(C) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula:$W = MB(\cos \theta_1 - \cos \theta_2)$Given:Magnetic

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

(C) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula:$W = MB(\cos \theta_1 - \cos \theta_2)$Given:Magnetic moment $M = 2 \times 10^4 \, JT^{-1}$Magnetic field $B = 6 \times 10^{-4} \, T$Initial angle $\theta_1 = 0^{\circ}$ (parallel to the field)Final angle $\theta_2 = 60^{\circ}$Substituting the values:$W = (2 \times 10^4) \times (6 \times 10^{-4}) \times (\cos 0^{\circ} - \cos 60^{\circ})$$W = 12 \times (1 - 0.5)$$W = 12 \times 0.5 = 6 \, J$Thus,the work done is $6 \, J$.

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

Medium

$A$ bar magnet having a magnetic moment of $2 \times 10^4 \, JT^{-1}$ is free to rotate in a horizontal plane. $A$ horizontal magnetic field $B = 6 \times 10^{-4} \, T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60^{\circ}$ from the field is.....$J$

AIPMT 2009 All AIPMT

A
$2$
B
$4$
C
$6$
D
$8$

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

AIPMT 2009 Paper All AIPMT years →

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

DifficultAP EAMCET 2012

View Solution

$A$ donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$. When we put two back-to-back magnets on the rod,the upper one will float. At what height $z$ does it float?

Medium

View Solution

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

Medium

View Solution

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

DifficultAP EAMCET 2007

View Solution

Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on position (axial position) is $9 \ Gauss$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on position (equatorial position)?

Medium

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

Similar Questions

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

$A$ donut-shaped permanent magnet (magnetization parallel to the axis) can slide frictionlessly on a vertical rod. Treat the magnets as dipoles with mass $m_d$ and dipole moment $M$. When we put two back-to-back magnets on the rod,the upper one will float. At what height $z$ does it float?

$A$ bar magnet of magnetic moment $10^4\,J/T$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times 10^{-5}\,T$ to a direction $60^\circ$ from the field will be... $J$.

Due to a small magnet,the magnetic intensity at a distance $x$ in the end-on position (axial position) is $9 \ Gauss$. What will be the intensity at a distance $\frac{x}{2}$ in the broadside-on position (equatorial position)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module