question_answer

A short bar magnet with its north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by 90o in the horizontal plane, the net magnetic induction at P is (Horizontal component of earth?s magnetic field = \[{{B}_{H}}\])                                     [EAMCET (Engg.) 2000]

A)            0    

B)            2 BH 

C)            \[\frac{\sqrt{5}}{2}{{B}_{H}}\]                                          

D)            \[\sqrt{5}\,{{B}_{H}}\]

Correct Answer: D

Solution :

                   Initially                    Neutral point obtained on equatorial line and at neutral point \[\left| {{B}_{H}} \right|=\left| {{B}_{e}} \right|\]                    where BH = Horizontal component of earth?s magnetic field, Be = Magnetic field due to bar magnet on it?s equatorial line                    Finally                    Point P comes on axial line of the magnet and at P, net magnetic field \[B=\sqrt{B_{a}^{2}+B_{H}^{2}}\]            \[=\sqrt{{{(2{{B}_{e}})}^{2}}+{{({{B}_{H}})}^{2}}}=\sqrt{{{(2{{B}_{H}})}^{2}}+B_{H}^{2}}=\sqrt{5}\ {{B}_{H}}\]