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)

$0.021$T

Given that,

Magnetic moment of a magnetic needle, $m=5.8\times {10}^{-2}{m}^2$

Moment of inertia of magnetic needle, $I=7.8\times {10}^{-6}kg{m}^2$

Time period of oscillations, $T=\frac{6}{12}=0.5s^{-1}$

From $T=2\pi \sqrt{\frac{I}{mB}}$ we get,

Magnetic field , $B=\frac{4{\pi }^{2}I}{m{T}^2}=\frac{4\times (3.14{)}^2\times 7.8\times {10}^{-6}}{5.8\times {10}^{-2}\times (0.5{)}^2}=0.021T$