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

We are given a magnetic needle that executes small angular oscillations in a uniform magnetic field. The standard result for such a magnet is that the time period $$T$$ of one complete oscillation is related to its moment of inertia $$I$$, its magnetic moment $$M$$, and the magnetic-field magnitude $$B$$ by

$$T \;=\; 2\pi\sqrt{\dfrac{I}{M\,B}}$$

This follows from the equation of a physical torsional pendulum, where the restoring torque is $$M B \sin\theta \approx M B\,\theta$$ for small angles, giving simple harmonic motion.

The magnet makes 10 complete oscillations in 5 s, so the time period of one oscillation is

$$T \;=\; \dfrac{\text{total time}}{\text{number of oscillations}} \;=\; \dfrac{5\ \text{s}}{10} \;=\; 0.5\ \text{s}.$$

The numerical data are

$$I = 5\times10^{-6}\ \text{kg m}^2, \qquad M = 9.85\times10^{-2}\ \text{A m}^2.$$

First we isolate $$B$$ from the period formula. Beginning with

$$T = 2\pi\sqrt{\dfrac{I}{M B}},$$

divide by $$2\pi$$:

$$\dfrac{T}{2\pi} = \sqrt{\dfrac{I}{M B}}.$$

Now square both sides:

$$\left(\dfrac{T}{2\pi}\right)^2 = \dfrac{I}{M B}.$$

Cross-multiplying gives an explicit expression for $$B$$:

$$B = \dfrac{I}{M}\;\dfrac{4\pi^{2}}{T^{2}}.$$

The question supplies the approximation $$\pi^{2}=9.85$$, so

$$4\pi^{2} = 4 \times 9.85 = 39.4.$$

Calculate the numerator:

$$I \;(4\pi^{2}) = \bigl(5\times10^{-6}\bigr)\times39.4 = 197\times10^{-6} = 1.97\times10^{-4}.$$

Calculate the denominator $$M\,T^{2}$$ step by step:

$$T^{2} = (0.5)^{2} = 0.25,$$

$$M\,T^{2} = \bigl(9.85\times10^{-2}\bigr)\times0.25 = 2.4625\times10^{-2} = 0.024625.$$

Substituting into the expression for $$B$$:

$$B = \dfrac{1.97\times10^{-4}}{0.024625} = \dfrac{1.97}{2.4625}\times10^{-2}\ \text{T}.$$

The ratio $$\dfrac{1.97}{2.4625}$$ is exactly 0.8, so

$$B = 0.8\times10^{-2}\ \text{T} = 0.008\ \text{T}.$$

Because $$1\ \text{T}=10^{3}\ \text{mT}$$, converting to millitesla gives

$$B = 0.008 \times 10^{3}\ \text{mT} = 8\ \text{mT}.$$

So, the answer is $$8\text{ mT}$$.