A compass needle is attached to an axle that permits it to turn freely in a hori

Question

A compass needle is attached to an axle that permits it to turn freely in a horizontal plane so that only the horizontal component of the magnetic field affects its motion. The magnetic moment of the needle is 9.0x10^-3 A.m^2, its moment of inertia is 2.0x10^-8 kg.m^2, and the horizontal component of the Earth's magnetic field is 1.9x10^-5 T.

(a) What is the torque on the needle as a function of the angle between the needle and the north direction?

(b) What is the frequency of small oscillations of the needle about the north direction? [Hint: Compare the equation of the rotational motion of the needle with the motion of a pendulum.]

Explanation / Answer

The magnetic moment of the needle, M = 9 * 10^-3 Am^2 Moment of inertia, I = 2 * 10^-8 kg m^2 Horizontal component of earth's magnetic field, BH = 1.9 * 10^-5 T a) The torque is given by the formula,                 = MBHsin                = (17.1 * 10-8)sin   b) The time period of the needle is given by the formula             t = 2 * sqrt(I/MBH)             t = 2.15 s The frequency, f = 1/t = 0.46 Hz             t = 2.15 s The frequency, f = 1/t = 0.46 Hz

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