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 B_{1} = B_{H} = 24 × 10^{-6}

And , B_{2} = B_{H} – B

$\large B_2 = B_H – \frac{\mu_0 i}{2 \pi r}$

$\large = 24 \times 10^{-6} – \frac{4\pi \times 10^{-7}\times 18}{2\pi \times 0.2}$

B_{2} = 6 × 10^{-6} T

$\large \frac{T_2}{0.1} = \sqrt{\frac{24 \times 10^{-6}}{6 \times 10^{-6}}} = 2 $

T_{2} = 0.2 sec