Question

A circular coil of 16 turns and radius 0.1 m carries a current of 0.75 A . Its plane is normal to the external magnetic field of strength 0.05T . When the coil is turned slightly, it oscillates with a frequency of 2 Hz. Calculate the moment of inertia of the coil about the axis of rotation

• A. $0.194 \times 10^{-4}\,kgm^2$
• B. $2.194 \times 10^{-4}\,kgm^2$
• C. $0.1 \times 10^{-4}\,kgm^2$
• D. $1.194 \times 10^{-4}\,kgm^2$

Answer 1

Answer: (D)

$1.194 \times 10^{-4}\,kgm^2$

Answer 2

$T=2 \pi \sqrt{I/ mB}$ or $I =\frac{mB}{4 \pi^2\,v^2}$ $=\frac{(NiA)B}{4 \pi^2\,v^2}=\frac{Ni \times \pi r^2\,B}{4 \pi^2\,v^2}\,\,\,(as \,A=\pi \,r^2)$ (Here, i stands for current adn I for nioment ofinertia) or $I=\frac{16 \times 0.75 \times (0.1)^2 \times 0.05}{4 \times 3.14 \times (2)^2}kgm^2 =1.194 \times 10^{-4} kgm^2$