A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30 ° . The number of oscillations per minute become 10 at another place Q of 60 ° dip. The ratio of the total magnetic field at the two places B Q : B P is:

Step-by-step solution

At point P the needle oscillates 20 times per minute. Therefore, time period T P = 60 20 =   3   s and at point Q , T Q = 60 10 = 6   s . The time period of oscillation magnetometer is given by, T = 2 π I B H M ⇒ T P = 3 = 2 π I B P cos 30 ° M And T Q = 6   s = 2 π I B Q cos 60 ° M Taking ratio, 3 6 = 1   B P 3 2 × B Q 2 ⇒ 3 4 = B Q B P ⇒ B Q : B P = 3 : 4