A magnetic compass needl oscillates 30 times per minute at a place where the dip is `45^(@)` and 40 times per minute where the dip is `30^(@)` if `B_(1)` and `B_(2)` are respectively the total magnetic field due to the earth at the two places then the ratio `B_(1)//B_(2)` is best given by

Options

A. 3.6
B. 1.8
C. 2.2
D. 0.7

Answer

D

Explanation

Frequency of oscillation for the magnet can be written as: `f = (1)/(2pi) sqrt((MB)/(I))`
Compass needle is placed horizontally hence only horizontal component (B cos `theta` ) of earth.s magnetic field is used for oscillation, where is angle of dip.
` f_1 = (1)/(2pi) sqrt( (MB_1 cos 45^@ )/(I) )`
` f_2 = (1)/(2pi) sqrt ( (MB_2 cos 30^@)/(I) )`
` rArr f_1/f_2 = sqrt( (B_1 cos 45^@ )/(B_2 cos 30^@) ) rArr 30/40 = sqrt( (B_1 xx 2)/(sqrt(2 xx B_2 xx sqrt3) )`
` rArr 9/16 = (B_1 xx sqrt2)/(B_2 xx sqrt3) rArr B_1/B_2 = (9 xx sqrt3)/(16 xx sqrt2) = 0.7`