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Question

Physics Question on Magnetism and matter

A magnetic compass needle oscillates 3030 times per minute at a place where the dip is 4545^{\circ}, and 4040 times per minute where the dip is 3030^{\circ}. If B1B_1 and B2B_2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2B_1/B_2 is best given by :

A

2.2

B

1.8

C

0.7

D

3.6

Answer

0.7

Explanation

Solution

f1=12πμB1cos45If2=12πμB2cos30If_{1} = \frac{1}{2\pi} \sqrt{\frac{\mu B_{1} \cos45^{\circ}}{I}} f_{2} = \frac{1}{2\pi } \sqrt{ \frac{\mu B_{2} \cos30^{\circ}}{I }}
f1f2=B1cos45B2cos30B1B2=0.7\frac{f_{1}}{f_{2}} = \sqrt{\frac{B_{1}\cos 45^{\circ}}{B_{2} \cos30^{\circ}}} \therefore \frac{B_{1}}{B_{2}} = 0.7