question_answer

A bar magnet of magnetic moment \[\overset{\to }{\mathop{M}}\,\]is placed in the magnetic field\[\overset{\to }{\mathop{B}}\,\]. The torque Acting on the magnet is:                                                                                                                                                 [BHU M-2003]

A)  \[\overset{\to }{\mathop{M\times }}\,\overset{\to }{\mathop{B}}\,\]                                 

B)  \[\overset{\to }{\mathop{M-}}\,\overset{\to }{\mathop{B}}\,\]

C)  \[\frac{1}{2}\overset{\to }{\mathop{M\times }}\,\overset{\to }{\mathop{B}}\,\]                            

D)  \[\overset{\to }{\mathop{M+}}\,\overset{\to }{\mathop{B}}\,\]

Correct Answer: A

Solution :

                 Key Idea: Torque is equal to instantaneous moment of deflecting couple. The torque acting is given by                 \[\tau =force\left( {{F}_{1}}={{F}_{2}} \right)\times \]perpendicular distance                 \[\tau =i\,\,B\,\,l\times b\,\,\sin \,\,\theta \] Where \[i\] is current, \[B\] is magnetic field, \[l\] the length and \[b\] the distance. The term \[ilb=\overset{\to }{\mathop{M}}\,=\]dipole moment \[\therefore \]                  \[\overset{\to }{\mathop{\tau }}\,=\overset{\to }{\mathop{M}}\,\overset{\to }{\mathop{B}}\,\,\sin \,\theta \]                                 \[\overset{\to }{\mathop{\tau }}\,=\overset{\to }{\mathop{M}}\,\times \overset{\to }{\mathop{B}}\,\]