Question

A magnetic needle of magnetic moment $6.7\times {10}^{-2}{\text{Am}}^2$ and moment of inertia $7.5\times {10}^{-6}{\text{kg m}}^2$ is performing simple harmonic oscillations in a magnetic field of $0.01\text{T}$. Time taken for $10$ complete oscillations is - (in second)

• A. 6.65
• B. 6.647
• C. 6.64

Answer 1

Answer: (A), (B), (C)

Given:
$M=6.7\times {10}^{-2}{\text{Am}}^2$
$I=7.5\times {10}^{-6}{\text{kg m}}^2$
$B=0.01\text{T}$
Time period of oscillation of needle is given by
$T=2\pi \sqrt{\frac{I}{MB}}$
$T=2\pi \sqrt{\frac{7.5\times {10}^{-6}}{6.7\times {10}^{-2}\times 0.01}}$
$T=2\pi \sqrt{\frac{75}{67\times 100}}$
$T=\frac{2\pi }{10}\times \sqrt{\frac{75}{67}}\times 10$
Time taken for for $10$ oscillation is
$t=10T=2\pi \sqrt{\frac{75}{67}}$
$T=6.28\times \sqrt{\frac{75}{67}}=6.65s$