A short magnet oscillates in an oscillation magnetometer with a time period of 0.10s where the earth's horizontal magnetic field is 24 PT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} {\mathrm{T}}_1=\sqrt[2\mathrm{\pi }]{\frac{\mathrm{T}}{{\mathrm{MB}}_{{\mathrm{H}}_1}}} \dots \left(\mathrm{i}\right) \\ \mathrm{Where} {\mathrm{B}}_{{\mathrm{H}}_1}=24\times {10}^{-6}\mathrm{T} \\ \mathrm{The} \mathrm{magnetic} \mathrm{field} \mathrm{produced} \mathrm{by}, \mathrm{wire} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}.\frac{\mathrm{i}}{\mathrm{r}} \\ =(2\times {10}^{-7})\times \frac{\left(18\right)}{0.20} \\ =1.8\times {10}^{-6}\mathrm{T} \\ \mathrm{Now} {\mathrm{B}}_{{\mathrm{H}}_2}={\mathrm{B}}_{{\mathrm{H}}_1}+\mathrm{B}=42\times {10}^{-6}\mathrm{T} \\ {\mathrm{T}}_2 =\sqrt[2\mathrm{\pi }]{\frac{\mathrm{I}}{{\mathrm{MBH}}_2}} \dots \left(\mathrm{ii}\right) \\ \mathrm{Using} \mathrm{equations} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right), \mathrm{and} \mathrm{substituting} \mathrm{the} \\ \mathrm{values}, \mathrm{we} \mathrm{get} \\ {\mathrm{T}}_2=0.076\mathrm{s} \end{gathered}$$