Question

A magnetic needle has magnetic moment $5.8 \times 10^{-2}\, A\, m^2$ and moment of inertia of $7.8 \times 10^{-6}\, kg\, m^2$, it performs $12$ complete oscillations in $6.0 \,s$. What is the magnitude of magnetic field?

• A. $0.011\, T$
• B. $0.021 \,T$
• C. $0.031 \,T$
• D. $0.041 \,T$

Answer 1

Answer: (B)

Given that,
Magnetic moment of a magnetic needle, $m =5.8 \times 10^{-2} m ^2$
Moment of inertia of magnetic needle, $I =7.8 \times 10^{-6} kgm ^2$
Time period of oscillations, $T =\frac{6}{12}=0.5 s ^{-1}$
From $T=2 \pi \sqrt{\frac{I}{ mB }}$ we get,
Magnetic field, $B =\frac{4 \pi^2 I }{ mT ^2}=\frac{4 \times(3.14)^2 \times 7.8 \times 10^{-6}}{5.8 \times 10^{-2} \times(0.5)^2}=0.021 T$