Question

A short magnet oscillates in an oscillation magnetometer with a time period of 0.10s where the earth's horizontal magnetic field is 24μT. A downward current of 18A is established in a vertical wire placed 20cm east of the magnet. Find the new time period.

Answer 1

B_H = 24 × 10^–6T, T₁ = 0.1'

= (24 –10) × 10^–6 = 14 × 10^–6

=> T₂² = (0.01 x 14)/24 => T₂ = 0.0766s