` from south to north. If `T_1 and T_2` be the time periods without and with the current.
`T_1 = sqrt(I/(MB_H)` and T_2 = sqrt(I/(M(B_H - B))`
or, `T_2/T_1 = sqrt(B_H)/(B_H - B) = sqrt(24 muT)/(6 muT) = 2`
or, `T_2 = 2T_1 = 0.20 s`.