(C) The work done $W$ by an external torque to rotate a magnetic dipole in a uniform magnetic field is given by the change in potential energy: $W = U

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

(C) The work done $W$ by an external torque to rotate a magnetic dipole in a uniform magnetic field is given by the change in potential energy: $W = U_f - U_i = -MB \cos \theta_f - (-MB \cos \theta_i) = MB(\cos \theta_i - \cos \theta_f)$.Given: Magnetic moment $M = 2 \text{ A m}^2$, Magnetic field $B = 0.3 \text{ T}$.Initially, the magnet is aligned with the field, so $\theta_i = 0^\circ$.Finally, the magnet is normal to the field, so $\theta_f = 90^\circ$.Substituting the values:$W = MB(\cos 0^\circ - \cos 90^\circ)$$W = 2 \times 0.3 \times (1 - 0)$$W = 0.6 \times 1 = 0.6 \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

Difficult

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

TS EAMCET 2023 All TS EAMCET

A
$0.15$
B
$0.3$
C
$0.6$
D
$1.2$

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 2023 Paper All TS EAMCET years →

Write the equation of torque on a magnetic needle placed in a uniform magnetic field and derive the expression for its time period $T = 2\pi \sqrt{\frac{I}{mB}}$.

Medium

View Solution

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

Medium

View Solution

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

DifficultJEE MAIN 2024

View Solution

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

Easy

View Solution

Two small bar magnets are placed in a line with like poles facing each other at a certain distance $d$ apart. If the length of each magnet is negligible as compared to $d$,the force between them will be inversely proportional to

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

$A$ bar magnet of magnetic moment $2 \text{ A m}^2$ lies aligned with the direction of a uniform magnetic field of $0.3 \text{ T}$. The amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction is (in $\text{ J}$)

Similar Questions

Write the equation of torque on a magnetic needle placed in a uniform magnetic field and derive the expression for its time period $T = 2\pi \sqrt{\frac{I}{mB}}$.

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

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

The magnetic field due to a short magnet at a point on its axis at a distance $X \, cm$ from the center of the magnet is $200 \, G$. The magnetic field at a point on the equatorial axis at a distance $X \, cm$ from the center of the magnet is.....$G$.

Two small bar magnets are placed in a line with like poles facing each other at a certain distance $d$ apart. If the length of each magnet is negligible as compared to $d$,the force between them will be inversely proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module