A short magnet in an oscillation magnetometer oscillates with a time period of 0.10 s at a location where the earth's horizontal magnetic field is 24 T . A downward current of 18 A is set up in a vertical wire positioned 20 cm east of the magnet. Determine the new time period.

Step-by-step solution

The magnetic field produced by the vertical wire at the position of the magnet is given by: B_ wire = ₀ I 2 r Substituting the given values: B_ wire = 4 10⁻⁷ 18 2 0.20 = 18 10⁻⁶ T = 18 T By the right-hand grip rule, for a downward current in a wire located east of the magnet, the magnetic field at the magnet's position points North. Since the Earth's horizontal magnetic field B_H also points North, the two magnetic fields add up. The net magnetic field is: B_ net = B_H + B_ wire = 24 T + 18 T = 42 T The time period