NEET 2024 — Physics solution

NEET2024Physics

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 $9.8 \times 1 0^{- 6}$ kgm $^{2}$ . If the magnitude of magnetic moment of the needle is $x \times 1 0^{- 5}$ Am $^{2}$ ; then the value of ' $x$ ' is:

A
$50 π^{2}$
B
$1280 π^{2}$
C
$5 π^{2}$
D
$128 π^{2}$

Solution

Period of a freely suspended magnet: $T = 2 π \frac{I}{m B}$ .

Here $T = 5/20 = 0.25$ s.

Squaring: $T^{2} = \frac{4 π ^{2} I}{m B} \Rightarrow m = \frac{4 π ^{2} I}{T ^{2} B}$ .

$m = \frac{4 π ^{2} \times 9.8 \times 1 0 ^{- 6}}{( 0.25 ) ^{2} \times 0.049} = \frac{4 π ^{2} \times 9.8 \times 1 0 ^{- 6}}{0.0625 \times 0.049}$ .

Numerically $m = 1280 π^{2} \times 1 0^{- 5}$ Am $^{2}$ , so $x = 1280 π^{2}$ .

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