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, \[{{B}_{2}}=\FRAC{{{B}_{1}}\,\sin \,{{\theta }_{1}}}{\sin \,{{\theta }_{2}}}=\frac{1.2\times {{10}^{-2}}\,\sin \,{{15}^{o}}}{\sin \,{{45}^{o}}}\] \[=\frac{1.2\times {{10}^{-2}}\times 0.2588}{0.7071}=4.4\times {{10}^{-3}}\,Tesla\]