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 = $0.5A{m}^2$)

• A. $0.07$J
• B. $0.08$J
• C. $0.09$J
• D. $0.1$J

Answer 1

Answer: (C)

$0.09$J

Given,

$\overrightarrow{\mu }=0.5A{m}^2,\overrightarrow{B}=0.09T$

Work Done$=$ Change in Potential energy

$=△U=U_{final}-U_{initial}$

Potential energy of this system is $U=-\overrightarrow{\mu }\cdot \overrightarrow{B}$

$\theta ={180}^{\circ }$, most unstable.

$\theta =0^{\circ }$, most stable.

$\therefore W=△U=-0.5\times 0.09\times \cos 180-(-0.5\times 0.09\times \cos 0)$

$=0.09J$