Question 12 Marks
$a$. 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 $T=2 \pi \sqrt{\frac{I}{m B}},$ where $I$ is a moment of inertia of the magnetic dipole.
$b$. Identify the following magnetic materials:
$i$. A material having susceptibility $\left(\chi_m\right)=-0.00015$.
$ii.$ A material having susceptibility $\left(\chi_m\right)=10^{-5}$.
$b$. Identify the following magnetic materials:
$i$. A material having susceptibility $\left(\chi_m\right)=-0.00015$.
$ii.$ A material having susceptibility $\left(\chi_m\right)=10^{-5}$.
Answer
View full question & answer→a. Let us consider a uniform magnetic field $\vec{B}$ exists in the region, in which a magnet of dipole moment $\vec{m}$ is placed.
The dipole is making small angle $\theta$ with the magnetic field.
The torque acts on the magnet is given by
$\vec{\tau}=\vec{m} \times \vec{B}$
$= mB \sin \theta$ In magnitude $, \tau= mB \sin \theta$
$=- mB \sin \theta\ (\because \theta\ \text { in small }) \ldots \text { (i) }$
Also the torque on dipole try to restore its initial position i.e., along the direction of magnetic field.
$(I =$ moment of inertia$)$ In equilibrium
$I \frac{d^2 \theta}{d t^2}=-m B \sin \theta \ldots( ii )$
Negative sign implies that restoring torque is in opposition to deflecting torque.
$\frac{d^2 \theta}{d t^2}=\frac{-m B}{I} \theta \ldots (iii)$
Comparing with equation of angular $\text{SHM}$
$\frac{d^2 \theta}{d t^2}=-\omega^2 \phi \ldots \ (iv)$
We have
$\omega^2=\frac{m B}{I} $
$\Rightarrow \omega=\sqrt{\frac{m B}{I}}$
$\Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{m B}{I}}$
$\Rightarrow \frac{T}{2 \pi}=\sqrt{\frac{I}{m B}}$
$T=2 \pi \sqrt{\frac{I}{m B}}$
$b. \ i$.Diamagnetic substance.
$ii$. Paramagnetic substance.
The dipole is making small angle $\theta$ with the magnetic field.
The torque acts on the magnet is given by
$\vec{\tau}=\vec{m} \times \vec{B}$
$= mB \sin \theta$ In magnitude $, \tau= mB \sin \theta$
$=- mB \sin \theta\ (\because \theta\ \text { in small }) \ldots \text { (i) }$
Also the torque on dipole try to restore its initial position i.e., along the direction of magnetic field.
$(I =$ moment of inertia$)$ In equilibrium
$I \frac{d^2 \theta}{d t^2}=-m B \sin \theta \ldots( ii )$
Negative sign implies that restoring torque is in opposition to deflecting torque.
$\frac{d^2 \theta}{d t^2}=\frac{-m B}{I} \theta \ldots (iii)$
Comparing with equation of angular $\text{SHM}$
$\frac{d^2 \theta}{d t^2}=-\omega^2 \phi \ldots \ (iv)$
We have
$\omega^2=\frac{m B}{I} $
$\Rightarrow \omega=\sqrt{\frac{m B}{I}}$
$\Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{m B}{I}}$
$\Rightarrow \frac{T}{2 \pi}=\sqrt{\frac{I}{m B}}$
$T=2 \pi \sqrt{\frac{I}{m B}}$
$b. \ i$.Diamagnetic substance.
$ii$. Paramagnetic substance.
