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 \[M=220\,A{{m}^{2}},\,B=0.25\,N/Am,\,\theta ={{30}^{o}}\] \[\therefore \] \[\tau =220\times 0.25\times \cos {{30}^{o}}\] \[=220\times 0.25\times \frac{\sqrt{3}}{2}\] \[=47.63\,Nm\]You need to login to perform this action.
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