The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, 1.0 × 10⁻² A m² and 10⁻⁶π² kg m². If it completes 10 oscillations in 10 s, the magnitude of the magnetic field is (a) 0.4 T (b) 4 T (c) 0.4 mT (d) 4 mT

Answer: (c). 10 oscillations in 10 s: T = 1 s. T = 2π√(I/MB) ⇒ B = (4π² I)/(MT²) B = 4π² × 10⁻⁶/π²10⁻² × 1 = 4 × 10⁻⁴ T = 0.4 mT So the answer is (c) 0.4 mT.

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Vibration Magnetometer

The magnetic moment and moment of inertia of a magnetic needle as shown are, respectively, $1.0 \times 10^{-2}\text{ A m}^{2}$ and $\frac{10^{-6}}{\pi^{2}}\text{ kg m}^{2}$ . If it completes 10 oscillations in $10\text{ s}$ , the magnitude of the magnetic field is

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Answer: (c)

10 oscillations in 10 s: $T = 1$ s.
$T = 2\pi\sqrt{\frac{I}{MB}} \Rightarrow B = \frac{4\pi^2 I}{MT^2}$
$B = \frac{4\pi^2 \times 10^{-6}/\pi^2}{10^{-2} \times 1} = 4 \times 10^{-4}\ \text{T} = 0.4\ \text{mT}$
So the answer is (c) 0.4 mT.

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