Two identical magnets with a length `10cm` and weights `50gf` each are arragned freely with their like poles facing in a vertical glass tube. The upper magnet hangs in air above the lower one so that the distance between the nearest poles of the magnets is `3mm`. Determine the pole strength of the poles of these magnets.

Answer

`6*64Am`

Explanation

Let `m_1=m_2=m`, `r=3mm=3xx10^-3m`
For hanging in air in balanced position,
`F=50g f=50xx10^-3xx9*8N`
As `F=(mu_0)/(4pi)(m_1m_2)/(r^2)`
`:. 50xx10^-3xx9*8=10^-7xx(m.m)/((3xx10^-3)^2)`
`m^2=9xx5xx9*8xx10^-1`, `m=6*64Am`.