1.

A bar magnet has a magnetic moment 5 xx 10^(-5) A-m^2. It is suspended in a magnetic field of 8xx 10^(-4) tesla. The magnet vibrates with a period of vibration equal to 15 s. The moment of inertia of the magnet is

Answer»

`22.5 KG m^2`
`11.25 kg m^2`
`5.62 kg m^2`
`7.16xx10^(-7) kg m^2`

Solution :`because T = 2PI sqrt((I)/(mB))`, hence moment of INERTIA of needle `I= (T^2 mB)/(4pi^2) = ((15)^2 xx 5 xx 10^(-5)xx 8pi xx 10^(-4))/(4pi^2)= 7.16 xx 10^(-7) Kg-m^2`


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