question_answer

A magnetic needle is placed in a uniform magnetic field and is aligned with the field. The needle is now rotated by an angle of \[60{}^\circ \] and the work is W. The torque on the magnetic needle at this position is:

A) \[2\sqrt{3}W\]               

B) \[\sqrt{3}W\]  

C) \[\frac{\sqrt{3}}{2}W\]                          

D) \[\frac{\sqrt{3}}{4}W\]

Correct Answer: D

Solution :

\[\theta =0{}^\circ ,\,\,\,~{{\theta }_{2}}=6{{0}^{P}}\] \[\operatorname{W}=P{{E}_{1}}-P{{E}_{2}}\] \[= MB cos 0{}^\circ  - MB cos60{}^\circ \] \[=MB-\frac{MB}{2}\] \[W=\frac{MB}{2}\]                  .....(1) \[\tau \left( \theta = 60{}^\circ  \right) = MBSm60{}^\circ  = \frac{\sqrt{3}}{2} MB\] using equation (1) in this we get \[\tau = \sqrt{3}\operatorname{W}\]