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 × \(10^{–6}\) kg \(\text{m}^2\) . If the magnitude of magnetic moment of the needle is x × \(10^{–5}\) \(Am^{2}\) ; then the value of 'x' is : 

(1) \( 5\ \pi^2\) 

(2) \( 128\ \pi^2\) 

(3) \( 50\ \pi^2\) 

(4) \(1280\ \pi^2\) 

Answer 1

Correct option is : (4) \( 1280\ \pi^2\) 

\(B = 0.049\ T,\ f = \frac{20}{5} = 4Hz\) 

\(I = 9.8\times10^{-6}\ kg - m^2\) 

\(M = x \times10^{-5}\ A - m^2\)

\(f = \frac{1}{2\pi}\sqrt{\frac{MB}{I}}\) 

\(M  = \frac{f^2I(4\pi^2)}{B} = \frac{16\times4\pi^2\times98\times10^{-7}}{49\times10^ {-3}}\)

\(x\times10^{-5} = 128\pi^2 \times10^{-4}\)

\(x = 1280\ \pi^2\)