(D) At equilibrium,the initial potential energy of the dipole is $U_{i} = -MB_{H} \cos 0^{\circ} = -MB_{H}$.The final potential energy of the dipole a

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

(D) At equilibrium,the initial potential energy of the dipole is $U_{i} = -MB_{H} \cos 0^{\circ} = -MB_{H}$.The final potential energy of the dipole after rotating by $60^{\circ}$ is $U_{f} = -MB_{H} \cos 60^{\circ} = -\frac{MB_{H}}{2}$.The work done $W$ is the change in potential energy: $W = U_{f} - U_{i} = -\frac{MB_{H}}{2} - (-MB_{H}) = \frac{MB_{H}}{2}$.From this,we get $MB_{H} = 2W$.The torque required to keep the magnet in this new position is $\tau = MB_{H} \sin 60^{\circ}$.Substituting $MB_{H} = 2W$,we get $\tau = (2W) \times \frac{\sqrt{3}}{2} = \sqrt{3} W$.

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

Medium

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

NEET 2016 All NEET

A
$\frac{\sqrt{3} W}{2}$
B
$\frac{2W}{\sqrt{3}}$
C
$\frac{W}{\sqrt{3}}$
D
$\sqrt{3} W$

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

NEET 2016 Paper All NEET years →

$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \, J$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be.....$J$

MediumAIPMT 2012

View Solution

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

DifficultTS EAMCET 2001

View Solution

The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

MediumTS EAMCET 2006

View Solution

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

Easy

View Solution

$A$ magnetic dipole of magnetic moment $ 6 \times 10^{-2} \text{ A m}^2 $ and moment of inertia $ 12 \times 10^{-6} \text{ kg m}^2 $ performs oscillations in a magnetic field of $ 2 \times 10^{-2} \text{ T} $. The time taken by the dipole to complete $ 20 $ oscillations is (assume $ \pi \simeq 3 $). (in $\text{ s}$)

MediumKCET 2017

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

$A$ bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by $60^{\circ}$ is $W$. Now,the torque required to keep the magnet in this new position is

Similar Questions

$A$ magnetic needle suspended parallel to a magnetic field requires $\sqrt{3} \, J$ of work to turn it through $60^{\circ}$. The torque needed to maintain the needle in this position will be.....$J$

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ Hm}^{-1}$)

The effect due to a uniform magnetic field on a freely suspended magnetic needle is as follows:

The rate of change of torque $\tau$ with respect to deflection $\theta$ is maximum for a magnet suspended freely in a uniform magnetic field of induction $B$ when $\theta = ........ ^\circ$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module