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

Question

Navjyot Kalra · Answer

Sol. T = 2π√I/MB base H Here I’ = 2I T base 1 = 1/40 min T base 2 = ? T base 1/T base 2 = √I/I’ ⇒ 1/40T base 2 = √1/2 ⇒ 1/1600T base 2^2 = 1/2 ⇒ T base 2^2 = 1/800 ⇒ T base 2 = 0.03536 min For 1 oscillation Time taken = 0.03536 min. For 40 Oscillation Time = 4 * 0.03536 = 1.414 = √2 min

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

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

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