question_answer

A bar magnet of magnetic moment \[\text{220 A}{{\text{m}}^{\text{2}}}\]is suspended in a magnetic field of intensity 0.25 N/Am. The couple required to deflect it through \[\text{3}{{\text{0}}^{\text{o}}}\]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 \[\text{\vec{B}}\]with its axis at an angle \[\theta \] with the direction of \[\text{\vec{B}}\]is given by\[\tau =iAB\,\sin \theta \] Here, magnitude of dipole moment,\[M=i\,A\] \[\therefore \] \[\tau =MB\sin \theta \] Putting the numerical values, we have \[M=220\,A{{m}^{2}},\,B=0.25\,N/Am,\,\theta ={{30}^{o}}\] \[\therefore \] \[\tau =220\times 0.25\times \cos {{30}^{o}}\] \[=220\times 0.25\times \frac{\sqrt{3}}{2}\] \[=47.63\,Nm\]