A magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by $19 \%$, By…
A magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by $19 \%$, By doing this, the time period of the magnet approximately
Increases by $11 \%$
Decreases by $19 \%$
Increases by $19 \%$
Decreases by $4 \%$
Solution
Time period of suspended magnet in uniform magnetic field is
$\mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}$
When the magnetic moment is reduced by $19 \%$, then
$\begin{aligned} & \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{I}}{0.81 \mathrm{MB}}}=\frac{1}{0.9}\left(2 \pi \sqrt{\frac{\mathrm{I}}{\mathrm{MB}}}\right)=\frac{\mathrm{T}_1}{0.9} \\ & \therefore \quad \mathrm{~T}_2=1.11 \mathrm{~T}_1\end{aligned}$
$\therefore \%$ increase in time period is
$\Delta \mathrm{T} \%=\left(\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{~T}_1} \times 100\right) \%$
$=\left(\frac{1.11 \mathrm{~T}_1-\mathrm{T}_1}{\mathrm{~T}_1}\right) \times 100=11 \%$