JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Magnetic Properties of Matter

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip

3 0^{\circ}

. The number of oscillations per minute become 10 at another place Q of

6 0^{\circ}

dip. The ratio of the total magnetic field at the two places

(B_{Q} : B_{P})

is:

Answer: (A)

Solution

Time period

T = 2 π \frac{I}{B _{H} M}

, where

B_{H} = B cos δ

is the horizontal component. At P:

T_{P} = 3

B_{H}^{P} = B_{P} cos 3 0^{\circ} = \frac{3}{2} B_{P}

. At Q:

T_{Q} = 6

B_{H}^{Q} = B_{Q} cos 6 0^{\circ} = \frac{1}{2} B_{Q}

. Taking ratio:

\frac{T _{P}}{T _{Q}} = \frac{B _{H}^{Q}}{B _{H}^{P}}

, so

\frac{1}{2} = \frac{B _{Q} /2}{3 B _{P} /2}

. Squaring:

\frac{1}{4} = \frac{B _{Q}}{3 B _{P}}

, giving

\frac{B _{Q}}{B _{P}} = \frac{3}{4}

, or

B_{Q} : B_{P} = 3 : 4

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option