Question

A magnetic needle has a frequency of 60 oscillations per minute in the earth horizontal field. When a field B supports the earths horizontal field its frequency increases by 30 oscillations per minute. Then the ratio of the field B to that of earth is :

• A. 4:7
• B. 7:4
• C. 5:4
• D. 4:5

Answer 1

Answer: (C)

5:4

$T=2\pi \sqrt{\frac{I}{MB}}$

So the frequency of oscillation is directly proportional to the square root of magnetic field

$\frac{f_2^2}{f_1^2}=\frac{B_2}{B_1}$

$B_2=B_h+B$

$B_1=B_h$

$\frac{f_2^2}{f_1^2}=\frac{B_h+B}{B_h}$

$\frac{8100}{3600}=\frac{B_h+B}{B_h}$

$\frac{B}{B_h}=5:4$