A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is `25 muT`. Another short magnet of magnetic moment `1.6 A m^2` is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north pole (a) towards north and (b) towards south.

Explanation

Here `T_(1) = 40` oscillations//minute
`B_(H) = 25mu T`
M of second magnet
`= 1.6A- m^(2)`
`d = 20cm = 0.2m`
(a)For north pole facing
`gamma_(1) = (1)/(2 pi) sqrt((MB_(H))/(1))`
`gamma_(2) = (1)/(2 pi) sqrt((M(B_(H) - B))/(1))`
`B = (mu_(0))/(4 pi) (m)/(d^(3))`
`= ((10^(-7) xx 1.6)/(8 xx 10^(-3)) = 20 mu T`
`(ga`
`= mma_(1))/(gamma_(2))= sqrt((B_(H))/(B_(H) - B))`
`rArr (40)/(gamma_(2)) = sqrt((25)/(5))`
rArr (40)/(gamma_(2)) = sqrt(5)`
`rArr gamma_(2)= (40)/(sqrt(5)) = 17.88`
`= 18 osci//min `
(b) For north pole facing south
`gamma_(1) = (1)/(2 pi) sqrt((MB_(H))/(1))`
`gamma_(2) = (1)/(2 pi) sqrt((M(B + B_(H))/(1))`
`(gamma_(1))/(gamma_(2)) = sqrt((B_(H))/(B +B_(H)))`
`rArr (40)/(gamma_(2)) = sqrt((25)/(45))`
`gamma_(2) = (40)/(sqrt(25 //45))) = 54 osci//min`