The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = 0.5 A m2)

Question

The work done in moving a dipole from its most stable to most unstable position in a 0.09 T uniform magnetic field is (dipole moment of this dipole = 0.5 A m2) (A) 0.07 J (B) 0.08 J (C) 0.09 J (D) 0.1 J

Tardigrade Admin · Accepted Answer

Since the most stable position is at θ=0 and the most unstable position is at θ=180°, then the work done is given by W=∫ limitsθ=0°θ=180° τ(θ) d θ=∫ limits0°180° m B sin θ d θ= -m B[ cos θ]0180° =-m B[ cos 180°- cos 0°]=-m B[-1-1] =-m B[-2]=2 m B Here, m=0.5 A m2 text and B=0.09 T ∴ W=2 × 0.50 × 0.09=0.09 J

Q. The work done in moving a dipole from its most stable to most unstable position in a $0.09 T$ uniform magnetic field is
(dipole moment of this dipole $= 0.5 A m^{2}$ )

$0.07 J$

$0.08 J$

$0.09 J$

$0.1 J$

Q. The work done in moving a dipole from its most stable to most unstable position in a 0.09T uniform magnetic field is (dipole moment of this dipole =0.5Am2)

0.07J

0.08J

0.09J

0.1J

Q. The work done in moving a dipole from its most stable to most unstable position in a $0.09 T$ uniform magnetic field is
(dipole moment of this dipole $= 0.5 A m^{2}$ )

$0.07 J$

$0.08 J$

$0.09 J$

$0.1 J$