A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with t

Q: Why do the hoop and cylinder oscillate when placed in the magnetic field?

When a magnetic dipole is placed at an angle to a uniform magnetic field, it experiences a restoring torque ($\vec{\tau} = \vec{M} \times \vec{B}$) that tries to align it with the field. This restoring torque causes it to perform simple harmonic motion.

Q: What is the moment of inertia of a hoop and a solid cylinder?

For a hoop (or ring) where all mass is at distance $R$, $I = mR^2$. For a solid cylinder (or disc) with uniform mass distribution, $I = \frac{1}{2}mR^2$.

Q: Does the mass or radius of the objects affect the final answer?

No! Because the problem states they have the *same* mass and radius, the terms $m$ and $R^2$ appear in both moments of inertia and completely cancel out when we take the ratio.

Q: Why did we divide $T_h$ by $T_c$?

Taking the ratio of two quantities is the most efficient way to compare them, especially when they share common constants like $2\pi$ and the external magnetic field $B$, which immediately cancel out.

Q: What if the magnetic field wasn't uniform?

If the magnetic field were non-uniform, the dipoles would experience a net translational force in addition to the torque. The simple harmonic motion formula $T = 2\pi \sqrt{\frac{I}{MB}}$ strictly applies only in a uniform magnetic field.

Question

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 hoop is twice of 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 hoop and cylinder are $T_h$ and $T_c$ respectively, then

Options:

(A) $T_h = 0.5 T_c$

(B) $T_h = T_c$

(C) $T_h = 2 T_c$

(D) $T_h = 1.5 T_c$

Sigma Percentile · Accepted Answer

### The Setup: Magnetic Oscillators

Imagine you are in a physics lab, and you have two objects: a hoop and a solid cylinder. Both have the exact same mass $m$ and radius $R$.

Now, here is the twist. Both of these objects are made of a permanent magnetic material, meaning they act like tiny bar magnets. Their magnetic moments are aligned perfectly with their central axes.

We place both of them in a uniform magnetic field $\vec{B}$. When we give them a slight nudge, tilting them by a small angle $	heta$, they don't just sit there. The magnetic field exerts a restoring torque on them, given by $\vec{	au} = \vec{M} 	imes \vec{B}$, causing them to oscillate back and forth like a pendulum!

### The Master Equation: Time Period of a Dipole

To understand how fast they oscillate, we need the master equation for the time period of a magnetic dipole in a uniform magnetic field.

The time period $T$ is given by:
$$T = 2\pi \sqrt{\frac{I}{MB}}$$

This beautiful equation tells us a story. The time period depends directly on the rotational inertia $I$ (how hard it is to spin the object) and inversely on the magnetic strength $MB$ (how strongly the field pulls it back).

### Analyzing the Contenders: Inertia and Magnetism

Let's analyze our two contenders one by one. First, the hoop. Because all of its mass is concentrated at the very edge (at distance $R$), its moment of inertia is maximum:
$$I_h = mR^2$$

The problem also gives us a crucial piece of intel: the magnetic moment of the hoop is twice that of the solid cylinder. So, $M_h = 2M_c$.

Next, let's look at the solid cylinder. Its mass is distributed uniformly throughout its volume, meaning a lot of its mass is closer to the axis of rotation. This makes it easier to spin! Its moment of inertia is exactly half that of the hoop:
$$I_c = \frac{1}{2}mR^2$$

Its magnetic moment is simply $M_c$.

### The Grand Cancellation: Finding the Ratio

We want to find the relationship between their time periods, $T_h$ and $T_c$. The most elegant way to do this in physics is to take a ratio.

Let's divide the time period of the hoop by the time period of the cylinder:
$$\frac{T_h}{T_c} = \frac{2\pi \sqrt{\frac{I_h}{M_h B}}}{2\pi \sqrt{\frac{I_c}{M_c B}}}$$

Notice how the $2\pi$ and the magnetic field $B$ completely cancel out! We are left with a much simpler expression:
$$\frac{T_h}{T_c} = \sqrt{\frac{I_h}{I_c} 	imes \frac{M_c}{M_h}}$$

Now, we carefully substitute the values we found earlier.

For the inertia ratio, we have:
$$\frac{I_h}{I_c} = \frac{mR^2}{\frac{1}{2}mR^2} = 2$$

For the magnetic moment ratio, we have:
$$\frac{M_c}{M_h} = \frac{M_c}{2M_c} = \frac{1}{2}$$

Let's plug these back into our square root:
$$\frac{T_h}{T_c} = \sqrt{2 	imes \frac{1}{2}} = \sqrt{1} = 1$$

This is a moment of pure physical poetry. The hoop is twice as hard to rotate, but it has a magnet that is twice as strong. The solid cylinder is twice as easy to rotate, but its magnet is half as strong.

Nature perfectly balanced the scales! The two effects completely cancel each other out, leaving us with the final conclusion:
$$T_h = T_c$$

Select Answer:

Visualized Solution

$τ = M \times B$

$T = 2 π \frac{I}{M B}$

$I_{h} = m R^{2}$

$I_{c} = \frac{1}{2} m R^{2}$

$\frac{T _{h}}{T _{c}} = \frac{I _{h}}{I _{c}} \times \frac{M _{c}}{M _{h}}$

$\frac{T _{h}}{T _{c}} = \frac{m R ^{2}}{\frac{1}{2} m R ^{2}} \times \frac{M _{c}}{2 M _{c}}$

$T_{h} = T_{c}$

$T \propto \frac{I}{M}$

The Sigma Insight: Bar Magnet

The Setup

The Master Equation

Analyzing the Contenders

The Grand Cancellation

Similar Questions

A thin rectangular magnet suspended freely has a period of oscillation equal to $T$ . Now, it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T^{'}$ , the ratio $T^{'} / T$ is

A magnetic needle of magnetic moment $6.7 \times 1 0^{- 2} Am^{2}$ and moment of inertia $7.5 \times 1 0^{- 6} kg m^{2}$ is performing simple harmonic oscillations in a magnetic field of $0.01 T$ . Time taken for 10 complete oscillations is

A magnet of total magnetic moment $1 0^{- 2} \hat{i} A-m^{2}$ is placed in a time varying magnetic field, $B \hat{i} (cos ω t)$ , where $B = 1 T$ and $ω = 0.125 rad/s$ . The work done for reversing the direction of the magnetic moment at $t = 1 s$ is

In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85 \times 1 0^{- 2} A/m^{2}$ and moment of inertia $5 \times 1 0^{- 6} kg-m^{2}$ . If it performs 10 complete oscillations in 5 s, then the magnitude of the magnetic field is ......... mT. [Take, $π^{2}$ as 9.85]

Two short magnetic dipoles $m_{1}$ and $m_{2}$ each having magnetic moment of $1 Am^{2}$ are placed at point $O$ and $P$ , respectively. The distance between $O P$ is $1 m$ . The torque experienced by the magnetic dipole $m_{2}$ due to the presence of $m_{1}$ is ...... $\times 1 0^{- 7} Nm$ .

A magnetic needle lying parallel to a magnetic field requires $W$ unit of work to turn it through $6 0^{\circ}$ . The torque needed to maintain the needle in this position will be

A magnetic needle is kept in a non-uniform magnetic field. It experiences

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τ=M×B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=2πMBI​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ih​=mR2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ic​=21​mR2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tc​Th​​=Ic​Ih​​×Mh​Mc​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tc​Th​​=21​mR2mR2​×2Mc​Mc​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Th​=Tc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T∝MI​​