A bar magnet of magnetic moment $M$ and moment of inertia $I$ is suspended in a uniform magnetic field $B$. If it is displaced slightly from its equilibrium position, its time period of oscillations for small angular displacement $\theta$ is $T_1$. Now, if another identical magnet is glued to the first magnet along its length with the like poles together, the new time period of oscillations for small angular displacement is $T_2$. The ratio $\frac{T_1}{T_2}$ is:

## CORE INFORMATION\n✓ **Answer**: C\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Time period of oscillation of a bar magnet in a magnetic field\n\t**Related Topics**: Magnetic moment, Moment of inertia, Simple Harmonic Motion\n\t**Prerequisites**: Understanding of SHM, Magnetic torque, Moment of inertia\n## SOLUTION PATHWAY\n**Given Information**:\n\tInitial magnetic moment = M\n\tInitial moment of inertia = I\n\tInitial time period = T1\n\tFinal magnetic moment = 2M\n\tFinal moment of inertia = 2I\n\tFinal time period = T2\n\tMagnetic field = B (uniform)\n\tSmall angular displacement\n**Method & Steps**:\n\t1. Identify the formula for the time period of a bar magnet in a magnetic field.\n\t\tReasoning: The time period of oscillation is derived from the torque acting on the magnet and the resulting angular SHM.\n\t\tFormula/Rule: $T = 2\pi \sqrt{\frac{I}{MB}}$\n\t2. Calculate the initial time period $T_1$.\n\t\tWorking: $T_1 = 2\pi \sqrt{\frac{I}{MB}}$\n\t\tKey Point: Use the given values of I, M, and B\n\t3. Calculate the new magnetic moment and moment of inertia.\n\t\tReasoning: When magnets are glued together with like poles, the magnetic moment is the sum of the individual moments. The moment of inertia of the new system is the sum of the individual moments of inertia as the magnets are identical and aligned along the length.\n\t\tWorking: New magnetic moment = 2M. New moment of inertia = 2I\n\t4. Calculate the new time period $T_2$.\n\t\tWorking: $T_2 = 2\pi \sqrt{\frac{2I}{2MB}} = 2\pi \sqrt{\frac{I}{MB}}$\n\t\tKey Point: Substitute the new values of magnetic moment and moment of inertia\n\t5. Calculate the ratio $\frac{T_1}{T_2}$.\n\t\tResult: $\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I}{MB}}}{2\pi \sqrt{\frac{I}{MB}}} = 1$\n\t\tVerification: Since both expressions for T1 and T2 are same, their ratio is 1.\n ## OPTION ANALYSIS\n[A] :x: This option is incorrect. The ratio is not 1/2.\n[B] :x: This option is incorrect. The ratio is not 2.\n[C] ✓: The ratio of the time periods is 1, so this option is correct.\n[D] :x: This option is incorrect. The ratio is not 1/4.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting that magnetic moments add up when magnets are joined like-pole to like-pole. [How to avoid]: Remember that magnetic moment is a vector quantity, and like poles add.\n\t\t2.[Error 2]: Forgetting that moment of inertia also changes when magnets are joined. [How to avoid]: Consider the changes in mass distribution.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Notice that the new moment of inertia and magnetic moment are both multiplied by 2, which cancels out in the time period formula.\n\t\t2.[Memory aid]: Remember that time period of oscillation of magnet is proportional to square root of moment of inertia and inversely proportional to square root of magnetic moment

A bar magnet of magnetic moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$. If the magnet is rotated by a small angle $\theta$ from its equilibrium position, the restoring torque acting on it is proportional to:

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Torque on a magnetic dipole in a magnetic field\n\t**Related Topics**: Magnetic moment, Equilibrium, Simple Harmonic Motion\n\t**Prerequisites**: Definition of torque, magnetic moment, small angle approximation\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Magnetic moment of the bar magnet: $\vec{m}$\n\t* Uniform magnetic field: $\vec{B}$\n\t* Small angle of rotation from equilibrium: $\theta$\n\t* Magnet is rotated from its equilibrium position.\n**Method & Steps**:\n\t1. The torque on a magnetic dipole in a magnetic field is given by the cross product of the magnetic moment and the magnetic field.\n\t\tReasoning: This is the general formula for the torque on a magnetic dipole.\n\t\tFormula/Rule: $\vec{\tau} = \vec{m} \times \vec{B}$\n\t2. The magnitude of the torque is given by $\tau = mB\sin\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.\n\t\tWorking: Taking the magnitude of the cross product, we get $|\vec{\tau}| = |\vec{m}||\vec{B}|\sin\theta$.\n\t\tKey Point: The torque is a restoring torque, so it acts in the opposite direction of the displacement.\n\t3. For small angles, we can use the small angle approximation, where $\sin\theta \approx \theta$.\n\t\tReasoning: This approximation is valid for small values of $\theta$ in radians.\n\t\tFormula/Rule: $\sin\theta \approx \theta$ (for small $\theta$)\n\t4. Therefore, the torque becomes $\tau \approx mB\theta$.\n\t\tResult: The restoring torque is proportional to the angle $\theta$.\n\t\tVerification: The result matches the derived formula using the approximation.\n ## OPTION ANALYSIS\n[A] :x: The torque is proportional to $\sin\theta$, which is not directly proportional to $\theta$ for larger angles. This is only true for small angles.\n[B] ✓: The torque is proportional to $\theta$ when using the small angle approximation $\sin\theta \approx \theta$.\n[C] :x: The torque is related to the sine of the angle, not the cosine.\n[D] :x: The torque is related to the sine of the angle, not the tangent.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting the small angle approximation. This would lead to selecting an option with $\sin\theta$. [How to avoid]: Remember that the small angle approximation is crucial for this problem.\n\t\t2.[Error 2]: Confusing sine and cosine in the torque formula. [How to avoid]: Remember that the torque is proportional to $\sin\theta$ not $\cos\theta$.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: When dealing with small angular displacements, use small angle approximations to simplify the expression.\n\t\t2.[Memory aid]: Remember that for small angle oscillations, the restoring force/torque is proportional to the displacement.

The correct Option is "$\theta$"

NEET Physics Magnetic Effects of Current and Magnetism Couple Acting On The Bar Magnet MCQs

Prepare for NEET Physics Magnetic Effects of Current and Magnetism (Couple Acting On The Bar Magnet) with MCQs & PYQs on NEET.GUIDE. Access free practice, previous year questions, and expert help to determine torques on magnets in external fields.

NEET Questions / Physics / Magnetic Effects of Current and Magnetism / Couple Acting On The Bar Magnet