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 × 10^–6 kg m² . If the magnitude of magnetic moment of the needle is x × 10^–5 Am² ; then the value of 'x' is :

The problem gives the following data (as shown in the diagram of the magnetic needle oscillating in a uniform field):

• Magnetic field magnitude, B = 0.049 T
• Moment of inertia of the needle, I = 9.8 × 10^‑6 kg·m²
• The needle completes 20 full oscillations in 5 s.

From the oscillation count we find the frequency:

$f\; =\; \backslash frac\{\backslash text\{number\; of\; oscillations\}\}\{\backslash text\{time\}\}\; =\; \backslash frac\{20\}\{5\}\backslash ;\backslash text\{Hz\}\; =\; 4\backslash ;\backslash text\{Hz\}$

The period is the reciprocal of the frequency:

$T\; =\; \backslash frac\{1\}\{f\}\; =\; \backslash frac\{1\}\{4\}\backslash ;\backslash text\{s\}\; =\; 0.25\backslash ;\backslash text\{s\}$

For small angular displacements a magnetic needle in a uniform field executes simple harmonic motion with angular frequency

$\backslash omega\; =\; \backslash sqrt\{\backslash frac\{\backslash mu\; B\}\{I\}\}$

where μ is the magnitude of the magnetic moment. The period of a simple harmonic oscillator is related to ω by

$T\; =\; \backslash frac\{2\backslash pi\}\{\backslash omega\}\; =\; 2\backslash pi\backslash sqrt\{\backslash frac\{I\}\{\backslash mu\; B\}\}$

Solving this expression for the magnetic moment μ gives

$\backslash mu\; =\; \backslash frac\{4\backslash pi^\{2\}\; I\}\{T^\{2\}\; B\}$

Substituting the known values:

$\backslash mu\; =\; \backslash frac\{4\backslash pi^\{2\}\backslash ,(9.8\backslash times10^\{-6\})\}\{(0.25)^\{2\}\backslash ,(0.049)\}$

Calculate the denominator first:

$(0.25)^\{2\}\backslash ,B\; =\; 0.0625\; \backslash times\; 0.049\; =\; 0.0030625$

Now the numerator:

$4\backslash pi^\{2\}\; I\; =\; 4\backslash pi^\{2\}\backslash ,(9.8\backslash times10^\{-6\})\; =\; 39.2\backslash times10^\{-6\}\backslash ,\backslash pi^\{2\}$

Therefore

$\backslash mu\; =\; \backslash frac\{39.2\backslash times10^\{-6\}\backslash ,\backslash pi^\{2\}\}\{0.0030625\}\; =\; \backslash frac\{39.2\}\{0.0030625\}\backslash times10^\{-6\}\backslash ,\backslash pi^\{2\}\; =\; 12800\backslash times10^\{-6\}\backslash ,\backslash pi^\{2\}\; =\; 1.28\backslash times10^\{-2\}\backslash ,\backslash pi^\{2\}\backslash ;\; \backslash text\{A·m\}^\{2\}$

Express μ in the required form x × 10^‑5 A·m²:

$1.28\backslash times10^\{-2\}\; =\; 1280\backslash times10^\{-5\}$

Hence

$\backslash mu\; =\; (1280\backslash pi^\{2\})\backslash times10^\{-5\}\backslash ;\backslash text\{A·m\}^\{2\}$

So the value of x is

$1280\backslash pi^\{2\}$