(A) The period of oscillation of a magnetic dipole in a uniform magnetic field $B$ is given by $T = 2\pi \sqrt{\frac{I}{\mu B}}$,where $I$ is the mome

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(A) The period of oscillation of a magnetic dipole in a uniform magnetic field $B$ is given by $T = 2\pi \sqrt{\frac{I}{\mu B}}$,where $I$ is the moment of inertia and $\mu$ is the magnetic moment.For a hoop,$I_h = MR^2$ and $\mu_h = 2\mu_0$.For a solid cylinder,$I_c = \frac{1}{2}MR^2$ and $\mu_c = \mu_0$.Taking the ratio of the periods:$\frac{T_h}{T_c} = \sqrt{\frac{I_h}{I_c} \cdot \frac{\mu_c}{\mu_h}}$Substituting the values:$\frac{T_h}{T_c} = \sqrt{\frac{MR^2}{\frac{1}{2}MR^2} \cdot \frac{\mu_0}{2\mu_0}}$$\frac{T_h}{T_c} = \sqrt{2 \cdot \frac{1}{2}} = \sqrt{1} = 1$Therefore,$T_h = T_c$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

Medium

$A$ hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of the hoop is twice that of the solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of the hoop and cylinder are $T_h$ and $T_c$ respectively,then the relation between them is....

JEE MAIN 2019 All JEE MAIN

A
$T_h = T_c$
B
$T_h = 2T_c$
C
$T_h = 1.5T_c$
D
$T_h = 0.5T_c$

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Magnetism and Matter

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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$A$ bar magnet of length $6\,cm$ has a magnetic moment of $4\,J\,T^{-1}$. Find the strength of magnetic field at a distance of $200\,cm$ from the centre of the magnet along its equatorial line.

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 \ cm$ from its center is $1.5 \times 10^{-5} \ T \cdot m$. The magnetic moment of the dipole is . . . . . . $A \cdot m^2$. (Given: $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m \cdot A^{-1}$)

$A$ magnet of length $10 \text{ cm}$ and magnetic moment $1 \text{ Am}^2$ is placed along side $AB$ of an equilateral triangle $ABC$. If the length of the side $AB$ is $10 \text{ cm}$, find the magnetic induction at point $C$. (Given $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

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