Question
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 $9.8 \times 10^{- 6}\text{ }kg{\text{ }m}^{2}$. If the magnitude of magnetic moment of the needle is $x \times 10^{- 5}{Am}^{2}$, then the value of ' $x$ ' is :
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Solution
Sol. [Key Concept: Oscillating magnetic needle — period and field strength]
Time period of Oscillation, $T = 2\pi\sqrt{\frac{I}{MB}}$
$$\begin{matrix} & \ \Rightarrow \ \frac{1}{4} = 2\pi\sqrt{\frac{9.8 \times 10^{- 6}}{M \times 0.049}} \\ & \ \Rightarrow \ \frac{1}{16} = 4\pi^{2} \times \frac{9.8 \times 10^{- 6}}{M \times 49 \times 10^{- 3}} \\ & \ \Rightarrow \ M = \frac{4\pi^{2} \times 9.8 \times 10^{- 6}}{49 \times 10^{- 3}} \times 16 \\ & \ = \frac{4\pi^{2} \times 9.8 \times 16 \times 10^{- 3}}{49} \\ & \ = 12.8\pi^{2} \times 10^{- 3} \times 10^{- 2} \times 10^{2} \\ & \ = 1280\pi^{2} \times 10^{- 5}{Am}^{2} \end{matrix}$$
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