A bar magnet of length 5cm, width 3cm and height 2cm takes 5s to complete an oscillation in viberation magnetometer placed in a horizontal magnetic field of `20muT`. The mass of this bar magnet is 250g(a). Find the magnetic moment of the magnet. (b) If the magnet is put in the magnetormer with its 0.5cm edge horizontal, what would be the new time period?

Explanation

(a) Moment of inertia of magnet is given by,
`" " I=(m(l^(2)+b^(2)))/(12)`
where m= mass of magnet
`therefore " " I=(250(5^(2)+3^(2))xx10^(-4)xx10^(-3))/(12)=7.08xx10^(-5)kgm^(2)`
Also, `therefore " " I=(4(3.14)^(2)xx(7.08xx10^(-5)))/(20xx10^(-6)xx5xx5)=5.58Am^(2)`
(b) New moment of inertia is given by `I'=(m_(0)(l^(2)+h^(2)))/(12)`
`therefore " " T'=2pi sqrt((I')/(MH))`
`Rightarrow (T')/(T)=sqrt((I')/(I))= sqrt((l^(2)+h^(2))/(l^(2)+b^(2)))=sqrt(5^(2)+(0.5)^(2))/(5^(2)+3^(2)) Rightarrow (T')/(T)=0.36`
` therefore T'=Txx0.86=5xx0.86=4.30s`