Question

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85\times {10}^{–2}{\text{A/m}}^2$ and moment of inertia $5\times {10}^{–6}{\text{kg m}}^2$. If it performs $10$ complete oscillations in $5\text{seconds}$ then the magnitude of the magnetic field is $\text{mT}$.
[Take ${\pi }^2$ as $9.85$]

Answer 1

The time period for an oscillation of the magnetic needle in terms of moment of inertia and magnetic moment is given by,

$T=2\pi \sqrt{\frac{I}{MB}}$

So, solving for $B$ we can write,

$B=4{\pi }^2\frac{I}{M{T}^2}$

Here, $T=\frac{5}{10}=0.5\text{s}$

$B=4\times 9.85\times \frac{5\times {10}^{–6}}{9.85\times {10}^{–2}\times {0.5}^2}$

$B=8\times {10}^{-3}=8\text{mT}$

$\therefore \text{Correct answer}:8$