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.

Options

A. 0.1s
B. 0.089s
C. 0.076s
D. 0.057s

Answer

C

Explanation

Magnetic field due to downward conductor –
`B= (mu_(0))/(4pi) .(2i)/(a) = (10^(-7) xx 2 xx 18)/(0.2)= 18 mu T`
`T prop (1)/(sqrtB)`
`(T_(1))/(T_(2)) = sqrt((B + B_(H))/(B_(H))) rArr (T)/(T.)= sqrt((18+ 24)/(24)) = T= 0.076` sec