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प्रश्न
A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by this combination. Neglect any induced magnetism.
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उत्तर
Given :
Frequency of oscillations of the bar magnet in the oscillation magnetometer, `V = 40 "min"^-1`
Time period for which the bar magnet is placed in the oscillation magnetometer, T1 = `1/40 min`
The time period of oscillations of the bar magnet (T) is given by `T = 2pi sqrt(I/(MB_H)`
Here,
I = Moment of inertia
M = Magnetic moment of the bar magnet
BH = Horizontal component of the magnetic field
Now, let T2 be the time period for which the second demagnetised magnet is placed over the magnet.
As the second magnet is demagnetised, the combination will have the same values of Mand BH as those for the single magnet. However, variation will be there in the value of Ion placing the second demagnetised magnet.
`therefore` `T_1/T_2 = I_1/I_2`
⇒ `1/(40T_2) = sqrt(1/2)`
⇒ `1/(1600 T_2^2) = 1/2`
⇒ `T_2^2 = 1/800`
⇒ `T_2 = 0.03536 min`
For 1 oscillation,
Time taken = 0.03536 min
For 40 oscillations,
Time taken = 0.03536 × 40
= `1.414 min = sqrt(2) min`
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