A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle 180° , then the work is (M = magnetic dipole moment of bar magnet)

Question

A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle 180° , then the work is (M = magnetic dipole moment of bar magnet) (A) MB (B) 2MB (C) (MB/2) (D) zero

Tardigrade Admin · Accepted Answer

W = MB(1-cos θ) and = MB(1 -cos 180°) = 2MB

Q. A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle $18 0^{\circ}$ , then the work is (M = magnetic dipole moment of bar magnet)

MB

2MB

$\frac{MB}{2}$

zero

W = $MB (1 - cos θ)$ and = $MB (1 - cos 18 0^{\circ}) = 2 MB$

Q. A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle 180∘, then the work is (M = magnetic dipole moment of bar magnet)

MB

2MB

2MB​

zero

W = MB(1−cosθ) and = MB(1−cos180∘)=2MB

Q. A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If it is rotated through an angle $18 0^{\circ}$ , then the work is (M = magnetic dipole moment of bar magnet)

$\frac{MB}{2}$

W = $MB (1 - cos θ)$ and = $MB (1 - cos 18 0^{\circ}) = 2 MB$