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 x 10^–6 kg m². If the magnitude of magnetic moment of the needle is x × 10^–5 Am² , then the value of ‘x’ is :

The correct option is 1280π².

1. Analyzing the Image and Given Parameters:
From the provided image, we observe a magnetic needle suspended vertically by a thread. The needle is placed in a uniform horizontal magnetic field, represented by the dashed lines pointing to the right and labeled $B$. The poles of the needle are marked as $N$ (North) and $S$ (South). When perturbed, the needle undergoes angular simple harmonic motion (oscillations) in this magnetic field.

The parameters given in the problem are:
- Magnetic field strength, $B=0.049\text{T}$
- Number of oscillations, $N=20$
- Total time, $t=5\text{s}$
- Moment of inertia of the needle, $I=9.8\times {10}^{-6}{\text{kg m}}^2$
- Magnetic moment of the needle, $M=x\times {10}^{-5}{\text{A m}}^2$

2. Determining the Time Period of Oscillation ($T$):
The time period $T$ is the time taken to complete one full oscillation:
$T=\frac{t}{N}=\frac{5}{20}=0.25\text{s}$

3. Formula for the Oscillation of a Magnetic Needle:
The time period of oscillation for a magnetic dipole suspended in a uniform magnetic field is given by:
$T=2\pi \sqrt{\frac{I}{MB}}$

Squaring both sides of the equation to isolate $M$:
$T^2=4{\pi }^2\left(\frac{I}{MB}\right)$

Rearranging the equation to solve for the magnetic moment $M$:
$M=\frac{4{\pi }^{2}I}{T^{2}B}$

4. Step-by-Step Calculation:
Substitute the given values into the rearranged equation:
$x\times {10}^{-5}=\frac{4{\pi }^2\times \left(9.8\times {10}^{-6}\right)}{{\left(0.25\right)}^2\times 0.049}$

Since ${\left(0.25\right)}^2=0.0625=\frac{1}{16}$, we can rewrite the denominator:
$x\times {10}^{-5}=\frac{4{\pi }^2\times 9.8\times {10}^{-6}\times 16}{0.049}$

Simplifying the division of the decimal values:
$\frac{9.8}{0.049}=200$

Substitute this simplified value back into the expression:
$x\times {10}^{-5}=4{\pi }^2\times 200\times {10}^{-6}\times 16$

Multiplying the constant terms:
$x\times {10}^{-5}=12800{\pi }^2\times {10}^{-6}$

Expressing the right side in terms of ${10}^{-5}$:
$x\times {10}^{-5}=1280{\pi }^2\times {10}^{-5}$

Comparing both sides, we get:
$x=1280{\pi }^2$