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 $\mathrm{B}_{1}$ and $\mathrm{B}_{2}$ 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 :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Magnetic Properties of Matter

A magnetic compass needle oscillates 30 times per minute at a place where the dip is

4 5^{\circ}

, and 40 times per minute where the dip is

3 0^{\circ}

. 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 :

Answer: (D)

A compass needle oscillates with frequency

f = \frac{1}{2 π} \frac{μ B _{H}}{I}

, where the restoring field is the horizontal component

B_{H} = B cos δ

. Taking the ratio for the two places,

\frac{f _{1}}{f _{2}} = \frac{B _{1} cos 4 5 ^{\circ}}{B _{2} cos 3 0 ^{\circ}}

, and squaring with

f_{1} : f_{2} = 30 : 40

gives

\frac{9}{16} = \frac{B _{1}}{B _{2}} \cdot \frac{cos 4 5 ^{\circ}}{cos 3 0 ^{\circ}}

. Hence

\frac{B _{1}}{B _{2}} = \frac{9}{16} \cdot \frac{cos 3 0 ^{\circ}}{cos 4 5 ^{\circ}} = \frac{9}{16} \cdot \frac{3 /2}{1/ 2} \approx 0.7

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option