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