The frequency of oscillation of the magnet in an oscillation magnetometer in the earth's magnetic field is 40 oscillations per minute. A short bar magnet is placed to the north of the magnetometer, at a separation of 20cm from the oscillating magnet, with its north pole pointing towards north (figure). The frequency of oscillation is found to increase to 60 oscillations per minute. Calculate the magnetic moment of this short bar magnet. Horizonatl component of the earth's magnetic field is `24 muT`.

Explanation

Let the magnetic field due to the short magnet have magnitude `B` at the site of the oscillating magnet. From the figure, this magnetic field will be towards north and hence the resultant horizontal field will be `B_H + H`. Let `M and M'` denote the magnetic moments of the oscillating magnet and the other magnet respectively. If `v and v'` be the frequencies wihtout and with the other magnet, we have
`v = 1/(2pi) sqrt(MB_H)/I` and
`v' = 1/(2pi) sqrt(M(B_H + B)/(I)`
or, `v'^2/(v^2) = (B_H + B)/B_H)`
or, `(60/40)^2 = 1 + B/B_H`
or, `B/B_H = 1.25`
or,
B= 1.25 xx 24 muT = 30 xx 10^(-6) T`.
The oscillating magnet is in end-on position of the short magnet. Thus, the field `B` can be written as
`B = (mu_0)/(4pi) (2M')/d^3`
or, `M' = (2pi)/(mu_0) Bd^3`
`0.5 xx 10^7 A m^(-1) xx (30 xx 10^(-6) T) xx (20 xx 10^(-2) m)^3`
`1.2 A m^2`.