Question

The magnetic moment of two bar magnets of same size are in
ratio 1:2..When they are placed one over the other with their similar poles together,then their period of oscillation in a magnetic field is 3s. If one of the magnets is reversed , then the period of oscillation in the same field will be.

Answer 1

Given that,

The magnetic moment of two bar magnets of same size are in  ratio 1:2.

Time period = 3 s

We know that,

The time period of oscillation is defined as,

$T=2\pi\sqrt{\dfrac{I}{MB}}$

Where, I = moment of inertia

B = magnetic field

When they are placed one over the other with their poles together, then magnetic moment is

$M=x+2x$

$M=3x$

Where, x = magnetic moment

The time period of oscillation is

$T_{1}=2\pi\sqrt{\dfrac{I}{M_{1}B}}$

Put the value into the formula

$3=2\pi\sqrt{\dfrac{I}{3xB}}$.....(I)

When polarity is reversed than the net magnetic moment is

$M_{2}=x-2x=x$

The time period of oscillation is

$T_{2}=2\pi\sqrt{\dfrac{I}{M_{2}B}}$

Put the value into the formula

$T_{2}=2\pi\sqrt{\dfrac{I}{xB}}$.....(II)

We need to calculate the period of oscillation in the same field

Using equation (I) and (II)

$\dfrac{3}{T_{2}}=\dfrac{2\pi\sqrt{\dfrac{I}{3xB}}}{2\pi\sqrt{\dfrac{I}{xB}}}$

$\dfrac{9}{T_{2}^2}=\dfrac{4\pi^2x}{4\pi^2\times3x}$

$\dfrac{T_{2}}{9}=\dfrac{3x}{x}$

$T_{2}^2=27$

$T_{2}=\sqrt{27}$

$T_{2}=3\sqrt{3}\ sec$

Hence, The period of oscillation in the same field is $3\sqrt{3}\ sec$

Answer 2

Answer:

the period of oscillation in the same feild will be=3√3 sec.