In a uniform magnetic field of , a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is . If the magnitude of magnetic moment of the needle is , then the value of is [2024]

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Solution

Q.
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 [2024]

1 $5 π^{2}$

2 $128 π^{2}$

3 $50 π^{2}$

4 $1280 π^{2}$

Ans.
(4)

Time period of the needle, $T = \frac{5}{20} = \frac{1}{4}$ $s$

Given, $I = 9.8 \times 10^{- 6} kg$ $m^{2}, M = x \times 10^{- 5}$ ${Am}^{2}, B = 0.049$ $T$

Since, $T = 2 π \sqrt{\frac{I}{M B}}$

$\Rightarrow \frac{1}{4} = 2 π \sqrt{\frac{9.8 \times 10^{- 6}}{x \times 10^{- 5} \times 0.049}} \Rightarrow \frac{1}{8 π} = \sqrt{\frac{9.8 \times 10^{- 6}}{x \times 10^{- 5} \times 0.049}}$

$\Rightarrow \frac{1}{64 π^{2}} = \frac{9.8 \times 10^{- 6}}{x \times 10^{- 5} \times 0.049}$

$\Rightarrow x = 64 π^{2} \times \frac{9.8 \times 10^{- 1}}{0.049} = 1280 π^{2}$

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