Q.A magnetic needle placed in a uniform magnetic field has a magnetic moment of $2\times10^{-2}$ A m$^2$ and a moment of inertia of $7.2\times10^{-7}$ kg m$^2$. It performs 10 complete oscillations in 6 s. What is the magnitude of the magnetic field?

[!TLDR] With $T_{period}=0.6$ s, $B=\dfrac{4\pi^2I}{mT_{period}^2}\approx39.48\times10^{-4}$ T. [!AN …

The needle performs 10 oscillations in 6 s, so its time period is $T_{period}=6/10=0.6$ s. From $T_{period}=2\pi\sqrt{I/(mB)}$, rearranged for B: $B=\dfrac{4\pi^2I}{mT_{period}^2}$. Substituting: $4\pi^2\approx39.478$; numerator $=39.478\times7.2\times10^{-7}=2.842\times10^{-5}$; denominator $=m\times T_{period}^2=(2\times10^{-2})\times(0.6)^2=(2\times10^{-2})\times0.36=7.2 …

Convert "10 oscillations in 6 s" to a period $T_{period}=0.6$ s, then solve $T_ …

Using the oscillation count/frequency directly in the period formula instead of first converting to the actual time period …