In an oscillation magnetometer, a short magnet completes 40 oscillations per minute at a location where the earth's horizontal magnetic field is 25 T . A second short magnet, possessing a magnetic moment of 1.6 A m ^2 , is positioned 20 cm east of the oscillating magnet. Calculate the new frequency of oscillation assuming the second magnet has its north pole oriented towards the north.

Step-by-step solution

The initial frequency of the oscillating magnet is f₁ = 40 oscillations/min in the earth's horizontal magnetic field B_H = 25 T . The second magnet is placed 20 cm east of the oscillating magnet, with its north pole pointing north. This means its magnetic moment M points north, and the oscillating magnet lies on its equatorial line. The magnetic field produced by the second magnet at the position of the oscillating magnet is given by the equatorial field formula: B_ eq = ₀ 4 M d^3 Substituting the given values ( M