A compass needle oscillates 20 times per minute at a place where the dip is `45^@` and 30 times per minute where the dip is `30^@`. Compare the total magnetic field due to the earth at the two places.

Explanation

The time period of oscillation is given by
`T = 2pi sqrt((I)/(MB_H))`
The time period at the first place is `T_1 = 1/20 minute = 3.0 s` and at the second place it is `T_2 = 1/30` minute `= 2.0 s`.
If the total magnetic field at the first place is `B_1`, the horizontal component of the field is
`B_H1 = B_2 cos 30^@ = B_2 sqrt(3/2)`
We have,
`T_1 = 2pi sqrt(I/(MB_H1))` and `T_2 = 2pi sqrt(I)/(MB_H2))`
Thus, `T_1/T_2 = sqrt(B_(H2)/(B_(H1)` or, `B_(H2)/B_(H1) = T^2_1/T^2_2)`
or,
(B_2 sqrt(3/2))/(B_1)/(sqrt(2) = T^2_1/T^2_2`
or `B_2/B_1` = sqrt(2/3) T^2_1/T^2_2) = sqrt(2/3) xx 9/4 = 1.83`.
Once we know how to measure `MB_H`, we can easily compare magnetic moments `M_1 and M_2` by measuring `M_1 B_H` and `M_2 B_H`. Similarly, we can compare the horizontal components of the earth's magnetic field at two places.