question_answer

A bar magnet of magnetic moment \[220\,\,A{{m}^{2}}\] is suspended in a magnetic field of intensity\[0.25\,\,N/Am\]. The couple required to deflect it through \[{{30}^{o}}\] is:

A) \[27.5\,\,Nm\]                 

B) \[20.25\,\,Nm\]

C) \[47.63\,\,Nm\]                               

D) \[12\,\,Nm\]

Correct Answer: C

Solution :

Key Idea: A bar magnet suspended in a uniform magnetic field sets itself with its axis parallel to the field. A magnet placed in the magnetic field experiences a torque which rotates the magnet to a position in which the axis of the magnet is parallel to the field. The magnitude of torque acting on a current loop placed in a magnetic field \[\overset{\to }{\mathop{\mathbf{B}}}\,\] with its axis at an angle \[\theta \] with the direction of \[\overset{\to }{\mathop{\mathbf{B}}}\,\] is given by\[\tau =iAB\sin \theta \] Here, magnitude of dipole moment,\[M=i\,\,A\]                 \[\tau =MB\sin \theta \] Putting the numerical values, we have \[M=220\,\,A{{m}^{2}},\,\,B=0.25\,\,N/Am,\,\,\theta ={{30}^{o}}\] \[\therefore \]  \[\tau =220\times 0.25\times \cos {{30}^{o}}\]                 \[=220\times 0.25\times \frac{\sqrt{3}}{2}\]                 \[=47.63\,\,Nm\]