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 × 10⁻⁶ kg m². If the magnitude of magnetic moment of the needle is x × 10⁻⁵ Am², then the value of ' x ' is: (a) 5 π² (b) 128 π² (c) 50 π² (d) 1280 π²

Answer: (d). 20 oscillations in 5 s: T = 0.25 s. M = (4π² I)/(BT²) = 4π² × 9.8 × 10⁻⁶0.049 × 0.0625 M = 4π² × 3.2 × 10⁻³ = 1280π² × 10⁻⁵ A m² So x = 1280π², answer (d).

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Vibration Magnetometer

In a uniform magnetic field of $0.049\text{ T}$ , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is $9.8 \times 10^{-6}\text{ kg m}^{2}$ . If the magnitude of magnetic moment of the needle is $x \times 10^{-5}\text{ Am}^{2}$ , then the value of ' $x$ ' is:

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Answer: (d)

20 oscillations in 5 s: $T = 0.25$ s.
$M = \frac{4\pi^2 I}{BT^2} = \frac{4\pi^2 \times 9.8 \times 10^{-6}}{0.049 \times 0.0625}$
$M = 4\pi^2 \times 3.2 \times 10^{-3} = 1280\pi^2 \times 10^{-5}\ \text{A m}^2$
So $x = 1280\pi^2$ , answer (d).

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