question_answer

A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is \[60{}^\circ \] and one of the fields has a magnitude of \[1.2\times {{10}^{-2}}T\]. If the dipole comes to stable equilibrium at an angle of \[15{}^\circ \] with this field, what is the magnitude of other field?

A) \[4.4\times {{10}^{-3}}\,tesla\]

B) \[5.2\times {{10}^{-3}}\,tesla\]

C) \[3.4\times {{10}^{-3}}\,tesla\]

D) \[7.8\times {{10}^{-3}}\,tesla\]

Correct Answer: A

Solution :

[a] Given that : \[{{B}_{1}}=1.2\times {{10}^{-2}}\,T\], orientation of dipole with the field \[{{B}_{1}},{{\theta }_{1}}={{15}^{o}}\] Hence, orientation of dipole with \[{{B}_{2}}\], \[{{\theta }_{2}}={{60}^{o}}-{{15}^{o}}={{45}^{o}}\] (figure) As the dipole is in equilibrium, therefore, the torque on the dipole due to the two fields must be   equal   and opposite.         If M be the magnetic dipole moment of the dipole, then \[{{\tau }_{1}}={{\tau }_{2}}\]    or    \[M{{B}_{1}}\,\sin \,{{\theta }_{1}}=M{{B}_{2}}\,\sin \,{{\theta }_{2}}\] or,      \[{{B}_{2}}=\frac{{{B}_{1}}\,\sin \,{{\theta }_{1}}}{\sin \,{{\theta }_{2}}}=\frac{1.2\times {{10}^{-2}}\,\sin \,{{15}^{o}}}{\sin \,{{45}^{o}}}\] \[=\frac{1.2\times {{10}^{-2}}\times 0.2588}{0.7071}=4.4\times {{10}^{-3}}\,Tesla\]