Q: A short magnet oscillation is a vibration magnetometer with a time period of 0.1 s where the horizontal component of earth’s magnetic field is 24 μT. An upward current of 18 A is established in the vertical wire placed 20 cm east of the magnet. Find the new time period ?
Sol: $\large \frac{T_2}{T_1} = \sqrt{\frac{B_1}{B_2}}$ ; Where B1 = BH = 24 × 10-6
And , B2 = BH – B
$\large B_2 = B_H – \frac{\mu_0 i}{2 \pi r}$
$\large = 24 \times 10^{-6} – \frac{4\pi \times 10^{-7}\times 18}{2\pi \times 0.2}$
B2 = 6 × 10-6 T
$\large \frac{T_2}{0.1} = \sqrt{\frac{24 \times 10^{-6}}{6 \times 10^{-6}}} = 2 $
T2 = 0.2 sec