A short magnet in an oscillation magnetometer oscillates with a...

The magnetic field produced by the vertical wire at the position of the magnet is given by:

$B_{\text{wire}} = \dfrac{\mu_0 I}{2\pi r}$

Substituting the given values:

$B_{\text{wire}} = \dfrac{4\pi \times 10^{-7} \times 18}{2\pi \times 0.20} = 18 \times 10^{-6} \text{ T} = 18 \mu\text{T}$

By the right-hand grip rule, for a downward current in a wire located east of the magnet, the magnetic field at the magnet's position points North. Since the Earth's horizontal magnetic field $B_H$ also points North, the two magnetic fields add up.

The net magnetic field is:

$B_{\text{net}} = B_H + B_{\text{wire}} = 24 \mu\text{T} + 18 \mu\text{T} = 42 \mu\text{T}$

The time period of an oscillation magnetometer is given by $T = 2\pi \sqrt{\dfrac{I}{MB}}$, which implies $T \propto \dfrac{1}{\sqrt{B}}$.

Therefore, the new time period $T_2$ is:

$\dfrac{T_2}{T_1} = \sqrt{\dfrac{B_1}{B_2}}$

$T_2 = 0.10 \times \sqrt{\dfrac{24}{42}} = 0.10 \times \sqrt{\dfrac{4}{7}}$

$T_2 \approx 0.10 \times 0.756 = 0.0756 \text{ s} \approx 0.076 \text{ s}$