A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is $24 \mu…
A short magnet oscillates with a time period 0.1 s at a place where horizontal magnetic field is $24 \mu \mathrm{~T}$. A downward current of 18 A is established in a vertical wire kept at a distance of 20 cm east of the magnet. The new time period of oscillations of the magnet is
0.1 s
0.089 s
0.076 s
0.057 s
Solution
$\mathrm{T}_1=0.1 \mathrm{~s}, \mathrm{~B}_{\mathrm{H}}=24 \mu \mathrm{~T}=24 \times 10^{-6} \mathrm{~T}, \mathrm{I}=18 \mathrm{~A}$,
$\mathrm{r}=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
Magnetic field due to vertical wire,
$\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}=\frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 20 \times 10^{-2}}=18 \mu \mathrm{~T}$
The time period of oscillation of magnet is
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}} \Rightarrow \mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{~B}}} \\
& \therefore \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{B}_{\mathrm{H}}}{\mathrm{~B}+\mathrm{B}_{\mathrm{H}}}}=\sqrt{\frac{24}{18+24}}=\frac{2}{\sqrt{7}} \\
& \therefore \mathrm{~T}_2=\frac{2}{\sqrt{7}} \mathrm{~T}_1=\frac{2}{\sqrt{7}} \times 0.1=0.076 \mathrm{~s}
\end{aligned}$