In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 seconds . Th

Question

In a uniform magnetic field of 0.049 T , a magnetic needle performs 20 complete oscillations in 5 seconds . The moment of inertia of the needle is 9.8 × 10^-6 kg m^2 . If the magnitude of magnetic moment of the needle is x × 10^-5 A m^2 , then the value of ' x ' is: ![image](https://pub-7faa453bd1ec

Accepted Answer

The time period of oscillation of a magnetic needle in a magnetic field is:

T = 2 π \frac{I}{MB}

Given:Number of oscillations

n = 20

, Time

t = 5 s

\Rightarrow T = \frac{t}{n} = \frac{5}{20} = \frac{1}{4} s

B = 0.049 T

I = 9.8 \times 1 0^{- 6} kg m^{2}

Squaring the time period formula:

T^{2} = \frac{4 π ^{2} I}{MB} \Rightarrow M = \frac{4 π ^{2} I}{T ^{2} B}

Substituting the values:

M = \frac{4 π ^{2} \times 9.8 \times 1 0 ^{- 6}}{( \frac{1}{4} ) ^{2} \times 0.049} = \frac{4 π ^{2} \times 9.8 \times 1 0 ^{- 6} \times 16}{0.049}

M = \frac{4 π ^{2} \times 16 \times 9.8 \times 1 0 ^{- 6}}{49 \times 1 0 ^{- 3}} = \frac{64 π ^{2} \times 0.2 \times 1 0 ^{- 3}}{1}

M = 12.8 π^{2} \times 1 0^{- 3} = 1280 π^{2} \times 1 0^{- 5} A m^{2}

Comparing with

x \times 1 0^{- 5}

, we get

x = 1280 π^{2}

.

In a uniform magnetic field of $0.049 T$ , a magnetic needle performs $20$ complete oscillations in $5 seconds$ . The moment of inertia of the needle is $9.8 \times 1 0^{- 6} kg m^{2}$ . If the magnitude of magnetic moment of the needle is $x \times 1 0^{- 5} A m^{2}$ , then the value of ' $x$ ' is:

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In a uniform magnetic field of 0.049T, a magnetic needle performs 20 complete oscillations in 5seconds. The moment of inertia of the needle is 9.8×10−6kg m2. If the magnitude of magnetic moment of the needle is x×10−5A m2, then the value of 'x ' is: