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Grade: upto college level
        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.
3 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							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

						
3 years ago
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