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In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}$ kg m$^2$. If the magnitude of magnetic moment of the needle is $x \times 10^{-5}$ Am$^2$, then the value of 'x' is;
A
$50\pi^2$
B
$1280\pi^2$
C
$5\pi^2$
D
$128\pi^2$
Detailed Solution
$T = \frac{5}{20}$ s, $I = 9.8 \times 10^{-6}\ kg\,m^2$, $m = x \times 10^{-5}\ Am^2$, B = 0.049 T
$T = 2\pi\sqrt{\frac{I}{mB}}$
$\frac{5}{20} = 2\pi\sqrt{\frac{9.8 \times 10^{-6}}{x \times 10^{-5} \times 0.049}} = 2\pi\sqrt{\frac{20}{x}}$
$\frac{25}{400} = 4\pi^2\left(\frac{20}{x}\right)$
$x = 16 \times 4\pi^2 \times 20 = 1280\pi^2$