Question

A bar magnet of magnetic moment 1.5 J/T is aligned with the direction of a uniform magnetic field of 0.22 T. The work done in turning the magnet so as to align its magnetic moment opposite to the field and the torque acting on it in this position are respectively :

• A. 0.33J, 0.33N-m
• B. 0.66J, 06.66N-m
• C. 0.33J, 0
• D. 0.66J, 0

Answer 1

Answer: (D)

0.66J, 0

$m=1.55Am$
$B=0.22T$
$\mu =\overrightarrow{m}.\overrightarrow{B}$

${\mu }_1=-\frac{1.5\times 2.2}{10}=-0.33T$

${\mu }_2=\frac{1.54\times 2.2}{10}=-0.33T$

$Work-done=\Delta \mu$
${\mu }_2-{\mu }_1$$=2{\mu }_1$
$=0.66J$
$\overrightarrow{z}=\overrightarrow{m}\times \overrightarrow{B}$
$=mBsin\theta$
$=0(\because \theta =\pi )$