Question

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45^{\circ }$ and $40$ times per minute where the dip is $30^{\circ }$ . If $B_1$ and $B_2$ are the net magnetic fields due to the earth at the two places respectively, then the ratio $B_1/B_2$ is approximately equal to

• A. $3.6$
• B. $1.8$
• C. $1.2$
• D. $0.7$

Answer 1

Answer: (D)

Let $f_1$ and $f_2$ be the respective oscillation frequencies respectively for $45^{\circ }$ and $30^{\circ }$ dip. Then,
$f_1=\frac{1}{2\pi }\sqrt{\frac{M{B}_{1}cos45^{\circ }}{I}}$
$f_2=\frac{1}{2\pi }\sqrt{\frac{M{B}_{2}cos30^{\circ }}{I}}$
$\frac{f_1}{f_2}=\sqrt{\frac{B_{1}cos45^{\circ }}{B_{2}cos30^{\circ }}}$
$\Rightarrow \frac{30^2}{40^2}=\frac{B_{1}cos45^{\circ }}{B_{2}cos30^{\circ }}$
$\Rightarrow \frac{B_1}{B_2}=\frac{900\times cos30^{\circ }}{1600\times cos45^{\circ }}$
$\Rightarrow \frac{B_1}{B_2}=0.688\approx 0.7$