Question

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.

Answer 1

The time taken for 40 oscillations by this combination is 1.414 minutes.

We know,

T = 2П*[√(I/MB)]

where,

I is the moment of inertia,

M is the magnetic moment of the magnet

B is the horizontal component of the magnetic field

T is the time period of oscillation.

Initially, there are 40 oscillations per minute.

So, T = 1/frequency = 1/40 min⁻¹

Let the final time period be T'

After demagnetisation, the parameters M, B remains constant, only I changes.

I', which is the final moment becomes twice the initial.

So, I' = 2I

Thus,

T/T' = √(I/I')

⇒ 1/40T' = √(I/2I)                 [As T = 1/40 min⁻¹]

⇒ 1/40T' = √(1/2)

⇒ 1/1600T'² = 1/2                 [Squaring both sides]

⇒ 1600T'² = 2

⇒ T'² = 2/1600 = 1/800

⇒ T' =√(1/800) = 0.0356 minute

So, it takes 0.0356 minutes for one oscillation after demagnetisation.

For 40 oscillations, it would take (40*0.0356) minutes = 1.414 minutes = √2 minutes.

Answer 2

The time taken for 40 oscillations by this combination is  $\sqrt{2} \mathrm{min}$

Explanation:

Step 1:

Here in this question it is given that  

The bar magnet in the oscillation magnetometer and its frequency of oscillation is given by

V= 40  

The time in which the bar magnet is located in the oscillation magnetometer  

$\mathrm{T} 1=\frac{1}{40} \mathrm{min}$

Step 2:

The formula for the time period of oscillation of the bar magnet (T) is said to be  

$\mathrm{T}=\sqrt[2 \pi]{\frac{I}{M B H}}$  

In this

I = Moment of Inertia

M= Magnetic moment of the bar magnet

$B_H$ = Horizontal component of the magnetic field

Step 3:

If the time period for which the second demagnetized magnet is placed above the magnet then we need to find out the value of $T_2$

If demagnetized is done for second magnet, on merging will have the same values of M and $B_H$ as those for single magnet. However, variation will be there in the value of ion placing the second demagnetized magnet

Therefore   $\frac{T 1}{T 2}=\frac{I 1}{I 2}$

 Step 4:

On squaring both the sides

 $\frac{1}{1600 T 2^{2}}=\frac{1}{2}$

 $T 2^{2}=\frac{1}{800}$

$T_2$  = 0.03536

Here for one oscillation  

Time taken = 0.03536

Time taken for 40 oscillation = $40 \times 0.03536$

                                                = 1.4144   min

                                                = $\sqrt{2}$min