(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$.The most stable position

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) The potential energy of a magnetic dipole in a magnetic field is given by $U = -\vec{M} \cdot \vec{B} = -MB \cos \theta$.The most stable position is at $\theta_1 = 0^\circ$,where $U_1 = -MB \cos(0^\circ) = -MB$.The most unstable position is at $\theta_2 = 180^\circ$,where $U_2 = -MB \cos(180^\circ) = MB$.The work done $W$ is the change in potential energy: $W = U_2 - U_1 = MB - (-MB) = 2MB$.Given $M = 2.5 \text{ Am}^2$ and $B = 4 \times 10^{-5} \text{ T}$.$W = 2 \times 2.5 \times 4 \times 10^{-5} \text{ J} = 20 \times 10^{-5} \text{ J}$.

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

Medium

$A$ short bar magnet of magnetic moment $2.5 \text{ Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \text{ T}$. The work done in moving the magnet from its most stable position to its most unstable position is:

TS EAMCET 2025 All TS EAMCET

A
$40 \times 10^{-5} \text{ J}$
B
$25 \times 10^{-5} \text{ J}$
C
$10 \times 10^{-5} \text{ J}$
D
$20 \times 10^{-5} \text{ J}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Magnetism and Matter

Subtopic

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

Previous Year

TS EAMCET 2025 Paper All TS EAMCET years →

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?

Medium

View Solution

Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.

Easy

View Solution

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.

Medium

View Solution

Two short magnets with their axes horizontal and perpendicular to the magnetic meridian are placed with their centres $40 \, cm$ east and $50 \, cm$ west of a magnetic needle. If the needle remains undeflected,the ratio of their magnetic moments $M_1:M_2$ is

Easy

View Solution

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

MediumMHT CET 2020

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ short bar magnet of magnetic moment $2.5 \text{ Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \text{ T}$. The work done in moving the magnet from its most stable position to its most unstable position is:

Similar Questions

$A$ small magnetized needle $P$ is placed at point $O$,and the arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations) of another identical magnet $Q$. In which configuration is the system $NOT$ in equilibrium?

Two small magnets, each of magnetic moment $10 \, A \cdot m^2$, are placed in an end-on position at a distance of $0.1 \, m$ apart from their centers. The force acting between them is .... $N$.

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.

Two short magnets with their axes horizontal and perpendicular to the magnetic meridian are placed with their centres $40 \, cm$ east and $50 \, cm$ west of a magnetic needle. If the needle remains undeflected,the ratio of their magnetic moments $M_1:M_2$ is

$A$ torque of $1.732 \times 10^{-5} \text{ Nm}$ is required to hold a magnet at $90^{\circ}$ with the horizontal component of the Earth's magnetic field. The torque required to hold it at $60^{\circ}$ will be $\left[\sin 90^{\circ}=1, \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right] [\sqrt{3} = 1.732]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module