A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10…

Question

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places B_Q:B_P is:

MARKS Content Team · Accepted Answer

Correct answer: (A) √3:4. At point P the needle oscillates 20 times per minute. Therefore, time period T_P=60/20= 3 s and at point Q, T_Q=60/10=6 s. The time period of oscillation magnetometer is given by, T=2π√IB_HM ⇒T_P=3=2π√(IB_Pcos30°M) And T_Q=6 s=2π√(IB_Qcos60°M) Taking ratio, 3/6=√(1 B_P3)2×B_Q2 ⇒√34=B_QB_P ⇒B_Q:B_P=√3:4

A compass needle of oscillation magnetometer oscillates $20$ times per minute at a place $P$ of dip $30 °$ . The number of oscillations per minute become $10$ at another place $Q$ of $60 °$ dip. The ratio of the total magnetic field at the two places $(B_{Q} : B_{P})$ is:

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Relevant Questions

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places BQ:BP is:

Solution

Relevant Questions

A compass needle of oscillation magnetometer oscillates $20$ times per minute at a place $P$ of dip $30 °$ . The number of oscillations per minute become $10$ at another place $Q$ of $60 °$ dip. The ratio of the total magnetic field at the two places $(B_{Q} : B_{P})$ is: