Question

A bar magnet has a magnetic moment equal to $5\times 10^{-5}$ $Wbm$ . It is suspended in a magnetic field which has a magnetic induction $B=8\pi \times 10^{-4}\text{T}$ . The magnet vibrates with a period of vibration equal to $\text{15 s}$ . The moment of inertia of magnet is

• A. $4.54\times 10^{4}kg{m}^2$
• B. $4.54\times 10^{-5}\text{kg }{\text{m}}^2$
• C. $4.54\times 10^{-4}\text{kg }{\text{m}}^2$
• D. $4.54\times 10^5\text{kg}{\text{m}}^2$

Answer 1

Answer: (D)

Here, magnetic moment is given in weber-meter, which is the unit of ${\mu }_{0}M$ .
$\therefore {\mu }_{0}M=5\times 10^{-5}$ Wb-m
$M=\frac{5\times 10^{-5}}{{\mu }_0}A{m}^2$
Also, $B=8\pi \times 10^{-4}={\mu }_{0}H$
$\therefore H=\frac{8\pi \times 10^{-4}}{{\mu }_0}$
$T=2\pi \sqrt{\frac{I}{MH}}$
$I=\frac{MH{T}^2}{4{\pi }^2}=\frac{5\times 10^{-5}\times 8\pi \times 10^{-4}\times 15^2}{4{\pi }^2{\mu }_0^2}$
$=\frac{5\times 10^{-5}\times 8\pi \times 10^{-4}\times 225}{4{\pi }^2{\left(4\pi \times 10^{-7}\right)}^2}$
$I=\frac{2250\times 10^{-9}}{\pi \left(16{\pi }^2\right)\times 10^{-14}}=4.54\times 10^5\mathrm{kg}{\mathrm{m}}^2$