A) \[60{}^\circ \]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) None
Correct Answer: A
Solution :
[a] In \[tan\,A\] position, \[\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{d}^{3}}}={{B}_{H}}\tan 30{}^\circ =\frac{{{B}_{H}}}{\sqrt{3}}\] ??..(1) Magnetic moment of second magnet, \[{M}'=(3m)(2\times 2\ell )=6M\] In tan B position, \[\frac{{{\mu }_{0}}}{4\pi }\frac{6M}{{{d}^{3}}}={{B}_{H}}\tan \theta \] ?.....(2) dividing eq. (2) by (1) we get \[\frac{6}{2}=\frac{\tan \theta }{1/\sqrt{3}}\] or \[tan\,\theta =\sqrt{3}\] or \[\theta =60{}^\circ \]You need to login to perform this action.
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