A thin uniform bar magnet of mass 30 g and length 10 cm is...

The time period of oscillation of a bar magnet in a uniform magnetic field is given by:

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

where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B$ is the magnetic field.

The moment of inertia of a thin uniform bar magnet about an axis passing through its centre and perpendicular to its length is:

$I = \dfrac{mL^2}{12}$

Given:
$m = 30 \text{ g} = 30 \times 10^{-3} \text{ kg}$
$L = 10 \text{ cm} = 0.1 \text{ m}$

$I = \dfrac{30 \times 10^{-3} \times (0.1)^2}{12} = \dfrac{30 \times 10^{-5}}{12} = 2.5 \times 10^{-5} \text{ kg m}^2$

We are also given:
$T = \dfrac{\pi}{2} \text{ s}$
$B = 0.2 \text{ T}$

Substituting these values into the time period formula:

$\dfrac{\pi}{2} = 2\pi \sqrt{\dfrac{2.5 \times 10^{-5}}{M \times 0.2}}$

$\dfrac{1}{4} = \sqrt{\dfrac{2.5 \times 10^{-5}}{0.2 M}}$

Squaring both sides:

$\dfrac{1}{16} = \dfrac{2.5 \times 10^{-5}}{0.2 M}$

$0.2 M = 16 \times 2.5 \times 10^{-5}$

$0.2 M = 40 \times 10^{-5}$

$M = \dfrac{40 \times 10^{-5}}{0.2} = 200 \times 10^{-5} = 2.0 \times 10^{-3} \text{ A m}^2$

Answer: $2.0 \times 10^{-3} \text{ A m}^2$