(a) Show that the time period (T) of oscillations of a freely suspended magnetic dipole of magnetic moment (m) in a uniform magnetic field (B) is given by T=2\pi \sqrt{\frac{I}{mB}}, where I is a moment of inertia of the magnetic dipole.

(b) Identify the following magnetic materials :

(i) A material having susceptibility(\chi_{m})=-0.00015.

(ii) A material having susceptibility(\chi_{m})=10^{-5}.

 

 
 
 
 
 

Answers (1)
S safeer

(a)  \tau =\vec{M}\times \vec{B}

        \tau =MB\sin \theta

For small ang \theta, we have 

        \tau =MB\theta

\alpha=\frac{d^2\theta}{dt^2}

        \alpha=\left ( \frac{MB}{I} \right )\theta

Now, compairing with angular SHM, we get

        \omega ^{2}=\frac{mB}{I}

T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{I}{mB}}

Where m=magnetic moment

             I= moment of inertia

            B= magnetic field intensity

(b) Such that,

\because x_{m}< 0, diamagnetic

\because x_{m}> 0, paramagnetic

\because x_{m}> > 1, 1ferromagnetic

Where, x_{m} is the magnetic susceptibility

(i) x_{m}=-0.00015

hence, it is diamagnetic

(ii) x_{m}=10^{-5}> 0

hence, it is paramagnetic

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