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,
$f_{1}=\frac{1}{2 \pi }\sqrt{\frac{M B_{1} c o s 45 ^\circ }{I}}$
$f_{2}=\frac{1}{2 \pi }\sqrt{\frac{M B_{2} cos 30 ^\circ }{I}}$
$\frac{f_{1}}{f_{2}}=\sqrt{\frac{B_{1} c o s 45 ^\circ }{B_{2} c o s 30 ^\circ }}$
$\Rightarrow \frac{30^{2}}{40^{2}}=\frac{B_{1} c o s 45 ^\circ }{B_{2} c o s 30 ^\circ } \, $
$\Rightarrow \frac{B_{1}}{B_{2}}=\frac{900 \times c o s 30 ^\circ }{1600 \times c o s 45 ^\circ }$
$\Rightarrow \frac{B_{1}}{B_{2}}=0.688\approx0.7$