The magnetic moment of a magnetic needle of mass 3.2 g is 980 CGS units . From which point should the needle be hung so that it will remain horizontal in the magnetic meridian ? Horizontal component of the earth's magnetic field at that place is 0.32 Oe and angle of dip `=45^(@)N.` [ g = `980"cm".s^(-2)` ]

Explanation

Let the magnetic needle be hung from a point at a distance x from its centre of gravity (towards the north pole ) [Fig ]
In equilibrium ,
`mVxx2l=Wxxx`
`"or, " V.p_(m)=W.x[becausep_(m)=m.2l]`
`"or, " x=(V.p_(m))/(W)=(Htantheta.p_(m))/(W)[becauseV=H tantheta]`
`=(0.32xxtan45^(@)xx980)/(3.2xx980)=0.1"cm"`